Solutions to Modern Algebra (John R. Durbin, 5E) Jason Rosendale
[email protected] December 21, 2011 This work was done as an undergraduate student: if you really don’t understand something in one of these proofs, it is very very possible that it doesn’t make make sense because it’s wrong. Any questions questions or corrections corrections can be directed to
[email protected].
1 Mappings and Operations 1.26 Let α : S S be a mapping that is one-to-one but not onto. Because the mapping is not onto, we cannot directly define an inverse mapping α−1 because it would be undefined for some elements in S . Howeve However, r, we can define a second function β : S S as:
→
→
β (s) =
α
−1
(s)
iff s iff s
s
∈ α(S ) ∈ α(S )
By showing that β is onto but not one-to-one, the “only if” half of the proof will be demonstrated. First, proof that the function β is onto: α is well-defined ( a S )( )( b S )(α )(α(a) = b) ( a S )( )( b S )(β )(β (b) = a) β is onto
→∀ ∈ ∃ ∈ →∀ ∈ ∃ ∈ →
assumed definiti definition on of wellwell-defi defined ned defin definit itio ion n of β of β definition of onto
Proof that the function β is not one-to-one: α is not onto ( b S )( )( a ( b S )( )( a
assumed definition of onto definition of image
→ ∃ ∈ ∀ ∈ S )(α )(α(a) = b) )(α(a) =b∧b∈ → ∃ ∈ ∀ ∈ S )(α α(S ))
Because α is one-to-one, we know that α(b) has an image in S . And becau because se b that α(b) = b. So:
)(α(a) =b∧b∈ → (∃b, c ∈ S )()(∀a ∈ S )(α α(S ) ∧ α(b) = c ∧ c = b) → (∃b, c ∈ S )()(∀a ∈ S )(β )(β (b) = b ∧ β (c) = b ∧ c = b) → β is not one-to-one
∈ α(S ),), we know
definition of β definition of one-to-one
Now, assume that α is some function that is onto but not one-to-one. Because the mapping is not one-toone, we can’t directly define an inverse mapping because it would be overdefined for some elements of S . However, we can define a function γ : S S as:
→
γ (s) = min r
{ ∈ S : α(r) = s}
This function maps each element s onto the smallest element r such that α(r) = s. The function γ is one-to-one. If we assume that γ (s1 ) = γ (s2 ), then by the definition of γ this means that min r1 S : α(r1 ) = s1 = min r2 S : α(r2 ) = s2 . This equality equality just means that the two minimum minimum
{ ∈
}
{ ∈
}
1
elements are the same: r1 = r2 . So ther theree is some some r such that α(r) = s1 and α(r) = s2 . Becaus Becausee α is well-defined, this shows that s1 = s2 . Therefore γ is one-to-one. The function γ is not onto. Because α is not one-to-one, there is some r, s S such that α(r ) = α(s) even though r = s. But the the definiti definition on of γ causes there to be no element that maps to the larger of the two elements r, s. This larger element is not in the image of γ (S ), ), and therefore γ is not onto.
∈
This shows that a function that is one-to-one but not onto can be used to construct a function that is onto but not one-to-one, and vice-versa. 1.27
1.28
x
∈ α(A ∪ B) )(α(s) = x) ↔ (∃s ∈ A ∪ B)(α )(α(s1 ) = x) ∨ (∃s2 ∈ B )(α )(α(s2 ) = x) ↔ (∃s1 ∈ A)(α ↔ x ∈ α(A) ∨ x ∈ α(B) ↔ x ∈ α(A) ∪ α(B)
assumed definition defin definit itio ion n definition definition
∈ α(A ∩ B) )(α(s) = x) ↔ (∃s ∈ A ∩ B)(α → (∃s1 ∈ A)(α )(α(s1 ) = x) ∧ (∃s2 ∈ B )(α )(α(s2 ) = x)
assumed definition of the domain of α definit definition ion of set inter intersec sectio tion n
x
of of of of of
the domain of α set set unio unions ns the domain of α set unions
Note that the previous step can not be justified in the other direction without assuming that s2 = s1 .
↔ x ∈ α(A) ∧ x ∈ α(B) ↔ x ∈ α(A) ∩ α(B)
definition of the domain of α
1.29 As noted in the previous problem, the proof is bidirectional if we can assume that α(s1 ) = x implies that s1 = s2 . 1.30 Let R represent the infinite subset of S . If R is infinite, then there is a mapping α : R one-to-one but not onto. Define a new function β : S S to be:
→
β (s) =
α(s) s
iff s iff s
∧ α(s2) = x
→ R that is
∈R ∈ R
This function is one-to-one. Assume that a = b:
case i) If a If a R and b R, then β (a) = α(a) and β (b) = α(b). And because α is one-to-one, the fact that a = b implies that α(a) = α(b) and therefore β (a) = β (b).
∈
∈
case ii) If a
∈ R and b ∈ R, then β(a) = a and β (b) = b. So the fact that a = b implies that β(a) = β(b). case iii) If a ∈ R and b inR, inR, then β (a) = α(a) and β (b) = b. And the ranges of these functions functions means ∈ ∈ that α(a) R and b R. So α(a) = b, and therefore β (a) = b. case iv) Similar to case (iii)
So in all cases, a = b β (a) = β (b). By contrapositiv contrapositive, e, β is one-to-o one-to-one. ne. It can also also be shown shown that β is not onto: because α is not onto, there is some r R that is not in the domain of α(R). But the image of ˜ ): so this r is not in the image of β . β is (α(R) R
→
∪
Therefore β : S
∈
→ S is a function that is one-to-one but not onto: therefore, S is an infinite set.
2 Composition 2.20 Let α : A B be an invertible function with an inverse β : B A. Then, by the definition of invertible, α β = ιB and β α = ιA . But this is exactly exactly the definition definition for a function β : B A that is invertible with an inverse α : A B.
→
◦
◦
→
→
→
2.23 The domains and codomains of β and γ are equal by definition. definition. We need only show show that β (t) = γ (t) for every t T . T . Proof by contradiction:
∈
2
( t ( ( ( ( β
∈ T )( T )(β β (t) = γ (t)) hypothesis of contradiction T )(β β (t) = γ (t)) ∧ (∀t1 ∈ T )( T )(∃s ∈ S )(α )(α(s) = t1 ) defin definiti ition on of onto onto ∃t ∈ T )( ∃t ∈ T )( T )(β β (t) = γ (t)) ∧ (∃s ∈ S )(α )(α(s) = t) choose t1 = t T )(β β (α(s)) = γ (α(s))) algebraic replacement ∃t ∈ T )( ∃t ∈ T )( T )(β β ◦ α(s) = γ ◦ α(s)) definition of ◦ definition of mapping equality ◦ α = γ ◦ α And the last statement is false, since we are told that β ◦ α = γ ◦ α. ∃ → → → → →
2.24 The domains and codomains of β and γ are equal by definition. We need only show that β (s) = γ (s) for every s S .
∈
α β = α γ ( s S )(α )(α β (s) = α γ (s)) ( s S )(α )(α(β (s)) = α(γ (s))) ( s S )(β )(β (s) = γ (s)) β = γ
◦ ◦ →∀ ∈ ◦ →∀ ∈ →∀ ∈ →
◦
given definit definition ion of mappin mappingg eequa qualit lity y defin definit itio ion n of compo composi siti tion on α is one-to-one definition of mapping equality
2.27 (a) If both b oth functions are invertible invertible,, they are b oth one-to-one one-to-one and onto (theorem (theorem 2.2). Therefore, Therefore, by 2.1, their composition composition is both one-to-one and onto. onto. Therefore Therefore,, by theorem 2.2, their composition composition is invertibl invertible. e. 2.27 (b) The truth of this statement follows directly from theorem 2.1 parts (b) and (d).
9 Equivalence, Congruence, Divisibility 9.5a 9.5a
reflexi reflexivit vity y (x, y) R R (x, y) R R (x, y) R R (x, y) (x, y)
assumed definition definition of cartesian cartesian product reflexivity of = definition of
symmetry (x1 , y1 ) (x2 , y2 ) (x1 , y1 ), (x2 , y2 ) R (x1 , y1 ), (x2 , y2 ) R (x2 , y2 ) (x1 , y1 )
assumed definition of symmetry of = definition of
∈ × → ∈ × ∧y ∈ R → ∈ × ∧y =y → ∼ → → →
∼
∼
∈ × R ∧ y1 = y2 ∈ × R ∧ y2 = y1
transitivity (x1 , y1 ) (x2 , y2 ) (x2 , y2 ) (x3 , y3 ) (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) R R y1 = y2 (x1 , y1 ), (x3 , y3 ) R R y1 = y3 (x1 , y1 ) (x3 , y3 )
→ → →
∼
∼
∼
∧ ∼ ∈ × ∧ ∈ × ∧
∼ ∼
∧ y1 = y3
assumed definition of transitivity of = definition of
∼ ∼
9.5b Each equivalence class represents a horizontal line of the form y = c for some constant c 9.5c (x, y)
{
∈ R × R : x = 0}
9.6 It is not transitive: (1, (1, 2)
∼ (1, ∼ (3, (1, 3) and (1, (1, 3) ∼ (3, (3, 3), but (1, (1, 2) (3, 3).
3
∈ R.
9.7 9.7
refle reflexi xivi vitty a R a=a a = a
assumed reflexivity of = definition of absolute value
symmetry a = b a =b a = b (a = b a = b) (a = b (b = a b = a) ( b = a (b = a ) ( b = a ) b = a
assumed definition of absolute value definition of absolute value symmetry of = definition of absolute value definition of absolute value
∈ → →| | | |
|| || →| | ∨| | − → ∨− ∨ − ∨ −a = −b) → ∨ − ∨ − ∨ −b = −a) → ||∨− || →| | | |
transitivity a = b b = c ( a = b a = b) (b = c (a =b= c) (a = b= c) (a =b= c) (a = b= c) (a = c) (a = c) (a = It cannot be the case that a = (a = c)
|| → → → → →
| |∧| | | | | ∨ | | − ∧ | | ∨ −b = |c|) | | | ∨ | | − | | ∨ (|a| = b ∧ −b = |c|) ∨ (|a| = −b ∧ b = |c|) | | | ∨ | | − | | ∨ (|a| = b ∧ b = −|c|) ∨ (−|a| = b ∧ b = |c|) | | | ∨ | | | | ∨ | | −|c|) ∨ (−|a| = |c|) | | −|c| or −|a| = |c| unless a = 0 = c,so: | || One set of equivalence class representatives is {x ∈ R : x ≥ 0}. 9.8 9.8
| | | | |
refle reflexi xivi vitty a N a=a a = 1a a = 100 a a a
reflexivity of = multiplicative identity of N x0 = 1 by definition definition of
symmetry a b ( n Z)(a )(a = 10 n b) ( n Z)(10−n a = b) ( n Z)(a )(a10−n = b) b a
assumed definition of existence of multiplicative inverses in R commutativity of multiplication in R definition of
∈ → → → → ∼
∼ ∼
transitivity a b b c ( m, n Z)(a )(a = 10 n b b = 10m c ( m, n Z)(a )(a = 10 n 10m c ( m, n Z)(a )(a = 10 n+m c a c
∼ ∈ ∈ ∈
transitivity of =
∼
∼ →∃ ∈ →∃ ∈ →∃ ∈ → ∼ ∼ ∧ →∃ →∃ →∃ → ∼
assumed definition of absolute value logica log icall distri distributi butivit vity y
∧
assumed definition of algebraic replacement
∼
definition of
∼ 0(mod 10)}. One set of equivalence class representatives is {x : x = 9.9 The first equivalenc equivalencee relation relation satisfies symmetry and transitivit transitivity y, but not reflexivit reflexivity: y: 0 0. The seco second nd relation satisfies reflexivity and symmetry, but not transitivity: 1 0 and 0 1, but 1 1.
∼ ∼ − 9.10 One set of equivalence classes would be the set of lines {y = tan(θ tan(θ)x : − 2 < θ < 2 } ∪ {x = 0}. 9.11 One set of equivalence classes would be {θ : − 2 ≤ θ ≤ 2 }. The range range of the inve inverse rse sine sine function function is ∼
∼−
π
π
π
π
usually restricted to this set of equivalence classes in order to guarantee that the function is well-defined and one-to-one.
9.12 One set of equivalence classes would be θ : 0
{
≤ θ ≤ π}. 4
9.15 9.15
reflexi reflexivit vity y x R x x=0 x x
assumed algebra definition of
symmetry x y ( n Z)(x )(x y = n) ( n Z)( 1(x 1(x y ) = ( n Z)(y )(y x = n) y x
assumed definition of algebra algebra definition of
transitivity x y y z ( m, n Z)(x )(x ( m, n Z)(x )(x ( m, n Z)(x )(x ( m, n Z)(x )(x x z
assumed defin definit itio ion n of alge algebr braa algebraic replacement algebra definition of
∈ → − → ∼
∼ →∃ ∈ − → ∃ ∈ − − −n) →∃ ∈ − − → ∼ ∼ ∧ →∃ →∃ →∃ →∃ → ∼
∼ ∈ ∈ ∈ ∈
∼ ∼ ∼
− y = m ∧ y − z = n) − y = m ∧ y = n + z) − (n + z) = m) − z = m + n)
∼ ∼
Each equivalence class is a set of real numbers that are equivalent to one another modulus 1 (i.e., they have identical identical digits following following the decimal decimal point). point). One set of equivalence equivalence class representa representative tivess is the half-open half-open interval (0, (0, 1]. 9.16 9.16
reflexi reflexivit vity y (x, y) R R x R x x Z (x, y) (x, y)
∈ × → ∈ → − ∈ → ∼
assumed definition definition of cartesian cartesian product product reflexivity of the relation in 9.15 definition of
symmetry (a, b) (x, y) a x Z x a Z (x, y) (a, b)
assumed definition of symmetry of the relation in 9.15
∼ → − ∈ → − ∈ → ∼
transitivity (a, b) (m, n) (m, n) (x, y) a m Z m x Z a x Z (a, b) (x, y)
∼ ∧ ∼ → − ∈ ∧ − ∈ → − ∈ → ∼
∼ ∼
assu assume med d definition of transitivity of the relation in 9.15 definition of
∼ ∼
(0, (0, 0) belongs to the equivalence class consisting of all points on vertical lines of the form x = c, c One set of equivalence class representatives is (x, 0) : 0 < x 1 .
{
≤ }
5
∈ Z.
9.17 9.17
reflexi reflexivit vity y (x, y) R R x R y R x x= Z y (x, y) (x, y)
assumed definition of cartesian product reflexivity of the relation in 9.15 definition of
symmetry (a, b) (x, y) a x Z b x a Z y (x, y) (a, b)
assumed definition of symmetry of the relation in 9.15 definition of
transitivity (a, b) (m, n) (m, n) (x, y) a m Z m x Z b n a x Z b y Z (a, b) (x, y)
assumed definition of transitivity of the relation in 9.15 definition of
∈ × → ∈ ∧ ∈ → − ∈ ∧ −y ∈R → ∼
∼
∼ → − ∈ ∧ −y ∈Z → − ∈ ∧ −b∈ Z → ∼
∼ ∼
∼ ∧ ∼ → − ∈ ∧ − ∈ ∧ − ∈ Z∧n−y ∈Z → − ∈ ∧ − ∈ → ∼
∼ ∼
(0, (0, 0) belongs to the equivalence class of intersections of vertical lines of the form x = c, c Z and horizontal lines of the form y = d, d Z. This This can be though thoughtt of as the set of all points points at which which the coordinate coordinate lines intersect intersect on a piece of graph paper. One set of equivalence equivalence class represen representativ tatives es would be (x, y) : 0 < x 1, 0 < y 1 . This This can be though thoughtt of as the set of all points points in one unit square square on a piece of graph paper.
∈
∈
{
≤
≤ }
9.18 The mapping α : S S defined as α(x) = x is a bijective function for any set S . From theore theorem m 2.2, this means that S S : so is reflexi reflexive ve.. If S T , T , then the definition of invertibility guarantees that T S : so is symmetric. If S T and T U , U , then from theorem 2.1 parts (a) and (c), S U : U : so is transitive.
∼
∼
→ ∼
∼
∼
∼
∼
∼
∼
9.19 a) Let a be any element of S . For every every b S , there is some permutation group (a ( a b) S n . So a b for every b. This shows shows that there is only one equivalenc equivalencee class in this case: that equivalenc equivalencee class contains every element of S .
∈
b) The equiv equivale alence nce classe classess are [1] = representatives is 1, 2, 4 .
{
}
{
}
∈
{1}, [2] = [3] = {2, 3}, [4] [4]
=
∼
{4}. One set of equiv equivale alence nce class class
c) The equivalence classes are [1] = [2] = [3] = 1, 2, 3 , [4] = 4 , [5] = 5 . One set of equivalence class representatives is 1, 4, 5 .
{
}
{}
{}
9.20 The properties of are direct directly ly inheri inherited ted from from the propert properties ies of =, which which is trivia trivially lly an equiv equivale alence nce relation. relation. One set of equivalence equivalence class representa representative tivess is the set 0 + a1 x + a2 x2 + . . . + an xn : ai R of polynomials with a zero constant term.
∼
{
∈ }
9.21 a) . . . ( a
∃ ∈ S )(a ∼ a) )(a b) . . . (∃a, b ∈ S )(a ∼ a) )(a ∼ b ∧ b c) . . . (∃a,b,c ∈ S )(a ∼ c) )(a ∼ b ∧ b ∼ c ∧ a
9.22 This proof is assuming that there is some x, y such that x y . If S consisted of only the element x, this set would be trivially symmetric symmetric and transitive. transitive. But it might not be the case that x x (for instance, if were defined as “is greater than”).
∼
∼
10 The Division Algorithm 10.9 This equivalence relation is equivalent to congruence modulo 10. 10.10 This is equivalent to congruence modulo 10 k . 10.13
ab
| ∧ b |c → (∃m, n ∈ Z)(am )(am = b ∧ bn = c) )(amn = c) → (∃m, n ∈ Z)(amn → (a|c)
definit definition ion of divisi divisibil bilit ity y algebraic replacement definition of divisibility 6
∼
10.14
ab
| ∧ b |a )(am = b ∧ bn = a) definition of divisibility → (∃m, n ∈ Z)(am → (∃m, n ∈ Z)(am )(am = b ∧ bn = a ∧ amn = a) algebraic replacement )(am = b ∧ bn = a ∧ mn = 1) uniqueness of identity element → (∃m, n ∈ Z)(am → (∃m, n ∈ Z)(am )(am = b ∧ bn = a ∧ [(m [(m = 1 = n) ∨ (m = −1 = n)]) )]) given iven )([a(1) = b ∧ b(1) = a] ∨ [a(−1) = b ∧ b(−1) = a]) alge algebr brai aicc repl replac acem emen entt → (∃m, n ∈ Z)([a → (a = b) ∨ (a = −b) → a = ±b 10.15 a ≡ b(mod n) ∧ c ≡ b(mod n) given (∃u, v ∈ Z)([a )([a = b + un] un] ∧ [c = d + vn]) vn ]) defin definit itio ion n of ≡ (∃u, v ∈ Z)(a )(a + c = b + d + n(u + v)) add both both sides a + c ≡ b + d(mod n) definition of ≡ 10.16 a ≡ b(mod n) ∧ c ≡ b(mod n) given (∃u, v ∈ Z)([a )([a = b + un] un] ∧ [c = d + vn] vn ]) definitio tion of ≡ 2 (∃u, v ∈ Z)(ac )(ac = bd + bvn + dun + n uv) multip iply ly both both side sidess uv) mult 2 (∃u, v ∈ Z)(ac )(ac = bd + nbv + ndu + n uv) uv) comm commutativ utativity ity of multiplica multiplication tion on R (∃u, v ∈ Z)(ac )(ac = bd + n(bv + du + nuv)) nuv)) dist distri ribu butiv tivit ity y of R ac ≡ bd(mod bd(mod n) definition of ≡ 0 ≡4 4. 10.17 Let a = 2, b = 4, n = 4. Then 22 ≡4 0 ≡4 42 , but 2 ≡4 2 = 10.18 a ≡ b(mod n) )(a = b + un) un) definition of ≡ → (∃u ∈ Z)(a 2 2 2 2 )(a = b + 2bun 2bun + u n ) squa square re both both side sidess → (∃u ∈ Z)(a 2 2 2 2 )(a = b + n2bu + n u ) commu com mutati tativit vity y of multip multiplic licati ation on on R → (∃u ∈ Z)(a )(a2 = b2 + n(2bu (2bu + nu2 )) distri distribut butivi ivity ty of R → (∃2u ∈ 2Z)(a → a ≡ b (mod n) 10.19 a ≡ b(mod n) ∧ n|a → (∃u, k ∈ Z)([a )([a = b + un] un] ∧ [nk = a]) defin definit itio ions ns of ≡ and divisibility )(nk = b + un) un) algebraic replacement → (∃u, k ∈ Z)(nk → (∃u, k ∈ Z)(n )(n(k − u) = b) algebra definition of divisibility → n|b 10.20 m|n ∧ a ≡ b(mod n) → (∃u, k ∈ Z)(mk )(mk = n ∧ a = b + un) un) defin definit itio ions ns of ≡ and divisibility )(a = b + umk) umk) algebraic replacement → (∃u, k ∈ Z)(a → (∃u, k ∈ Z)(a )(a = b + m(uk)) uk)) comm commut utat ativ ivit ity y of multi ultipl plic icat atio ion n on R definition of ≡ → a ≡ b(mod m) 10.21 a ≡ b(mod n) definition of equivalence modulo n ↔ n|(a − b) )(a − b = un) un) definit definition ion of divisi divisibil bilit ity y ↔ (∃u ∈ Z)(a )(a = b + un) algebr braa ↔ (∃u ∈ Z)(a un) alge The second part of the problem is true because a − b = un implies implies both a = un + b and b = un + a. n
n
n
n
n
n
n
m
10.22
a,b,n Z n > 0 ( q, r Z)(a )(a = nq + r) ( s, t Z)(b )(b = ns + t) ( q,r,s,t Z)(a )(a b = n(q s) + (r ( r t) ( q,r,s,t Z)(a )(a (r t) = b + n(q s)) ( q,r,s,t Z)(a )(a + (t (t r) = b + n(q s)) ( t, r Z)(a )(a + (t (t r) b(mod n)) ( x Z)(a )(a + x b(mod n))
→ → → → → →
∈ ∧ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈
∧∃ ∈ − − − − − − − − − ≡ ≡
divisi division on alg algori orithm thm algebra algebra algebra definition of n (t r) Z
− ∈
10.23 Let a = 0, b = 1, n = 10. There is no x such that 0x 0x = 1(mod 10).
7
≡
10.24
a is an odd integer ( k Z)(a )(a = 2k + 1) ( k Z)(a )(a2 = 4k 4 k 2 + 4k 4k + 1) 2 ( k Z)(a )(a = 4(k 4( k )(k )(k + 1) + 1) (k)(k+1) 2 ( k Z)(a )(a = 8( ) + 1) 2 2 a 1(mod 8)
→ → → → →
∃ ∃ ∃ ∃
∈ ∈ ∈ ∈ ≡
definition of odd algebra alge algebr braa (k)(k+1) Z because (k (k)(k )(k + 1) is even 2 definition of n
∈
≡
10.25 (a) 2 (b) 3 (c) 4 (d) k+1 10.26 (a) 2 (b) 3 (c) k+1 10.27 (a) The set of negative integers has no least element.
{ 1 : x ∈ N} has no least element. 10.28 i) (∃a, b ∈ Z) ∧ b > 0 )(a = bq + r ∧ 0 ≤ r < b) b) division algorithm → (∃!q, r ∈ Z)(a ii) (∃a, b ∈ Z) ∧ b < 0 )(a = (−b)q + r ∧ 0 ≤ r < −b) divisi division on alg algori orithm thm → (∃!q, r ∈ Z)(a )(a = b(−q ) + r ∧ 0 ≤ r < −b) algeb lgebrra → (∃!q, r ∈ Z)(a So, in either case, ( ∃!q, r ∈ Z)(a )(a = bq + r, 0 ≤ r < |b|). 10.29a Let S = {x : 10 ≡ 1(mod 9). Proof by induction that S = N: (b)
x
x
S is nonempty: 10 = 1 + 9 101 = 1 + 9(1) 101 1(mod 9) 1 S
definition definition of S of S
induction: n S 10n 1(mod 9) ( u Z)(10n = 1 + 9u 9u) n+1 ( u Z)(10 = 10 + 90u 90u) n+1 ( u Z)(10 = 1 + 9 + 90u 90u) n+1 ( u Z)(10 = 1 + 9(1 + 10u 10 u)) 10n+1 1(mod 9) n + 1 S
hypothesis of induction definition of membership in S definition of n algebra algebra alge algebr braa definition of n definition of membership in S
→ → ≡ → ∈ ∈ → →∃ →∃ →∃ →∃ → →
≡ ∈ ∈ ∈ ∈ ≡ ∈
Therefore S = N, which means 10 n
≡ ≡
≡ 1(mod 9) for all positive integers.
10.29b Each base-10 integer k can be expressed in expanded decimal form as k = a0 (100 ) + a1 (101 ) + . . . + an (10n ) where ai is ith digit to the left of the decimal decimal point. From the conclusion conclusion in part (a (a) of this problem, we Z conclude that there exist u0 , u1 , . . . , un such that:
∈
k = a0 (1 + 9u 9u0 ) + a1 (1 + 9u 9u1 ) + . . . + an (1 + 9u 9un ) k = (a0 + a1 + . . . + an ) + 9(a 9(a0 u0 + a1 u1 + . . . + an un ) k = (a0 + a1 + . . . + an )(mod 9)
→ →
11 Integers Modulo n 11.7 The fact that it’s a group is given, so we need only prove commutativity: [a] [b] = [a + b] = [b [ b + a] = [b [ b] [a]
⊕
⊕
defin definit itio ion n of operation commutativity of addition on Z definition of operation
⊕ ⊕
8
11.8
[a] [b] = [ab] ab] = [ba] ba] = [b] [a]
defin definit itio ion n of operation commutativity of multiplicatio tion on Z definition of operation
[a] ([b ([b] [c]) = [a [a] ([b ([b + c]) = [a(b + c)] = [ab + ac] ac] = [ab] ab] [ac] ac] = ([a ([a] [b]) ([a ([a] [c])
11.9
⊕
⊕ ⊕ [a] + [ −a] = [a + (−a)]
11.10
defin definiti ition on of operation definition of operation distributivity of Z definition of operation definition of operation
⊕ ⊕
defin definit itio ion n of operation property of negatives in Z
⊕
= [0]
11.11 Proof by contradiction: ( a Z)([a )([a] [0] [0] = [1]) [1]) ( a Z)([a )([a0] = [1]) [1]) ([0] = [1])
∃ ∈ →∃ ∈ →
assu assume med d defin definit itio ion n of operation property of 0 in Z
The conclusion is only true for Z1 : in any other case, our assumption must be false. 11.12 Associativit Associativity y: [a] ([b ([b] [c]) = [a [a] = [a [ a(bc) bc)] = [(ab [( ab))c] = [ab [ ab]] [c] = ([ ( [a] [b]) [c]
([bc]) ]) ([bc
defin definit itio ion n of operation definition of operation associativity of multiplication on Z definition of operation definition of operation
Commutativity: Commutativity: [a] [b] = [ab] ab] = [ba [ ba]] = [b [ b] [a]
defin definit itio ion n of operation commutati tativ vity of multip tiplication in Z definition of operation
Existence of an identity element: element: [a] [1] = [a [ a1] = [a [ a] = [1a [1 a] = [1] [a]
definition of multi ultipl pliicati cativ ve multi ultipl plic icat ativ ivee defin definit itio ion n of
operation
iden identi titty in Z iden identi titty in Z operation
11.13 Associativit Associativity y, comm commutativ utativity ity,, nonemptines nonemptiness, s, and the existence existence of an identity identity element were shown shown in problem 11.12. We need only show closure under the operation , which can be done with a Cayley table:
[1] [1] [2]
[1] [2]
[2] [2] [1]
11.14 Associativit Associativity y, comm commutativ utativity ity,, nonemptines nonemptiness, s, and the existence existence of an identity identity element were shown shown in # # problem 11.12. But Z4 is not closed, since [2] [2] = [0], and [0] / Z4 .
∈
11.15 The problem doesn’t specify which operation it wants us to prove this for, but it’s only a group under . The subgroup subgroup inherits associativit associativity y from Z6 and its nonemptine nonemptiness ss is given. given. We need to show show closure under , the existence existence of a unique identity identity element, and the existence existence of an inverse inverse for each each element. element. The following Cayley table shows that all of these requirements are met:
⊕
⊕
⊕
[0] [2] [4]
[0] [0] [2] [4]
9
[2] [2] [4] [0]
[4] [4] [0] [2]
11.16 As in the previous problem, we need only to show closure, the existence of a unique identity element, and the existence of an inverse for each element. [0] [0] [3] [6] [9]
⊕
[0] [3] [6] [9]
[3] [3] [6] [9] [0]
[6] [6] [9] [0] [3]
[9] [9] [0] [3] [6]
11.17(a) Proof that there is no more than 1 nonidentity element that is its own inverse: [a] is its own inverse ([a ([a] [a] n [0]) ([a ([a + a] n [0]) ( u Z)(a )(a + a = 0 + un) un) ( u Z)(2a )(2a = un) un) ( u Z)(a )(a = u n2 ) ( u, k Z)((a )((a = u n2 ) (u = 2k u = 2k + 1)) ( u, k Z)((a )((a = u n2 u = 2k 2 k ) (a = u n2 u = 2k + 1)) ( k Z)((a )((a = 2k n2 ) (a = (2k (2k + 1) n2 ) ( k Z)((a )((a = kn) kn) (a = kn + n2 ) ([a ([a] n [0]) ([a ([a] n [ n2 ])
→ → → → → → → → → →
⊕ ≡ ≡ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ≡ ∨
∧ ∧ ∨ ∨ ≡
∨ ∨
∧
assumed definition of inverse definition of operation definition of modular equivalence algebra algebra all integers are either even or odd logica log icall distr distribu ibutiv tivit ity y algebraic replacement algebra definition of modular equivalence
⊕
So if [a [a] is its own inverse, there are only two possible values it may take. One is the additive identity, and the other is [ n2 ]. And it’s only possible for [a [ a] = [ n2 ] if n2 is an integer, which is only true if n is even. But this only proves that there is at most one nonidentity element of Z2k that is its own inverse. To show that there is always at least one if n is even: (n is even)
assumed definition of even definition of Zn
n
→ 2 ∈Z → [2] ∈ Z → [ 2 + 2 ] = [n] ≡ → [ 2 ] ⊕ [ 2 ] ≡ [0] n n
n
n
n
n
n
[0]
n
definition of
⊕ operation
11.17(b) (see previous proof) 11.17(c)
[x] = [0] [1] . . . [n 1] [x] = [0 + 1 + . . . + n 1] [x] = [ n(n2−1) ] ( u Z)(x )(x = nu + n(n2−1) ) ( u Z)(x )(x = n(u + 12 n 12 ))
→ → →∃ ∈ →∃ ∈
⊕ ⊕ ⊕ − −
−
assumed definition of operation rules for series summation defin definit itio ion n of modul modular ar equi equiv valen alence ce alge algebr braa
⊕
If n is even, this last statement implies: ( u, k Z)(x )(x = n(u + 12 n 12 ) n = 2k ) ( u, k Z)(x )(x = n(u + k 12 )) ( u, k Z)(x )(x = n(u + k 1 + 12 )) ( u, k Z)(x )(x = n(u + k 1) + n2 ) [x] n [ n2 ]
∃ → → → →
∃ ∃ ∃
∈
≡
∈ ∈ ∈
− ∧ − − −
Otherwise, if n is odd, the same statement implies: ( u, k Z)(x )(x = n(u + 12 n 12 ) n = 2k + 1) ( u, k Z)(x )(x = n(u + k ) [x] n [0]
∃ ∈ →∃ ∈ → ≡
− ∧
So [x] is equivalent mod n to either [0] or [ n2 ]. 11.17(d) The modulus of the sum of the series is equal to the modulus of the median (centermost) element of the series if n is odd, and [0] otherwise. 11.18 [0]
∼ [1] and [−1] ∼ [−1]. But [0] [−1] = [−1] and [1] [−1] = [0], so is not well-defined. 10
12 Greatest Common Divisors 12.12 Theorem 12.2 proves that there is at least one choice of m, n such that (a, (a, b) = am + bn. bn. And for every number of choices for m, n such that x Z, (a(m + xb) xb) + b(n xa)) xa)) = am + bn. bn. So there are an infinite number (a, b) = am + bn. bn.
∈
12.13
x = gcd(a, gcd(a, b) x a x b (y a y b y x) ( pi Z)(xp )(xp1 = a xp2 = b (yp 3 = a yp 4 = b yp 5 = x)) ( pi Z)(xcp )(xcp1 = ac xcp2 = bc (ycp3 = ac ycp4 = bc ycp 5 = xc)) xc)) xc ac xc bc (yc ac yc bc yc xc) xc) xc = gcd(ac gcd(ac,, bc) bc) gcd(a, gcd(a, b)c = gcd(ac gcd(ac,, bc) bc)
→ → → → → →
12.14
−
| ∧ |∧ | ∧ | → | ∃ ∈ ∧ ∧ ∃ ∈ ∧ ∧ | ∧ | ∧ | ∧ | → |
1 = gcd( ad , db ) ( m, n Z)(1 = ad m + db n) ( m, n Z)(d )(d = am + bn) bn) ( m, n Z)(d )(d = am + bn) bn) gcd(a, gcd(a, b) a gcd(a, gcd(a, b) b ( m,n,r,s Z)(d )(d = am + bn a = gcd(a, gcd(a, b)r b = gcd(a, gcd(a, b)s) ( m,n,r,s Z)(d )(d = gcd(a, gcd(a, b)rm + gcd(a, gcd(a, b)sn) sn) ( m,n,r,s Z)(d )(d = gcd(a, gcd(a, b)(rm )(rm + sn) sn)) gcd(a, gcd(a, b) d gcd(a, gcd(a, b) d d a d b gcd(a, gcd(a, b) d d gcd(a, gcd(a, b) d = gcd(a, gcd(a, b)
assumed corollary of theorem 12.2 algebra property of gcd definit definition ion of divisi divisibil bilit ity y algebraic replacement distributivity definition of divisibility given given in the problem problem descriptio description n da db d gcd(a, gcd(a, b)
∃ ∃ ∃ ∃ ∃ ∃
∈ ∈ ∈
∈ ∈ ∈ | | ∧ | ∧ | | ∧ |
| ∧ | | ∧ | ∧ | ∧ | ∧
∧
∨ ∨ | ∧ ∨ | ∧
| |
| ∧
∧
∧
|
∧ ∈ ∈ ∈
| ∧ | → |
partial definition of gcd definition of p p prime logica log icall distri distribut butivi ivity ty algebr alg ebraic aic replac replaceme ement nt p a was given
|
definition of x from first step
gcd(a, gcd(a, c) = 1 gcd(b, gcd(b, c) = 1 ( m, n Z)(1 = ma + nc) nc) ( r, s Z)(1 = rb + sc) sc) ( m,n,r,s Z)(1 = (ma (ma + nc)( nc)(rb rb + sc) sc)) ( m,n,r,s Z)(1 = ab( ab(mr) mr) + c(ams + bnr + cns)) cns)) gcd(ab,c gcd(ab,c)) = 1
→∃ →∃ →∃ →
12.18
∈ ∈
x = gcd(a, gcd(a, p) x a x p p x a (x = 1 x = p) (x a x = 1) (x a x = p) (1 a x = 1) ( p a x = p) (1 a x = 1) x=1 gcd(a, gcd(a, p) = 1
→ → → → → → →
12.17
→
assumed theorem 12.2 the fractions are defined because d a, d b. corollary of theorem 12.2
→ → → → → → → → → →
12.16
∧
→
d = gcd(a, gcd(a, b) ( m, n Z)(d )(d = am + bn) bn) a ( m, n Z)(1 = d m + db n) 1 = gcd( ad , db )
→∃ →∃ →
12.15
∧
assume gcd(a, b) is defined definition of gcd definition of divisibility alge algebr braa definition of divisibility definition of gcd (x = gcd(a, gcd(a, b))
∧∃ ∈
gcd(ab,c gcd(ab,c)) = 1 ( m, n Z)(1 = (ab (ab))m + (c (c)n) ( m, n Z)((1 = (ab (ab))m + (c (c)n) ( m, n Z)((1 = (a (a)bm + (c (c)n) gcd(a, gcd(a, c) = 1 gcd(b, gcd(b, c) = 1
theo theore rem m 12 12.2 .2 multiply both equations algebra theorem 12.2 assumption theorem 12.2 p p p associativit associativity y and comm commutativi utativity ty theorem 12.2
→∃ ∈ →∃ ∈ ∧ (1 = (ab → ∧ (ab))m + (c (c)n)) →∃ ∈ ∧ (1 = (b(b)am + (c(c)n)) → ∧ c|ab ∧ gcd(a, gcd(a, c) = d assumed )(cr = ab) ab) ∧ (∃s, t ∈ Z)(d )(d = as + ct) ct) definit definition ion of divi divisib sibili ility ty and and theorem theorem 12.2 12.2 → (∃r ∈ Z)(cr )(cr = ab ∧ bd = abs + bct) algebra → (∃r,s,t ∈ Z)(cr bct) )(bd = crs + bct) bct) algebraic replacement → (∃r,s,t ∈ Z)(bd → (∃r,s,t ∈ Z)(bd )(bd = c(rs + bt) bt)) algebra definition of divisibility → c|bd 11
12.19
gcd(a, gcd(a, b) = gcd(bq gcd(bq + r, b) = gcd(bq gcd(bq + r + ( q)b, b) = gcd(r, gcd(r, b)
−
12.20
gcd(a, gcd(a, b) = 1 c a ( m,n,p Z)(1 = ma + nb cp = a) ( m,n,p Z)(1 = m(cp) cp) + nb) nb) ( m,n,p Z)(1 = c(mp) mp) + b(n)) gcd(c, gcd(c, b) = 1
∧| ∈ ∈ ∈
→∃ →∃ →∃ →
12.21
defin definit itio ion n of a of a gcd(a, b) = gcd(a gcd(a + nb,b), nb,b), from the Euclidean algorithm algebra
∧
gcd(a, gcd(a, m) = 1 ( x, s Z)(1 = ax + ms) ms) ( x, s Z)(ax )(ax = ( s)m + 1) ( x Z)(ax )(ax 1(mod m))
assumed theor theorem em 12 12.2 .2 and and gcd gcd algebraic replacement comm commut utat ativ ivit ity y and and asso associ ciat ativ ivit ity y theorem 12.2
assumed
→∃ ∈ theorem 12.2 alge algebr braa →∃ ∈ − →∃ ∈ ≡ definitio tion of mod m 12.22 n − − (mod ) 12.23 Let D = {x ∈ N : x is a common divisor of a,b, and c}. We know that this set is nonempty, since 1 ∈ D. n n
p
p
We also know that it is finite, since x must be less than the smallest of a, b, or c. Becaus Becausee D is a finite, nonempty set of positive integers, we know that it must have a greatest element. So we know that a,b, and c have a greatest common divisor. To prove that this divisor can be expressed as a linear combination: x = gcd(a,b,c gcd(a,b,c)) (x a x b x c) ((y ((y a y b y c) y x) ((x ((x a x b) x c) (((y (((y a y b) y c) y x) ((x ((x gcd(a, gcd(a, b)) x c) ((y ((y gcd(a, gcd(a, b)) y c) yx x = gcd(gcd(a, gcd(gcd(a, b), c) ( m, n Z)(x )(x = gcd(a, gcd(a, b)m + cn) cn) ( m,n,r,s Z)(x )(x = (ar + bs) bs)m + cn) cn) ( m,n,r,s Z)(x )(x = arm + bsm + cn) cn)
→ → → → → → →
assumed definition of gcd associ ociativ tivity definition of gcd definition of gcd Theorem 12.2 Theorem 12.2 algebra
| ∧ |∧ | ∧ | ∧ |∧ | → | | ∧ | ∧ | ∧ | ∧ | ∧ | → | | ∧ | ∧ | ∧ | → | ∃ ∈ ∃ ∈ ∃ ∈ 12.24 12.24 (a) One of a, b is nonzero, so one of 0 + a, 0 − a, b + 0, 0, or −b + 0 must be a positive integer. (b) We know that S has a least element by the well-ordering principle. We know that this least element can be express as d = am + bn because that’s the definition of membership in S . (c) We know that a, b S because a = 1a + 0b 0b and b = 0a + 1b 1b. So if we prove that d divides everything in S, we will of course have proven that it divides a and b.
∈
(d) We can assume that at least one such k exists b ecause ecause S is nonempty. nonempty. We can assume assume the existence existence of the integers q and r from the division algorithm algorithm of chapter chapter 10. The remaining remaining steps of part (d) are basic algebra. (e) r = 0 because we assumed that 0 r < d and then showed that it was a member of S . But we’v we’vee already assumed that d is the smallest positive element of S of S , so r cannot be a smaller positive element: it must be 0. So the fact that each each element element k S can be written k = dq + r actually tells us that k = dq + 0, or that d k for all k S .
≤
|
∈
(f ) Let d = gcd(a, gcd(a, b): c a c b (d = gcd(a, gcd(a, b)) ( m,n,r,s Z)(cn )(cn = a cm = b ( m,n,r,s Z)(d )(d = cnr + cms) cms) ( m,n,r,s Z)(d )(d = c(nr + ms) ms)) cd
| ∧|∧ →∃ →∃ →∃ → |
∈ ∈ ∈
∈
∧
bs) ∧ d = ar + bs)
assumed definition definition of divisi divisibilit bility y, Theore Theorem m 12.2 12.2 algebraic replacment algebra definition of divisibility
13 Factorization, Euler’s Phi Function In all of the following problems, exponents of 0 are allowed in the standard form. For example:
12
7
×
7 = pe1 pe2 pek 7 = 20 30 50 71 110 . . . f f f 6 = p1 p2 pk 6 = 21 31 50 70 110 . . . 6 = pe1 +f pe2 +f pek +f 42 = 21 31 50 71 110 . . . 1
1
2
1
2
1
2
··· ··· 2
k
k
···
→ → k
k
→
Because each pi always represents the ith prime, this makes it easier to compare or multiply two standard form numbers. 13.5
mn ( ( ( (
| )(mu = n) ↔ ∃u ∈ Z)(mu ↔ ∃u ∈ Z)( p )( p1 p2 · · · p )u = ( p ( p1 p2 · · · p − − ↔ ∃u ∈ Z)u = p1 p2 · · · p − ↔ ∀i ∈ N)(s )(s ≤ t )
13.15
s1 s2
t1
i
t2 k
s2
··· ···
k
k
2
2
··· 2
− 2
··· ··· ···
k
···
k
2
k
···
k
k
2
k
2
assumed FToA, p1 = 2 algebra φ is well-defined theorem 13.3 theo theore rem m 13 13.1 .1 algebra proper perty of the identity element 0 x = 1 by definition definition of n from first step
n is even ( u Z)(n )(n = 2u) n = 2( pe1 pe2 pek ) 2n = 2 2 ( pe1 pe2 pek ) 2n = ( p ( pe1 +2 pe2 pek ) φ(2n (2n) = φ( pe1 +2 pe2 pek ) φ(2n (2n) = φ( pe1 +2 )φ( pe2 pek ) φ(2n (2n) = pe1 +2 (1 1/2)φ 2)φ( pe2 pek ) φ(2n (2n) = 2 pe1 φ( pe2 pek ) e e φ(2n (2n) = 2φ( p1 p2 pek ) φ(2n (2n) = 2φ(n) 1
··· ··· ···
2
1
1
k
2
k
2
k
1
2
1
··· 2
−
1
1
2
1
2
··· ···
k
···
k
2
k
∃ ∃ ∃ ∃ ∃
···
k
k
gcd(a, gcd(a, b) = 1 a m b m ( r)(gcd(a, )(gcd(a, b) = 1 ar = m ( r)(gcd(a, )(gcd(a, b) = 1 ar = m ( r)(gcd(a, )(gcd(a, b) = 1 ar = m ( r, s)(ar )(ar = m bs = r) ( r, s)(abs )(abs = m) ab m
→ → → → → →
definition of divisibility defin definit itio ions ns of m of m and n algebra standard form cannot have negative exponents
tk k )
i
2
∃ ∈ → → → → → → → → →
13.17
s1
t1 t2
n is odd n = 20 pe2 pek 2n = 2 1 pe2 pek φ(2n (2n) = φ(21 pe2 pek ) e φ(2n (2n) = φ(21 )φ( p2 pek ) φ(2n (2n) = 2(1 1/2)φ 2)φ( pe2 pek ) φ(2n (2n) = φ( pe2 pek ) (2n) = φ(1 pe2 φ(2n pek ) φ(2n (2n) = φ(20 pe2 pek ) φ(2n (2n) = φ(n)
→ → → → → → → → →
13.16
sk k t2 s 2
∧ | ∧| ∧ ∧ ∧ ∧
∧ b|m) ar) ∧ b|ar) ∧ b|r)
|
assumed definition of even FToA algebra p1 = 2 φ is well-defined theorem 13.3 theo theore rem m 13 13.1 .1 algebra theorem 13.3 definition of n from first step assumed definiti definition on of divisi divisibil bilit ity y algebr alg ebraic aic replac replaceme ement nt lemma 13.1 definition of divisibility algebraic replacement definition of divisibility
13.18(a) Proof by contradiction: (n = pe1 pe2 pek ) ( i N)(e )(ei 2) e e e −2 2 (n = pi ( p1 p2 pi pek ) ( i p2i n n is not square free 1
→ → | →
2
··· 1
k
2
∧∃∈ ≥ )(e ≥ 2) ··· · · · ∧ ∃ ∈ N)(e i
k
i
assumption alge algebr braa definition of divisibility definition of square free
So, by contrapositive, if n is square free then there is no ei direction, we use another proof by contradiction:
13
≥ 2.
To prove prove that this holds in the other
n is not square free ( x Z)(x )(x2 n) ( u, x Z)(n )(n = x2 u) ( u, x Z)(n )(n = x2 ( pe1 pe2 ( u, x Z)(n )(n = x2 ( pe1 pe2 ( u, x Z)(n )(n = (x ( x2 pe1 pe2 n = ( pe1 pe2 pek ) ( i
→ → → → → →
assumption definition of square free definition of divisibility FToA excl exclud uded ed midd middle le FToA
∃ ∈ | ∃ ∈ ∃ ∈ · · · p )) ∃ ∈ · · · p ) ∧ [(∀ p )(x )(x = p ) ∨ (∃ p )(x )(x = p )]) +2 ∃ ∈ · · · p ) ∨ n = ( p · · · p )) ( p1 p2 · · · p )(e ≥ 2) · · · ∧ ∃ ∈ N)(e 13.18(b) (∃x ∈ Z)(n )(n = x2 ) assumption ( p1 p2 · · · p )2 FToA ↔ n = ( p ↔ n = ( p ( p21 p22 · · · p2 ) algebra ↔ each exponent of n, in standard form, is even 13.18(c) Let n = ( p1 p2 · · · p ). Let q be the product of every element in the set { p 5 1 1
1
2
1
2
1
2
ek k ek k ek k
k
2
ek k
e1 e2 e1
e2
e1 e2
i
e1 e2
i
ei i
i
ek k
i
i
ek k
ek k
: ei is odd (note that if pk is an element of n, it is only pk that is an element of q). Each Each element element of q is raised only to the first n power, power, so it is square-fre square-freee from 13.18 13.18(a). (a). Also, q has only even exponents, so it is a perfect square from n 13.18(b). And since n = q q , this means n is the product of a perfect square and a square-free integer.
}
i
√n is rational assumption √ definition of rational → (∃q, r ∈ N) n = → (∃q, r ∈ N)n2= 2 algebra algebra → (∃2q, r 2∈ N)r n2 = q 2 2 2 )( p1 p2 · · · p ) = ( p1 p2 · · · p ) FToA, FToA, definit definition ion of square squaress → ( p12 +p2 ·2· · p+ )( p 2 2 2 2 + ) = ( p1 p2 · · · p ) algebra → ( p1 p2 · · · p )(2r + n = 2q ) properties of exponents → (∀i ∈ N)(2r → (∀i ∈ N)(n )(n = 2(q 2(q − r )) algebra definition of divisibility → (∀i ∈ N)(2|n ) → (∀i ∈ N)(n )(n is even) definition of even 13.18(b) → p1 p2 · · · p is a perfect square definition of n → n is a perfect square √ 13.20 A general proof for all n: √n is rational assumption √ definition of rational → (∃q, r ∈ N) n = algebra → (∃q, r ∈ N)n = → (∃q, r ∈ N)r n = q algebra )( p1 p2 · · · p ) = ( p1 p2 · · · p ) FToA, FToA, definiti definition on of square squaress → ( p1 +p2 · · · p+ )( p + ) = ( p ( p1 p2 · · · p ) algebra → ( p1 p2 · · · p → (∀i ∈ N)(xr )(xr + n = xq ) properties of exponents )(n = x(q − r )) algebra → (∀i ∈ N)(n → (∀i ∈ N)(x )(x|n ) definition of divisibility )( p1 p2 · · · p = m ) 13.18(b) → (∃m ∈ N)( p )(n = m ) definition of n → (∃m ∈ N)(n √ √ This shows that n is rational only if n is a perfect cube. 2 is not a perfect cube, so 2 is not rational. 13.19
2
q r2
r1 r1
r2
n1
q r
rk k n2
r2
i
n1 n2 rk k
i
i
nk k
nk
q1
q1
q2
q2
qk k
qk k
i
i
i
i
i nk k
n1 n2
x
x
x
x
q rx
x
xr1 xr2
xr1 n1 xr2
q r
x
xrk k n2
i
n1 n2 xrk k
i
i
nk
nk k
xq1 xq2
xq1 xq2
xqk k
i
i
i n1 n2
x
i
nk k
x
3
3
13.21 (see problem 13.20)
14 Elementary Properties 14.11(b)
b = be = b(aa−1 ) = (ba) ba)a−1 = (ca) ca)a−1 = c(aa−1 ) = ce =c
xqk k
ba = ca was given
14
14.12
xa = b (xa) xa)a−1 = ba−1 x(aa−1 ) = ba−1 xe = ba−1 x = ba−1
→ → → →
14.13
axb = c a−1 axb = a−1 c a−1 axbb−1 = a−1 cb−1 (a−1 a)x(bb−1 ) = a−1 cb−1 exe = a−1 cb−1 x = a−1 cb−1
→ → → → →
14.14(a)
o(a) = m o(b) = n am = e bn = e amn = en bmn = em amn bmn = emn (ab) ab)mn = emn (ab) ab)mn = e
∧ ∧
→ → → → →
14.14(b)
∧
(ab) ab)mn = e o(ab) ab) mn o(ab) ab) o(a)o(b)
→ →
| |
assumed definition definition of o of o(x) algebra algebra commutativity, commutativity, from G being abelian
assumed from theorem 14.3 from from defin definiti ition on of o of o(a) = m, o(b) = n
14.14(c) In Z4 , o([2]) = 2, 2, o([2])o ([2])o([2]) = 4, 4, and o([2]
⊕ [2]) = 1.
14.15 Let a, b be permutation groups with a = (1 2), 2), b = (1 3). 3). Bo Both th of thes thesee group groupss are orde orderr 2. But But 4 4 (ab) ab) = (132) = (132). 14.16(a) Let G be an abelian group, and let S be S be the elements of finite order in G. To prove that S is S is a subgroup: e is in S: e1 = e o(e) = 1 e S
definition of o (x) definition of membership in S
→ → ∈
S is closed under its operation: a S b S ( m, n N)(o )(o(a) = m o(b) = n) ( m, n N)(a )(am = e bn = e) ( m, n N)(a )(amn = en bmn = em ) ( m, n N)(a )(amn = e bmn = e) ( m, n N)(a )(amn bmn = e) ( m, n N)((ab )((ab))mn = e) ( m, n N)(o )(o(ab) ab) mn) mn) ( m, n N)(o )(o(ab) ab) mn) mn) o(ab) ab) is finite ab S
∈ ∧ →∃ →∃ →∃ →∃ →∃ →∃ →∃ →∃ → → ∈
∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈
∧ ∧ ∧ ∧
|
≤
assumed defin definit itio ion n of me memb mber ersh ship ip in S definition of o o(x) alge algebr braa prop proper ertty of the the unit unity y elem elemen entt algebra from the commutativity of abel belian groups theorem 14.3 property of divisibility definition of membership in S
15
existence of inverses in S: a S ( m N)(o )(o(a) = m) ( m N)(a )(am = e) ( m N)(a )(am = e em = e) ( m N)(a )(am = e (aa−1 )m = e) ( m N)(a )(am = e am (a−1 )m = e) ( m N)(e )(e(a−1 )m = e) ( m N)((a )((a−1 )m = e) ( m N)(o )(o(a−1 ) m) ( m N)(o )(o(a−1 ) m) o(a−1 ) is finite a−1 S
∈ →∃ →∃ →∃ →∃ →∃ →∃ →∃ →∃ →∃ → →
∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈
∧ ∧ ∧
|
≤
assumed definition of memb ership in S definition of o (a) proper perty of the the identity element defin definit itio ion n of inv inverse ersess commu com mutat tativi ivity ty of abeli abelian an grou groups ps algebraic substitution theorem 14.3 property of divisibility
14.16(b) The only positive rational with a finite order is 1, with o(1) = 1. 14.18(a)
ab = ba (ab)( ab)(ab ab))−1 = ba( ba(ab) ab)−1 −1 e = ba( ba(ab) ab) (ba) ba)−1 e = (ba ( ba))−1 ba( ba(ab) ab)−1 (ba) ba)−1 = (ab) ab)−1 a−1 b−1 = b−1 a−1
assumed
↔ ↔ ↔ ↔ ↔
14.18(b)
14.19
ab = ba a(ab) ab) = a(ba) ba) a(ab) ab)b = a(ba) ba)b (aa)( aa)(bb bb)) = (ab)( ab)(ab ab)) 2 2 2 a b = (ab) ab)
property of inverses property of inverses theorem 14.1
assumed
→ → assoc associa iati tivi vity ty → → m ⊆ n ↔ (x ∈ m → x ∈ n) ↔ (∃ j ∈ N)(x )(x = m ) → (∃k ∈ N)(x )(x = n ) ↔ (∀ j ∈ N)(∃k ∈ N)(m )(m = n ) j
k
j
k
assumed definition of subset definit definition ion of mem members bership hip in m or n quantification of the conditional
When j = 1, this means that m = nk . So m must be an integral power of n. This This mean meanss that that a i necessary condition is that ( i N)(m )(m = n ). Continuing the proof: j
↔ (∀ j ∈ N)(∃k ∈ N)(m )(m )(n ↔ (∀ j ∈ N)(∃i, k ∈ N)(n
∃∈ = n ) ∧ (∃i ∈ N)(m )(m = n ) k
ij
i
= nk )
algebraic substitution
And this last condition can always be made true by choosing k = ij. ij . So the cond condit itio ion n that m = ni is both necessary and sufficient for m n.
⊆
14.22(a)
a = cb−1 a(bb−1 ) = cb−1 ab = c ab = (aa ( aa−1 )c b = a−1 c
↔ ↔ ↔ ↔
14.22(b)
(1 2 3)(1 2) = (1 3) (1 2)(1 3) = (1 2 3)
property of inverses right cancellation property of inverses left cancellation ab = c b−1 c = a
14.23 Let a represent the nonidentity element of the group. a2 = e aa = e aa = aa−1 a = a−1
↔ ↔ ↔
assumed definition of a2 property of inverses left cancellation 16
14.24 Assume G is a group with 2n 2n elements. From the definition of a group, each a G must have a unique inverse inverse.. Because Because the identity identity element element is its own inverse, inverse, this means that the other 2n 2 n 1 elements must have have a unique inverse inverse among themselves themselves.. Since a maximum of n 1 pairs can be formed from 2n 2n 1 elements, at least one element must be its own inverse by the pigeonhole principle.
∈
−
14.25
14.28
( a, b G)((ab )((ab))−1 = a−1 b−1 ) ( a, b G)((ab )((ab))−1 = (ba) ba)−1 ) ( a, b G)((ab )((ab))−1 (ab) ab) = (ba) ba)−1 (ab)) ab)) −1 ( a, b G)(e )(e = (ba) ba) (ab)) ab)) ( a, b G)((ba )((ba))e = (ba)( ba)(ba ba))−1 (ab)) ab)) ( a, b G)(ba )(ba = ab) ab) G is abelian
∀ ↔ ↔ ↔ ↔ ↔ ↔
∀ ∀ ∀ ∀ ∀
∈
∈ ∈ ∈ ∈ ∈
λ is onto G: b G b G a G b G a−1 G a−1 b G λ(a−1 b) = a(a−1 b) = b b Rng( Rng (λ)
∈ → ∈ ∧ ∈ → ∈ ∧ ∈ → ∈ → → ∈
λ is one-to-one: λ(x) = λ(y ) ax = ay x=y
−
−
assumed theorem 14.1
definition of abelian
assumed definition of a existence of inverses in groups G is closed under its operation definition of λ
assumed
→ definition of λ left cancellation → 14.29 (∀a, b ∈ G)(o )(o(a) = 2 ∧ o(b) = 2) assumed → (∀a, b ∈ G)(o )(o(a) = 2 ∧ o(b) = 2 ∧ ab ∈ G) G is closed )(o(a) = 2 ∧ o(b) = 2 ∧ (o(ab) ab) = 1 ∨ o(ab) ab) = 2)) ab eithe eitherr is or isn’ isn’tt the iden identit tity y → (∀a, b ∈ G)(o 2 2 2 2 → (∀a, b ∈ G)(a )(a = e ∧ b = e ∧ (ab) ab) = e) in either case, (ab) ab) = e 2 2 2 )(a b = e = (ab) ab) ) algebra → (∀a, b ∈ G)(a → (∀a, b ∈ G)(aabb )(aabb = abab) abab) )(abb = bab) bab) left cancellation → (∀a, b ∈ G)(abb )(ab = ba) ba) right cancellation → (∀a, b ∈ G)(ab → G is abelian defintion of abelian 14.30 We will prove the seemingly unrelated claim that o(a−1 ) ≤ o(a) and then show that this proves that −1 o(a) = o(a
).
Lemma: o(a−1 ) o(a): o(a) = m am = e am = e a0 = e am = e am−m = e am = e am a−m = e ea−m = e a−m = e (a−1 )m = e o(a−1 ) m o(a−1 ) m o(a−1 ) o(a)
≤
→ → → → → → → → → →
∧ ∧ ∧
|
≤ ≤
assumed definition of o(x) definition of a0 additive inverse additive inverse algebraic substitution algebraic substitution definition of negative exponents theorem 14.3 property of divisibility definition of m m
This shows not only that o(a−1 ) o(a), but also that o((a ((a−1 )−1 ) −1 −1 o(a ), which means that o(a) = o(a ).
≤
17
≤ o(a−1). So o(a−1) ≤ o(a) ∧ o(a) ≤
14.31
o(a−1 ba) ba) = m assumed −1 m (a ba) ba) = e definition of o(x) a−1 (baa−1 )m−1 ba = e associativity (If the justification for the previous step is unclear, see problem 14.32 for a clearer example.) a−1 (b)m−1 ba = e property of inverses −1 m−1 −1 −1 aa (b) baa = aea left and right multiplication (b)m−1 b = aa−1 cancellation of inverses m b =e property of inverses o(b) m
→ ↔ ↔ ↔ ↔ ↔ →
≤
And since m = o(a−1 ba), followed ba), this last step tell us that o(b) o(a−1 ba). ba). The steps of this proof can be followed −1 −1 in reverse to justify the claim that (o ( o(b) = n) (o(a ba) ba) o(b)). And, And, since since o(b) o(a ba) ba) o(b), −1 this mean that o(b) = o(a ba). ba).
→
14.32
o(ab) ab) = m (ab) ab)m = e a(ba) ba)m−1 b = e a−1 a(ba) ba)m−1 b = a−1 e (ba) ba)m−1 b = a−1 (ba) ba)m−1 ba = a−1 a (ba) ba)m = e o(ba) ba) m o(ba) ba) m o(ba) ba) o(ab) ab)
→ → → → → → → → →
|
≤ ≤
≤
≤
≤
≤
assumed definition of o(x) associativity left multiplication right multiplication theorem 14.3 property of divisibility definition of m
And of course, taking o(ba) ba) = n and following these same steps would prove that o(ab) ab) o(ab) ab) = o(ba). ba).
ba), so ≤ o(ba),
14.33 Every group G is a subgroup of itself, so if the claim were true then every group G would have a subgroup with G elements. elements. In other words, words, every group would would be a cyclic cyclic group. So every noncyclic noncyclic group group is a counterex counterexample ample.. One specific example example is S 3 , which is a subgroup of itself and has order 6 but has no elements of order 6.
| |
14.34 If e is the only element of G, then e = e = G and therefore G is cyclic. cyclic. Otherw Otherwise ise,, let a be a nonidentity element of G. Then a = e , since a1 a and a1 = e. And the only other subgroup of G is G itself, itself, so a = G and therefore G is cyclic.
14.35
G is ( ( ( ( ( ( ( (
→ → → → → → → →
{} { } ∈
finite k N)( G = k ) a G)( k N)( a k) a G)( k N)(o )(o(a) k ) a G)( k, m N)(a )(am = e m a G)( m N)(a )(am = e) a G)( m N)(aa )(aam−1 = e) a G)( m N)(a )(am−1 = a−1 ) a G)( b G)(b )(b = a−1 )
∃ ∀ ∀ ∀ ∀ ∀ ∀ ∀
∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈
| | ∃ ∈ | | ≤ ∃ ∈ ≤ ∃ ∈ ∧ ≤ k) ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈
given definition of finite every subset of a finite set is finite corollary to theorem 14.3 defin definit itio ion n of o of o(x) we don’t really care about the size of m uniq unique uene ness ss of inv inverse erse,, or left left multi ultipl plic icat atio ion n change of variable
So part (c) of theorem 7.1 is only necessary when analyzing an infinite group like N which fulfills requirements (a) and (b), but not part (c). 14.36 Let x =lcm[r, =lcm[r, s]. The fact that α, β are disjoint means that they commute: αβ = βα. βα .
18
α is an r-cycle β is an s-cycle αr = e β s = e αr = e β s = e r x s x ( m, n N)(α )(αr = e β s = e rm = x sn = x) ( m, n N)(α )(αrm = em β sn = en rm = x sn = x) αx = e β x = e αx β x = e (αβ) αβ )x = e o(αβ) αβ ) x o(αβ) αβ ) x o(αβ) αβ ) lcm[r, lcm[r, s]
→ ∧ → ∧ →∃ ∈ →∃ ∈ → ∧ → → → | → ≤ → ≤
∧
∧ | ∧ | ∧ ∧ ∧
∧
∧
∧
assumed definition of a cycle (p38) x =lcm[r, =lcm[r, s] definitio tion of divisibility algebraic replacement algebra α and β are commutative theorem 14.3 property of divisibility definition of x
So we’ve shown here that o(αβ) αβ ) is no greater than the least common multiple of [r, [ r, s]. No Now w we prov provee that it is a multiple of [r, [r, s]: o(αβ) αβ ) = y (αβ) αβ )y = e αy β y = e αy = e β y = e o(α) y o(β ) y o(α) o(αβ) αβ ) o(β ) o(αβ) αβ ) o(αβ) αβ ) is a multiple of both o(α) and o(β ) o(αβ) αβ ) is a multiple of both r and s
→ → → → → → →
∧ | ∧ | | ∧ |
assumed definition of o(x) commutativity of α, β disjointness of α, β theorem 14.3 algebraic replacement of o o(αβ) αβ ) = y definit definition ion of divisi divisibil bility ity definition of r, s
Since o(αβ) αβ ) is a common multiple of (r, ( r, s) and it’s less than or equal to the least such multiple, it must be equal to it. o(αβ) αβ ) =lcm[r, =lcm[r, s].
15 Generators, Direct Products 15.14 ∅ , by definition, is the intersection of all subgroups of G that contain ∅. A subgroup H contains ∅ iff H . And this is true for all sets. So ∅ represents the intersection of all subgroups of G. And since ∅ e is a subgroup of G and every subgroup of G must contain e , ∅ = e .
⊆ {}
{} {}
15.15 The necessary and sufficient condition is that S is a subgroup of G. Proof: If S is a group, rather than just an arbitrary set, then the smallest group that contains S is equal to S itself: i.e., S = S . So S being a group is a sufficient condition for S = S . If S is not a group, then at least one of the following three things must be true:
i) S is lacking the identity element of G. But S must contain the identity element, since it is a union of subgroups of G each of which contains the identity element. In this case, S = S .
ii) S does not contain the inverse for at least one element of S . But S is a union of subgroups of G that contain every element of S , so each subgroup must also contain the inverse of each element of S . Therefore S must contain the inverse of each element of S . In this case, S = S .
S . ther theree is some some a, b ∈ S such that ab ∈
iii) S is not closed under the operation of G: But S is a union of subgroups of G that are closed under the operation of G, so it cannot be the case that a, b S ab S . So S = S .
∈ ∧ ∈
In each of these three cases, we see that S = S . So S being a group is a necessary condition for S = S .
15.16 Let (a, (a, x), (b, y ), and (c, (c, z ) be elements of A ((a, ((a, x)(b, )(b, y))(c, ))(c, z ) = (ab,xy ( ab,xy)( )(c, c, z ) = (( ( (ab) ab)c, (xy) xy )z ) = (a ( a(bc) bc), x(yz)) yz )) = (a, ( a, x)(bc,yz )(bc,yz)) = (a, ( a, x)((b, )((b, y )(c, )(c, z ))
× B.
defin definit itio ion n of dire direct ct prod produc uctt oper operat atio ion n defin definit itio ion n of dire direct ct prod produc uctt oper operat atio ion n assoc ssocia iati tivi vitty of gro groups ups A and B defin definit itio ion n of dire direct ct prod produc uctt oper operat atio ion n definit definition ion of of direct direct produ product ct operat operation ion
19
15.17 15.17
existe existence nce of an ident identit ity y elemen element: t: eA A eB eB (eA , eB ) A eB
→
∈ ∧ ∈{ } ∈ ×{ }
necessary property of groups definition definition of cartesian cartesian product
(a, eB )(e )(eA , eB ) = (aeA , eB eB ) = (a, eB ) = (eA a, eB eB ) = (eA , eB )(a, )(a, eB ) completeness of inverses (a, eB ) A eB a A eB eB a−1 A eB eB −1 (a , eB ) A eB
∈ ×{ } → ∈ ∧ ∈{ } → ∈ ∧ ∈{ } → ∈ ×{ }
assumed definition definition of cartesian cartesian product existence of inverse elements definition definition of cartesian cartesian product
(a, eB )(a )(a−1 , eB ) = (aa ( aa−1 , eB eB ) = (e ( eA , eB ) = (a ( a−1 a, eB eB ) = (a ( a−1 , eB )(a, )(a, eB )
closed under the operation ((a, ((a, eB ) A eB ) ((x, ((x, eB ) A eB ) (a A eB eB ) (x A eB eB ) ax A eB eB eB (ax,eB eB ) A eB (a, eB )(x, )(x, eB ) A eB
→ → → →
∈ ×{ } ∧ ∈ ×{ } ∈ ∧ ∈{ } ∧ ∈ ∧ ∈{ } ∈ ∧ ∈{ } ∈ ×{ } ∈ ×{ }
15.18
A
15.20 15.20
existe existence nce of an ident identit ity y elemen element: t: eA A eB B (eA , eB ) A B
assumed definit definition ion of cartes cartesian ian product product closure property of A definition definition of cartesian cartesian product direct product operation
× B is abelian )(a, b)(x, )(x, y) = (x, y)(a, )(a, b) defin definiti ition on of abeli abelian an ↔ (∀a, x ∈ A,b,y ∈ B)(a, )(ax,by)) = (xa,yb ( xa,yb)) direct produ oduct oper peration ↔ (∀a, x ∈ A,b,y ∈ B)(ax,by ↔ (∀a, x ∈ A,b,y ∈ B)(ax )(ax = xa) xa) ∧ (by = yb) yb) defin definiti ition on of orde ordere red d pair pair equa equali lity ty definition of abelian ↔ A is abelian ∧B is abelian 15.19 [1]2 = Z2 and [1]4 = Z4 , so both of these groups groups are cyclic. cyclic. But there there is no element element of A × B that can generate both ([0], ([0], [2]) and ([1], ([1], [2]) So A × B is not a cyclic group. ∈ ∧ ∈ → ∈ ×
necessary property of groups definition definition of cartesian cartesian product
(a, b)(e )(eA , eB ) = (aeA , beB ) = (a, b) = (eA a, eB b) = (eA , eB )(a, )(a, b) completeness of inverses (a, b) A B a A b B a−1 A b−1 B (a−1 , b−1 ) A B
assumed definition definition of cartesian cartesian product existence of inverse elements definition definition of cartesian cartesian product
(a, b)(a )(a−1 , b−1 ) = (aa ( aa−1 , bb−1 ) = (e ( eA , eB ) = (a ( a−1 a, b−1 b) = (a ( a−1 , b−1 )(a, )(a, b)
direct product operation property of inverses property of inverses direct product operation
∈ × → ∈ ∧ ∈ → ∈ ∧ ∈ → ∈ ×
20
closed under the operation ((a, ((a, b) A B ) ((x, ((x, y) A B ) (a A b B ) (x A y B ) ax A by B (ax, ax, by) by) A B (a, b)(x, )(x, y) A B
assumed definiti definition on of cartes cartesian ian product product closure property of A and B definition definition of cartesian cartesian product direct direct product operation
∈ × ∧ ∈ × → ∈ ∧ ∈ ∧ ∈ ∧ ∈ → ∈ ∧ ∈ → ∈ × → ∈ ×
At this point, the only operation defined for the group Z has been addition, which causes the notation of properties like “x “x y” to differ from their notations under multiplicat multiplicative ive groups. groups. The following following table may make the next several problems clearer:
|
expr expres essi sion on xa x a x a, b x = gcd(a, gcd(a, b) x = gcd(a, gcd(a, b) x = (am )n
| ∈ ∈
↔ ↔ ↔ → → ↔
nota notati tion on in mult multip ipli lica cativ tivee grou groups ps:: ( m Z)(mx )(mx = a) ( m Z)(am )(am = x) ( m, n Z)(x )(x = am + bn) bn) xa xb ( m, n Z)(x )(x = am + bn) bn) mn x=a
∃ ∈ ∃ ∈ ∃ ∈ | ∧ | ∃ ∈
∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈
|
n
assumed ∈ a, b → (∃m, n ∈ Z)(x )(x = a b ) definition of x ∈ a, b )(x = a b ∧ d|a ∧ d|b) definition of d and the gcd → (∃m, n ∈ Z)(x )(x = a b ∧ d = a ∧ d = b) def. def. of divi divisib sibili ility ty in addi additiv tivee group group → (∃m,n,r,s ∈ Z)(x )(x = (d ) (d ) ) algebraic replacement → (∃m,n,r,s ∈ Z)(x )(x = d d ) algebra → (∃m,n,r,s ∈ Z)(x → (∃m,n,r,s ∈ Z)(x )(x = d ) algebra definition of d → x ∈ d x ∈ d assumed )(x = d ) definition of x ∈ a → (∃r ∈ Z)(x → (∃r ∈ Z)(x )(x = (gcd(a, (gcd(a, b)) ) definition of d )(x = (a b ) ) definition of gcd → (∃m,n,r ∈ Z)(x → (∃m,n,r ∈ Z)(x )(x = a b ) algebra definition of x ∈ a, b → x ∈ a, b This shows that a, b ⊆ d and d ⊆ a, b, so a, b = d. 15.24 x ∈ m assumed )(x = m ) definition of x ∈ a → (∃r ∈ Z)(x → (∃r ∈ Z)(x )(x = m ∧ a|m ∧ b|m) definition of m and the lcm )(x = m ∧ a = m ∧ b = m) definit definition ion of divisi divisibil bilit ity y → (∃r,s,t ∈ Z)(x → (∃r,s,t ∈ Z)(x )(x = (a ) ∧ x = (b ) ) algebraic replacement )(x = a ∧ x = b ) algebra → (∃r,s,t ∈ Z)(x definition of a, b → x ∈ a ∧ x ∈ b → x ∈ a ∩ b definition of intersection x ∈ a ∩ b assumed definition of intersection → x ∈ a ∧ x ∈ b )(x = a ∧ x = b ) → (∃r, s ∈ N)(x definition of divisibility → a|x ∧ b|x definition of m and the lcm → m|x → (∃r ∈ N)(m )(m = x) definition of divisibility definition of x ∈ a → x ∈ m This shows that a ∩ b ⊆ m and m ⊆ a ∩ b, so a ∩ b = m. 15.23
x
nota notati tion on in addi additi tive ve grou groups ps:: m ( m Z)(x )(x = a) ( m Z)(a )(am = x) ( m, n Z)(x )(x = am bn ) same, but note different meaning of ( m, n Z)(x )(x = am bn ) x = am
m n m n
m n
r m
r
s
s n
r m sn r m sn
r
r
m n r
mr nr
r r
r
s
t
s r
t r
r
s
r
t
r
s
r
21
15.25
1 = gcd(a, gcd(a, n) ( r, s N)(1 = ar ns ) ( r, s N)(1 = ar (mod n)) ( r N)([1]n = [ar ]n ) ( r N)([1]n = [a]nr ) [1]n [a]n [a] = Zn
assumed corollary of theorem 12.2 definiti definition on of of modula modularr equiv equivale alence nce change of notation [aaa] aaa] . . . = [a][a ][a][a ][a] . . .
↔∃ ∈ ↔∃ ∈ ↔∃ ∈ ↔∃ ∈ ↔ ∈ [1] is cyclic in all Z ↔ 15.26 x ∈ [a] )(x = [a [ a] ) definition of x ∈ [a] → (∃r ∈ N)(x → (∃r ∈ N)(x )(x = [a [ a] ∧ d|a) definition of d and gcd )(x = [a] ∧ d = a) defin definit itio ion n of divi divisi sibi bili litty → (∃r,s,t ∈ N)(x → (∃r,s,t ∈ N)(x )(x = [d ] ) algebraic replacement )(x = [d] ) algebra → (∃r,s,t ∈ N)(x definition definition of x of x ∈ [a] → x ∈ [d] x ∈ [d] assumed → (∃r ∈ N)(x )(x = [d [ d] ) definition of x ∈ [a] )(x = [d] ∧ d = a n ) theo theore rem m 12 12.2 .2 → (∃r,s,t ∈ N)(x )(x = [a n ] ) algebraic replacement → (∃r,s,t ∈ N)(x )(x = [a] [n] ) algebra → (∃r,s,t ∈ N)(x )(x = [a] [0] ) [n] = [0] → (∃r,s,t ∈ N)(x → (∃r,s,t ∈ N)(x )(x = [a] ) [0] is the identity in Z definition definition of x of x ∈ [a] → x ∈ [a] This shows that [a] ⊆ [d] and [d] ⊆ [a], so [a] = [d]. n
r r
r
s
s r sr
r
r
s
t
s t r sr
tr
sr
tr
sr
15.27
gcd(a, gcd(a, n) = gcd(b, gcd(b, n) ( x Z)(gcd(a, )(gcd(a, n) = x = gcd(b, gcd(b, n)) ( x Z)( [a] = [x] = [b] ) ( [a] = [b] )
→∃ ∈ →∃ ∈ →
n
assumed exercise 15.26 transitivity of =
For the second half of the proof, let x = gcd(a, gcd(a, n) and let y = gcd(b, gcd(b, n).
[a] = [b] assumed exercise 15.26 → [a] = [x] = [y] = [b] → [b] = [x] commutativity of = choice of specific element of [b] → [b] ∈ [x] )([b] = [x ]) definition of membership in [x] → (∃r ∈ Z)([b → (∃r, s ∈ Z)(b )(b = n x ) definition of modular equivalence And we know that x|n because x = gcd(a, gcd(a, n), so : → (∃r,s,t ∈ Z)(b )(b = n x ∧ x = n) definiti definition on of divisi divisibil bilit ity y )(b = (x (x ) x ) algebraic replacement → (∃r,s,t ∈ Z)(b → (∃r,s,t ∈ Z)(b )(b = x ) algebra definition of divisibility → x|b x = gcd(a, gcd(a, n) → x|b ∧ x|n → x| gcd(b, gcd(b, n) property of the gcd gcd(a, n)| gcd(b, gcd(b, n) definition of x → gcd(a, We know that gcd(a, gcd(a, n)| gcd(b, gcd(b, n). If we use the the same same steps steps with [a] = [y] instead of [b] = [x], it can also be shown that gcd(b, gcd( b, n)| gcd(a, gcd(a, n). In Z, this means that gcd(a, gcd(a, n) = gcd(b, gcd(b, n). r
s r
s r
t
t s r
ts r
22
15.28 15.28
existe existence nce of an ident identit ity y elemen elementt eA A (eA , eA ) A A (a, a)(e )(eA ) = (aeA , aeA ) = (a, ( a, a) = (eA a, eA a) = (e ( eA , a)(e )(eA , a)
∈ →
∈ ×
existence of inverse elements (a, a) A a A a−1 A (a−1 , a−1 ) A A −1 −1 −1 −1 (a, a)(a )(a , a ) = (aa , aa ) = (eA , eA ) = (a ( a−1 a, a−1 a) = (a−1 , a−1 )(a, )(a, a)
∈ ×→ ∈ →
∈ →
∈ ×
closure (a, a), (b, b) A A assumed a A b A definition of cartesian product ab A closure property of A (ab, definition of cartesian product ab, ab) ab) A A (a, a)(b, )(b, b) A A operation of direct products The diagonal subgroup of A represents the set of points in the cartesian plane along the line y = x.
∈ × → ∈ ∧ ∈ → ∈ → ∈ × → ∈ ×
15.29 a) Every Every group must have an identity identity element. element. If the a subgroup H of Z has only one element, then that element must be the additive identity. and 0 = 0 = H .
{}
b) If H has more than one element, element, it must contain contain some nonzero nonzero element a. And, because H is a group, it must also contain a. One of these two elements must be positive.
− c) If a = bq + r , then a − bq = r And since both a and b are elements of H , then r ∈ H . d) b is the least positive element of H and r ∈ H , so 0 ≤ r < b implies r = 0. e) By the division algorithm, all a ∈ H can be expressed as a = bq + r with 0 ≤ r < b. b . If r If r is always zero, this means that all a ∈ H are multiples of b. This means that H = b. 16 Cosets 16.9
a
∈ Hb (c) definition of Hb ↔ a ∈ {hb : h ∈ H } ↔ (∃h ∈ H )(a )(a = hb) hb) (b) −1 )(ab = h) ↔ (∃h−1∈ H )(ab (a) ↔ ab ∈ H definition of ∼ ↔a∼b transitivity of ∼ ↔a∼b∧b ∼a ↔ (∀x ∈ H )((x )((x ∼ a → x ∼ b) ∧ (x ∼ b → x ∼ a)) This last step is justified justified by the fact that ∼ is an equivalence relation. Because ∼ must be transitive, a ∼ b is both a necessary and sufficient condition for (x ( x ∼ a → x ∼ b). Similarly Similarly,, b ∼ a is both a necessary and sufficient condition for (x ( x ∼ b → x ∼ a). )(x ∼ a ↔ x ∼ b) ↔ (∀x ∈ H )(x )(xa−1 ∈ H ↔ xb−1 ∈ H ) definition of ∼ ↔ (∀x ∈ H )(xa ↔ (∀x ∈ H )(x )(x ∈ H a ↔ x ∈ Hb) Hb ) already shown that (a) ↔ (c) ↔ (H a ⊆ H b ∧ H b ⊆ H a) definition of subset ↔ (H a = H b) definition of equality, (d)
23
16.10 16.10
refle reflexi xiv ve a G a G a−1 a−1 a G a a
∈ → ∈ ∧ → ∈ → ∼
assumed inverse property of groups closure closure property property of groups groups
∈G
symmetric a b a−1 b H (a−1 b)−1 H closure closure property property of groups groups b−1 a H b a
∼ → ∈ → ∈ → ∈ → ∼
assumed definition of
definition of
∼ ∼
transitive a b b c a−1 b H b−1 c H (a−1 b)(b )(b−1 c) H a−1 c H a c
∼ ∧ ∼ assumed definition of ∼ → ∈ ∧ ∈ → ∈ closure closure property property of groups groups → ∈ definition of ∼ → ∼ 16.11 b ∈ aH (c) definition of aH ↔ b ∈ {ah : h ∈ H } ↔ (∃h ∈ H )(b )(b = ah) ah) (b) −1 )(a b = h) ↔ (∃−h1 ∈ H )(a (a) ↔ a b ∈ H ↔a∼b definition of ∼ transitivity of ∼ ↔a∼b∧b∼a ↔ (∀x ∈ H )((x )((x ∼ a → x ∼ b) ∧ (x ∼ b → x ∼ a)) a ∼ b is necessary and sufficient for (x ( x ∼ a → x ∼ b). This last step is justified justified by the fact that ∼ is an equivalence relation. Because ∼ must be transitive, a ∼ b is both a necessary and sufficient condition for (x ( x ∼ a → x ∼ b). Similarly Similarly,, b ∼ a is both a ∼ → ∼ necessary and sufficient condition for (x (x b x a). ↔ (∀x ∈ H )(x )(x ∼ a ↔ x ∼ b) )(a ∼ x ↔ b ∼ x) transitivity of ∼ ↔ (∀x ∈ H )(a ↔ (∀x ∈ H )(a )(a−1 x ∈ H ↔ b−1 x ∈ H ) definition of ∼ )(x ∈ aH ↔ x ∈ bH ) already shown that (a) ↔ (c) ↔ (∀x ∈ H )(x ↔ (aH ⊆ bH ∧ bH ⊆ aH ) definition of subset definition of equality, (d) ↔ (aH = bH ) 16.12 x ∈ H a assumed ↔ (∃h ∈ H )(x )(x = ha) ha) defin definit itio ion n of Ha )(x = ah) ah) com commu mutati tativit vity y of abelian abelian groups groups ↔ (∃h ∈ H )(x definition of aH ↔ x ∈ aH 16.16 Let H (x, y) be an arbitrary right coset of A × {e } in A × B . (m, n) ∈ H (x, y) ∩ {e } × B )(x, y) : (a1 , e ) ∈ H } ∧ (m, n) ∈ {e × B } definition of H (x, y) → (m, n) ∈ {(a1, e )(x, → (m, n) ∈ {(a1x, e y) : (a1, e ) ∈ H } ∧ (m, n) ∈ {e × B} direct product operation → ∃(a1 ∈ A, b ∈ B)((m, )((m, n) = (a ( a1 x, e y ) ∧ (m, n) = (e , b)) )(m = a1 x ∧ m = e ∧ n = e y ∧ n = b) → ∃(a1 ∈ A, b ∈ B)(m →m=e ∧n=e y → (m, n) = (e , e y) b
A
B
B
B
A
B
A
B
A
A
A
B
B
A
B
16.19 Let S be a subset of G. Let H, K be unique subgroups of G.
24
H = K assumed (H K ) (K H ) definition of set inequality ( h H )(h )(h / K ) ( k K )(k )(k / H ) defin definit itio ion n of set set me mem mber bership ship ( h H )(ha )(ha / K a) ( k K )(ka )(ka / H a) (H a Ka) Ka ) (Ka H a) definition of set inequality Ha = Ka So if H if H and K are unique subgroups, they have unique right cosets. Therefore S cannot be the right coset of two unique subgroups.
→ →∃ →∃ → →
⊆ ∨ ⊆ ∈ ∈ ∨∃ ∈ ∈ ∈ ∨∃ ∈ ⊆ ∨ ⊆
∈
∈
16.20 If H aK and H bK are not disjoint, then there is some x that that is a mem member ber of both groups. groups. So there there exists elements of H and K such that h1 ak1 = x = h2 bk2 . A bit of algebraic manipulation shows that this −1 −1 −1 1 implies a = (h− 1 h2 )b(k2 k1 ) and b = (h2 h1 )a(k1 k2 ). And this shows that these two sets are identical, −1 −1 since y H aK means that y = h3 (h1 h2 )b(k2 k1 )k3 which in turn implies that y H bK . And y HbK −1 1 means that y = h3 (h− HaK . HaK . 2 h1 )a(k1 k2 )k3 , which in turn implies that y
∈
∈
∈
∈
16 Lagrange’s Theorem Construct the lattice thingies for 11 and 17. 17.1
|S 3| = 3! = 6 and |(1 2)| = |{(1 2), 2), (1)}| = 2, so [S [S 3 : (1 2)] = 6/2 = 3. 17.2 3), (1 3 2), 2), (1)}| = 3, so [S [S 4 : (1 2 3)] = 24/ 24/3 = 8. |S 4| = 4! = 24 and |(1 2 3)| = |{(1 2 3), 17.3
|Z 1010| = 10 and |[2]| = |{[2], [2], [4], [4], [6], [6], [8], [8], [0]}| = 5, so [Z [Z 10 10/5 = 2. 10 : [2]] = 10/ 17.7 From Lagrange’s theorem we know that both 4 and 10 divide G . From theorem theorem 12.3, 12.3, this means means that G is divisible by 20, the least common common multiple of (4,10). We’re also given that G < 50. There There are only only two multiples of 20 that are less than 50, so we know that
| |
| |
| |
20,, 40} |G| ∈ {20 17.9 We’re told that H = 6 and that [G [G : H ] > 4. From this we can infer that
| |
[G : H ] > 4 G / H > 4 G /6 > 4 G > 24
| || | | | | |
(defi (defini niti tion on of the the ind index ex [G : H ]) ]) (we’re tol told that H = 6)
| |
We are also given an upper bound for G , so we know that 24 < G < 50. We can further further restri restrict ct the possible values of G with Lagrange’s Lagrange’s theorem: theorem: the fact that H = 6 tells us that G is a multiple multiple of 6. The set of all multiples of 6 within this interval is
| |
| |
| |
30,, 36 36,, 42 42,, 48} |G| ∈ {30 17.11 The subgroups are [0] , [0], [0], [2], [2], [4] , [0], [0], [3] , and Z6 .
{ }{
}{
}
25
| |
| |
17.13 The subgroups are (0, (0, 0) , (1, (1, 0), 0), (0, (0, 0) , (0, (0, 1), 1), (0, (0, 0) , (1, (1, 1), 1), (0, (0, 0) , and (1, (1, 1), 1), (1, (1, 0), 0), (0, (0, 1), 1), (0, (0, 0) .
{
} {
} {
} {
}
{
}
17.15 The subgroups are (0, 0)} {(0, (0, 0), 0), (0, (0, 1), 1), (0, (0, 2), 2), (0, (0, 3)} {(0, {(0, (0, 0), 0), (0, (0, 2)} (0, 0), 0), (1, (1, 1), 1), (0, (0, 2), 2), (1, (1, 3)} {(0,
(0, 0), 0), (1, (1, 0), 0), (0, (0, 1), 1), (1, (1, 1), 1), (0, (0, 2), 2), (1, (1, 2), 2), (0, (0, 3), 3), (1, (1, 3)} {(0, (0, 0), 0), (1, (1, 0)} {(0, {(0, (0, 0), 0), (1, (1, 0), 0), (0, (0, 2), 2), (1, (1, 2)} (0, 0), 0), (1, (1, 0), 0), (0, (0, 2), 2), (1, (1, 2)} {(0,
17.17 The subgroups are
[0] = [1] = [2] = [3] = [4] = [6] = [9] = [12] = [18] =
{[0]} {[0], [0], [2], [2], [4], [4], [6], [6], [8], [8], [10], [10], [12], [12], [14], [14], [16], [16], [18], [18], [20], [20], [22], [22], [24], [24], [26], [26], [28], [28], [30], [30], [32], [32], [34]} [0], [3], [3], [6], [6], [9], [9], [12], [12], [15], [15], [18], [18], [21], [21], [24], [24], [27], [27], [30], [30], [33]} {[0], {[0], [0], [4], [4], [8], [8], [12], [12], [16], [16], [20], [20], [24], [24], [28], [28], [32]} [0], [6], [6], [12], [12], [18], [18], [24], [24], [30]} {[0], [0], [9], [9], [18], [18], [27]} {[0], {[0], [0], [12], [12], [24]} [0], [18]} {[0],
Z36
17.20 From the definition of the index and the fact that M
| × N | = |M | · |N |, we know that |G × H | = |G| · |H | [G × H : A × B ] = |A × B| |A| · |B| From Lagrange’s theorem we know that |G| = |A| · [G : A] and |H | = |B | · [H : B ] and therefore this becomes |A| · [G : A] · |B| · [H : B] = [G : A] · [H : B] |A| · |B| 17.24 Let a be an arbitrary element of G other than the identity identity element. element. We are justified in assuming assuming this element element exists: if G if G contained only the identity element, then e = G. But we are told that G is not cyclic, so this can’t be the case. Therefore we are justified in choosing a to be any non-identity element of G. We now consider the generated subgroup a . By Lagrange’s theorem, the only possible sizes for subgroups of G are 1, 1, p, and p2 . We can rule out the possibility of a = 1, because this would force a = e and we’ve already assumed that a was not the identity element. We can also rule out the possibility of a = p2 , because this would imply that a = G and we’re told that G is not cyclic. So we are forced to conclude that a = p. Corollary 2 of theorem 14.3 tells us that o(a) = a , so it must be the case that o(a) = p. From this, we we p can directly show that a = e by appealing to the definition of “the order of an element a”: the definition (p78) tells us that p is the smallest integer such that a p = e. We have shown that a p = e for any arbitrary nonidentity element a. To complete the proof we need only to show that a p = e when a is the identity element, which is trivially true.
| |
| |
{} | |
| |
17.27 Proof by contradiction: assume that the intersection A B contains a non-identity element c and consider the generated subgroup c . Because c is an element of both A and B , Lagrange’s Lagrange’s theorem theorem tells us that c divides
∩
26
| |
both A and B . From the properties of the GCD, we know then that c must also divide gcd( A , B ) = 1. The only group of size one is e , so the fact that o(c) 1 proves that c = e and therefore therefore that c = e. We assumed that we could choose a nonidentity element of A B and proved that this element must be the identity: by contradiction, it must not be possible to choose a nonidentity element of A B .
| |
| |
{}
|
∩
| | {}
| || |
∩
17.34 We’re told that H is a subgroup of G, so by Lagrange’s theorem we know that G = H [G : H ] or, equivalently, G| that [G [G : H ] = ||H . We’re told that that K H and that K is a group, so K is a subgroup of H . By Lagrang Lagrange’s e’s | theorem we know that [H [H : K ] =
|H | . |K |
| | | |
⊆
We can now use these equivalencies to show that
[G : H ][H ][H : K ] =
|G| |H | = |G| = [G : K ] |H | |K | |K |
which is what we were asked to prove.
18 Isomorphisms 18.1 Let θ : (Z, +) ( 2m : m θ(x + y ) = 2x+y = 2 x 2y = θ (x) θ(y)
→{
x
∈ Z}, ∗) be defined as θ(x) = 2 .
∗
The proof that θ is a bijection is trivial. 18.2 Let θ : (Z
m n
× Z, +) → ({2
3 : m, n
m n
((x, y)) = 2 ∈ Z}, ∗) be defined as θ((x,
3 .
θ((a, ((a, b) + (x, ( x, y)) = θ ((a ((a + x, b + y )) a+x b+y =2 3 a b x y =2 3 2 3 = θ (a, b) θ(x, y)
∗
The proof that θ is a bijection is trivial. 18.3
* a b c d
a a b c d
b b c d a
c c d a b
d d a b c
The isomorphism for this group is θ : ( a,b,c,d , ) θ (a) = [0], [0], θ(b) = [1], [1], θ(c) = [2], [2], θ(d) = [3]
} ∗ → (Z4, +) defined as:
{
18.4
* a b c d a a b c d b b a d c c c d a b d d c b a The isomorphism for this group is θ : ( a,b,c,d , ) (Z2 Z2 , +) defined as: θ (a) = ([0], ([0], [0]), [0]), θ(b) = ([0], ([0], [1]), [1]), θ(c) = ([1][0]), ([1][0]), θ(d) = ([1], ([1], [1])
{
}∗ →
×
18.7 The proof of theorem 18.1 does not assume that the function G
→ H is one-to-one.
18.8 Let G = ( a,b,c , ) and let H = ( a,b,c , #). An isomorphic function can be defined as θ : G θ (a) = b, θ (b) = c, and θ(c) = a.
{
18.10 Let (θ (θ : G
}∗
{
}
((g, h)) = (h, (h, g ). × H → H × G) be defined as θ((g,
27
→ H with
θ((g, ((g, h)(i, )(i, j )) = θ((g ((g i, h# j)) j )) definit definition ion of of = (h ( h# j,g i) definition of = (h, ( h, g)( j,i )( j,i)) definition of = θ (g, h)θ (i, j ) definition of The proof that θ is a bijection is trivial.
∗
∗
18.11
θ is one-to-one θ (x1 ) = θ (x2 ) ex = ex ex −x = 1 ex −x = e0 x1 x2 = 0 x1 = x2
→ → → → − → 1
assumed definition definition of θ of θ division
2
1
2
1
2
direct direct produ product ct operat operation ion θ the direct product operation θ
θ is onto x R+ ln(x ln(x) R (ln(x)) = eln(x) = x θ(ln(x ( k R)(θ )(θ (k ) = x)
∈ → ∈ → →∃ ∈
assumed ln : R+ R
→
θ is homomorphic θ(x + y ) = ex+y = ex ey = θ (x)θ(y ) 18.12
⊕
[0] [1] [2] [3]
[1] [2] [3] [4]
[0] [0] [1] [2] [3] [1] [1] [2] [3] [4]
[1] [1] [2] [3] [0] [2] [2] [4] [1] [3]
[2] [2] [3] [0] [1] [3] [3] [1] [4] [2]
[3] [3] [0] [1] [2] [4] [4] [3] [2] [1]
An isomorphic function θ : (Z4 , ) [3].
#
[1], θ([1]) = [2], [2], θ([2]) = [4], [4], θ([3]) = ⊕ → (Z5 , ) can be defined as θ([0]) = [1],
18.14 Let θ : (Zmn , )
([a] ⊕ → (Z × Z , ×) be defined as θ([a m
θ is homomorphic θ([a ([a]mn [b]mn ) = θ ([a ([a + b]mn ) = ([ ( [a + b]m , [a + b]n ) = ([ ( [a]m [b]m , [a]n [b]n ) = ([ ( [a]m , [a]n ) ([b ([b]m , [b]n ) = θ ([a ([a]mn ) θ ([b ([bmn ])
⊕
⊕
×
×
⊕
n
mn )
= ([a ([a]m , [a]n ).
definition of definition of θ θ defin definit itio ion n of definit definition ion of dire direct ct produ product ct opera operatio tion n definition of θ θ
⊕ ⊕
28
θ is one-to-one θ([a ([a]mn ) = θ([b ([b]mn ) ([a ([a]m , [a]n ) = ([b ([b]m , [b]n )) ([a ([a]m = [b]m [a]n = [b [ b ]n ) ( r, s Z)(a )(a = mr + b a = ns + b) ( r, s Z)(a )(a b = mr a b = ns) ns) m (a b) n (a b) lcm(m, lcm(m, n) (a b) mn (a b) ( r Z)((a )((a b) = mnr) mnr) ( r Z)(a )(a = mnr + b) [a]mn = [b]mn
→ → → → → → → → → →
∧ ∃ ∈ ∃ ∈ − | − ∧ | − | − | − ∃ ∈ − ∃ ∈
∧ ∧ −
assumed definition of θ def. of ordered pair equality definit definition ion of modular modular equali equality ty algeb lgebrra definition of divisibility theorem 12.1, definition of lcm lcm of two primes is their product definition of divisibility algebra definition of modular equality
θ is onto gcd(m, gcd(m, n) = 1 ( r, s Z)(1 = mr + ns) ns) ( a, b Z)( r, s Z)((a )((a b) = mr( mr(a b) + ns( ns(a b)) ( a, b Z)( r, s Z)((a )((a b) = m(ra rb) rb) + n(sa sb) sb)) ( a, b Z)( r, s Z)(a )(a + m(rb ra) ra) = b + n(sa sb) sb)) ( a, b Z)( r,s,z Z)((z )((z = a + m(rb ra)) ra )) (z = b + n(sa ( a, b Z)( z Z)([z )([z ]m = [a [ a]m [z ]n = [b]n ) ( a, b Z)( z Z)(θ )(θ ([z ([z ]) = ([a ([a]m , [b]n )) ( ([a ([a]m , [b]n ) Zm Zn )( z Z)(θ )(θ ([z ([z ]) = ([a ([a]m , [b]n )) θ is onto Zm Zn
→ → → → → → → → →
18.15 18.15
∃ ∀ ∀ ∀ ∀ ∀ ∀ ∀
∈ ∈ ∈ ∈ ∈ ∈ ∈
∃ ∃ ∃ ∃ ∃ ∃
∈ ∈ ∈
− −
∈
∈ ∈ ∈ × ×
− ∧
− − −
∧
− − −
− sb)) sb))))
∃ ∈
given Theorem 12.2 corollary algebra algebra algebra alge algebr braa definition of modular equivalence definition of θ change of variable definition definition of onto
existe existence nce of an ident identit ity y elemen elementt eG G eH H eG G eH H eH B θ(eG ) = eH eH B θ(eG ) B eG A
necessary property of groups property of subgroups, lemma 7.1 theorem 18.2 algebraic replacement definition of A
existence of inverses a A (a G θ (a) B )) (a−1 G (θ(a))−1 (a−1 G (θ(a−1 )) a−1 A
assumed definition of A closure property of groups theorem 18.2 definition of A
∈ ∧ ∈ → ∈ ∧ ∈ ∧ ∈ → ∧ ∈ → ∈ → ∈ ∈ → ∈ ∧ ∈ → ∈ ∧ → ∈ ∧ → ∈
∈ B) ∈ B)
closure a A b A (a G θ (a) B ) (b G θ (b) (a G b G) (θ (a) B θ (b) (a b G) (θ(a)#θ )#θ(b) B ) a b G θ(a b) B a b A
∈ → → → → →
∧ ∈ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∗ ∈ ∧ ∈ ∗ ∈ ∧ ∗ ∈ ∗ ∈
∈ B) ∈ B)
defin definit itio ion n of A of A rearra rearrange ngemen mentt of terms terms closure property of groups homomorphism definition of A
18.16 This problem refers refers back back to several examples examples from previous previous chapters. chapters. Here are the definitions definitions of all the groups involves in this problem: 4.2 The function αa,b : R
→ R is defined as α (x) = ax + b for some given value of a and b. 5.8 The set A is defined as A = {α : a, b ∈ R}: the set of all possib possible le α functi functions ons.. This This set forms forms a a,b
a,b
group under the operation of function composition.
5.10 GL(2 nonzero determinants. determinants. This set is a group GL(2,, R) is defined as the set of all 2 2 matrices with nonzero under the operation operation of matrix multiplicati multiplication. on.
×
29
a b
18.16 G is defined to be the set of all matrices of the form
0
1
with a = 0. This is a subgroup of
GL(2 GL(2,, R). Now, let the function θ : G
→ A be defined as θ
a b
= αa,b . The proofs proofs that this functio function n are 0 1 one-to-one, onto, and isomorphic are straightforward but incredibly tedious to typeset, so to hell with that.
19 More on Isomorphism 19.11 From theorem theorem 19. 19.3, 3, any any group group of order order 59 is isomor isomorphi phicc to Z59 . Beca Becaus usee Z59 is cyclic cyclic,, any any group group isomorphic to it is cyclic. So there are no noncyclic groups of order 59. 19.12 39 is the product of distinct primes. Therefore, from the corollary of theorem 19.3 at the bottom of p99, every group of order 39 is isomorphic to Z39 . Because Z39 is cyclic, from theorem 19.1(c) this means that any group of order 39 is cyclic. 19.17 If G is cyclic, then ( a G)(G )(G = a = a0 , a1 , . . . , an−1 ). Define the function θ : a0 , a1 , . . . , an−1 Zn as θ (ax = [x]). The proof that this function is a bijection is trivial. To prove that this function preserves operations: θ (ax ay ) = θ(ax+y ) = [x [ x + y] = [x] [y ] = θ(ax ) θ (ay )
∃ ∈
{
}
·
{
⊕
}→
⊕
19.18 If G is an infinite cyclic group, then ( a G)(G )(G = a = an : n Z ). Define the function θ : an : n function is a bijection is trivial. trivial. To prove prove that this function Z Z as θ (ax = x). The proof that this function preserves operations: θ (ax ay ) = θ (ax+y ) = x + y = θ(ax ) + θ (ay )
∃ ∈
}→
{
∈ }
{
∈
·
19.19 Let θ represent the isomorphic function from G g = G.
G is cyclic
→ H . G is cyclic, so let g represent the element for which
assumed k
)(a = g ) → (∀a ∈ G)(∃k ∈ Z)(a )(θ (a) = θ (g )) → (∀a ∈ G)(∃k ∈ Z)(θ → (∀a ∈ G)(∃k ∈ Z)(θ )(θ (a) = θ (g ) ) )(θ(a) = θ(g) ) → (∀θ(a) ∈ θ(G))(∃k ∈ Z)(θ )(θ (a) = θ(g ) ) → (∀θ(a) ∈ H )()(∃k ∈ Z)(θ → H is cyclic k
k
k
k
θ is well-defined theorem 18.2(c) chang changee of variabl ariablee θ is onto H definition of cyclic
19.20 For any subgroup S of S of G G, we know that θ (S ) is a subgroup of H of H from 18.2(d). Because θ is an isomorphism, S θ(S ) and therefore o(S ) = o(θ(S )). )).
≈
19.21
19.22
G has a an element a of order n a is a cyclic subgroup of G of order n θ( a ) is isomorphic to a θ( a ) is a cyclic group of order n θ( a ) has an element of order n
→ → → → (∀x ∈ G)(x )(x = x−1 ) (∀x ∈ G)(xx )(xx = e ) (∀x ∈ G)(θ )(θ (xx) xx) = θ (e )) (∀x ∈ G)(θ )(θ (xx) xx) = e ) (∀x ∈ G)(θ )(θ (x)θ (x) = e ) (∀θ(x) ∈ H )(θ )(θ(x)θ (x) = e ) (∀y ∈ H )(yy )(yy = e ) (∀y ∈ H )(y )(y = y−1 ) G
G
H
H
H
H
assumed definition definition of o of o(a) 18.2(e) 19.1(a), 19.1(c) definition definition of cyclic cyclic
algebra θ is one-to-one 18.2(a) isomo somorrphis phism m prop proper ertty of θ f θ θ is onto change of variables uniqueness of inverses
19.23 The truth of this claim is a direct consequence of exercise 19.21 combined with the bijection property of θ. 30
19.25 Let θ, λ be arbitrary isomorophisms from G
→ G. Let Aut
G
existence of an identity element Let ig be the identity function iG (x) = x. Then for all λ
= f : f is an isomorphism G
{
→ G}.
∈ Aut
G:
(λ i)(x )(x) = λ(i(x)) = λ(x) = i(λ(x)) = (i (i λ)(x )(x)
◦
◦
existence of an inverse function Because the members λ AutG are isomorphisms, they are all bijective functions and therefore they all have a well-defined inverse function λ−1 .
∈
(λ λ−1 )(x )(x) = λ(λ−1 (x)) = x = λ−1 (λ(x)) = (λ (λ−1 λ)(x )(x)
◦
◦
closure (λ θ )(x )(x + y) = (λ ( λ(θ (x + y ))) = λ(θ (x) θ(y )) = λ(θ (x)) λ(θ(y)) = (λ (λ θ )(x )(x) (λ θ)(y )(y )
◦
∗
•
◦
• ◦
19.26 We know that this function is one-to-one and onto because, by the properties of groups, each a one unique inverse in G. So we need only prove that θ is operation preserving: θ (xy) xy ) = (xy) xy )−1 = y −1 x−1 = x−1 y −1 (from abelianism) = θ (x)θ (y )
31
∈ G has
19.2 19.27 7
lemm lemma a [a] is a generator for Zn [a] = Zn [a]k : k N = Zn [a]k : k N Zn [ak ] : k N Zn ( [x] Zn )( k N)([x )([x] = [a [ ak ]) each [x [x] Zn can be expressed as [a [ ak ] for some k
→ →{ → → →∀ →
∈ } ⊆{ ∈ } ⊆{ ∈ } ∈ ∃ ∈ ∈
definition of a generator definition of [a]
∈N
θ is one-to-one θ ([x ([x]) = θ ([y ([y ]) ( r, s N)([x )([x] = [ar ] [y ] = [as ] θ ([x ([x]) = θ ([y ([y ])) r s r ( r, s N)([x )([x] = [a ] [y ] = [a ] θ ([a ([a ]) = θ([a ([as ])) ( r, s N)([x )([x] = [ar ] [y ] = [as ] ([a ([ar a] = [as a])) r ( r, s N, v Z)([x )([x] = [a ] [y] = [as ] (ar+1 = nv + as+1 )) ( r, s N, v Z)([x )([x] = [ar ] [y] = [as ] (ar = n(va−1 ) + as )) ( r, s N)([x )([x] = [ar ] [y ] = [as ] [ar ] = [a [ as ]) [x] = [y ]
→ → → → → → →
∃ ∃ ∃ ∃ ∃ ∃
∈ ∈ ∈ ∈ ∈ ∈
∧ ∧ ∧
∈ ∈
∧
∧ ∧
∧ ∧ ∧ ∧
∧ ∧
definition of subset conclusion for people who skipped to the end
assumption lemma algebraic replacement definition of θ definit definition ion of modular modular equiv equivale alence nce −1 multi ultipl ply y by a definition of modular equivalence algebraic replacement
θ is onto (short, convincing, but possibly invalid proof) ( [y] Zn )(θ )(θ([ya ([ya −1 ] = [y]))
∀ ∈
θ is onto (longer, baffling, but valid proof) [y ] Rng(θ Rng(θ) [y] Zn [y] Zn [e] Zn [y] Zn θ([e ([e]) Zn [y] Zn [a] Zn [y] Zn [a−1 ] Zn [ya−1 ] Zn [ya−1 ] Zn θ([ya ([ya−1 ]) = [y [y ] −1 [ya ] Dom(θ Dom(θ ) θ ([ya ([ya−1 ]) = [y [y ] ( [x] Dom(θ Dom(θ ))(θ ))(θ([x ([x]) = [y [y]) θ is onto
→ → → → → → → → → →
∈
∈ ∈ ∈ ∈ ∈
∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∈ ∈ ∧ ∈ ∧ ∃ ∈
assumed definition of Rng(θ Rng(θ ) identity element of Zn definition of Rng(θ Rng(θ ) definition of θ inverse property of group Zn closure closure property property of group Zn definition of θ definition of Dom(θ ) quantification definition of onto
θ is isomorphic θ ([x ([x] [y ]) = θ ([x ([x + y ]) = [(x [(x + y )a] = [xa [ xa + ya] ya] θ ([x ([x]) θ([y ([y]) = [xa [xa]] [ya] ya] = [xa + ya] ya ]
⊕ ⊕
⊕
19.28 This can be proven by the process of elimination. There are only four possible functions from Z2 Z2 . Only two of the functions are bijections, and only one of those bijections maps the identity element onto itself.
→
21 Homomorphisms of Groups 21.7 Let θ : G
→ H be a homomorphism.
32
a, b θ(G) a θ (G) b θ (G) ab θ (G) ( x, y G)(θ )(θ (x) = a θ(y) = b ( x, y G)(θ )(θ (x) = a θ(y) = b ( x, y G)(θ )(θ (x) = a θ(y) = b ( x, y G)(θ )(θ (x) = a θ(y) = b ( x, y G)(θ )(θ (x) = a θ(y) = b ( x, y G)(θ )(θ (x) = a θ(y) = b ab = ba
→ → → → → → → →
∈ ∈ ∃ ∃ ∃ ∃ ∃ ∃
∈ ∈ ∈ ∈ ∈ ∈
∧ ∈
∧ ∈ ∧ ∧ ∧ ∧ ∧ ∧
∧ ab ∈ θ(G)) ∧ θ(x)θ(y) ∈ θ(G)) ∧ θ(xy) xy) ∈ θ (G)) xy) = θ (yx) yx )) ∧ θ(xy) ∧ θ(x)θ(y) = θ(y)θ(x)) ∧ ab = ba) ba)
assumed closure property: θ (G) is a group by 18.2(d) definition of a θ (G) alg algebr ebraic aic repla eplaccem emen entt homomorphism proper perty abelian property of G G homomo homomorph rphism ism propert property y algebraic replacement
∈
21.8 Let G = g = g0 , g1 , . . . .
{
}
y
∈ θ(G) → (∃x ∈ G)(θ )(θ(x) = y) )(x ∈ G ∧ θ (x) = y ) → (∃x ∈ G)(x → (∃x ∈ G)(x )(x ∈ {g : n ∈ Z} ∧ θ (x) = y ) )(x = g ∧ θ(x) = y) → (∃x ∈ G, n ∈ Z)(x )(θ(g ) = y) → (∃n ∈ Z)(θ )(θ(g) = y) → (∃n ∈ Z)(θ → y ∈ θ(g)
assumed definition of y
∈θ
n
n
n
n
21.9 21.9
definition of x x x algebraic replacement 18.2(c) definition of θ (g )
existe existence nce of iden identit tity y eG G eG A θ (eG ) θ(A) eH θ (A)
→ ∈ → ∈ → ∈
existence of inverse x θ (A) ( a A)(θ )(θ (a) = x) ( a A)(a )(a A θ(a) = x) ( a A)(a )(a−1 A θ(a) = x) ( a A)(a )(a−1 A θ(a)−1 = x−1 ) ( a A)(a )(a−1 A θ(a−1 ) = x−1 ) ( a−1 A)(θ )(θ (a−1 ) = x−1 ) x−1 θ(A)
∈ ∈ ∈ ∧ ∈ ∈ ∧ ∈ ∈ ∧ ∈ ∈ ∧ ∈ ∈
closed x θ (A) y θ (A) ( a, b A)(θ )(θ (a) = x θ(b) = y ) ( a, b A)(θ )(θ (a) = x θ(b) = y ab ( a, b A)(θ )(θ (a)θ (b) = xy ab A) ( a, b A)(θ )(θ (ab) ab) = xy ab A) ( ab A)(θ )(θ (ab) ab) = xy) xy ) xy θ (A)
∈ ∧ ∈ →∃ ∈ →∃ ∈ →∃ ∈ →∃ ∈ →∃ ∈ → ∈
true from group properties true from subgroup subgroup properties definition of images 18.2(a)
∈
∈ →∃ →∃ →∃ →∃ →∃ →∃ →
∈
∧ ∧
∧ ∈ A) ∧ ∈ ∧ ∈
assumed definition of θ (A) existence of inverses existence of inverses 18.2(c) change of variable definition of θ (A)
assumed definition of θ (A) clos closur uree prope propert rty y of A of A algebra homomorphism θ change of variable definition of θ (A)
33
21.10 21.10
existe existence nce of an ident identit ity y elemen elementt eH B eH B θ (eG ) = eH eH B θ (eG ) = eH eG G θ(eG ) B eG G eG θ−1 (B )
true by properties of subgroups 18.2(a) property property of groups groups algebraic replacement definition of θ −1
existence of an inverse a θ −1 (B ) a G θ (a) B a−1 G θ (a)−1 B a−1 G θ (a−1 ) B a−1 θ−1 (B )
assumed definition definition of θ of θ−1 inverse properties of groups G and B 18.2(c) definition of θ
∈ → ∈ ∧ → ∈ ∧ → ∈ ∧ ∈ → ∈
∧ ∈
∈ → → → →
∈ ∧ ∈ ∈ ∧ ∈ ∈ ∧ ∈ ∈ closure a ∈ θ−1 (B ) ∧ b ∈ θ−1 (B ) assumed a ∈ G ∧ θ(a) ∈ B ∧ b ∈ G ∧ θ(b) ∈ B definition definition of θ of θ−1 ab ∈ G ∧ θ (a) ∈ B ∧ θ (b) ∈ B closure property of G ab ∈ G ∧ θ (a)θ (b) ∈ B closure property of B ab ∈ G ∧ θ (ab) ab) ∈ B homomorphism → ab ∈ θ−1(B) definition of θ −1 21.11 (β ◦ α)(xy )(xy)) = β (α(xy) xy )) definition of composition = β (α(x) • α(y)) homomorphism of α α = β (α(x)) ∗ β (α(y )) homomorphism of β f β = (β ◦ α)(x )(x) ∗ (β ◦ α)(y )(y ) definit definition ion of com composi positio tion n 21.12a x ∈ ker(α ker(α) assumed → α(x) = e definition of the kernel → β (α(x)) = β (e ) β is well-defined )(x) = β (e ) definiti definition on of com composi positio tion n → (β ◦ α)(x )(x) = e 18.2(a) → (β ◦ α)(x ker(β ◦ α) definition of kernel → x ∈ ker(β 21.12b Let G = N, H = {2n : n ∈ N}, and K = {0}. Define Define the functio function n α : G → H to be α(x) = 2x and define the function β : H → K to be β (x) = 0. The kernel of α is {0} while the kernel of β ◦ α is N. 21.13 θ (a + b) = [k [ k (a + b)] = [ka [ka + kb] kb] = [ka [ ka]] ⊕ [kb] kb] = θ(a) ⊕ θ (b) H
H
H
K
21.17
o(a) = n an = eG θ(an ) = θ(eG ) θ(an ) = eH θ(a)n = eH o(θ(a)) n o(θ(a)) o(a)
→ → → → → →
21.18
| |
assumed theorem 14.3(a) θ is well-defined thoerem 18.2(a) thoerem 18.2(c) theorem 14.3(b) definition of n f n
θ([0]) = θ([3]) = (1) θ([1]) = θ([4]) = (1 2 3) θ([2]) = θ([5]) = (1 3 2) ker = [0], [0], [3]
{
}
21.19 False. Let G = S 3 . G is obviously a normal subgroup of itself, but (1 3 2)(1 2)(1 2 3) = (1 2).
34
( n N, g G)(g )(g −1 ng N ) N ) assumed −1 −1 ( n N, g G)((g )((g ng) ng) N ) existe existence nce on inve inverse rsess in group group N ( n N, g G)((gn )((gn−1 g −1 ) N ) And because every element in a group has a unique inverse, the set of all n−1 the set of all n N . N . So we can change the variables: ( n N, g G)((gng )((gng −1 ) N )
21.20
∀ ∈ ∈ →∀ ∈ ∈ →∀ ∈ ∈ →∀ ∈
∈
∈
∈
∈ ∈
∈ N is identical to
∈
The same justifications can be easily used to show the converse. 21.23 Let h be an arbitrary element of H and let m be an arbitrary element of θ (N ). N ). Proof that hmh 1 must then be in θ (N ): N ):
−
h
N ) ∈ H ∧ m ∈ θ(N ) → (∃g ∈ G)(θ )(θ (g ) = h) ∧ (∃n ∈ N )( N )(θθ (n) = m) N )(θθ (g ) = h ∧ θ (n) = m ∧ gng −1 ∈ N ) → (∃g ∈ G, n ∈ N )( → (∃g ∈ G, n ∈ N )( N )(θθ (g ) = h ∧ θ (n) = m ∧ θ(gng −1 ) ∈ θ(N )) N )) −1 N )(θθ (g )θ (n)θ(g ) ∈ θ (N ) N )) → (∃g ∈ G, n ∈ N )( −1 → (∃g ∈−1G, n ∈ N )( N )(θθ (g )θ (n)θ(g) ∈ θ (N ) N )) N ) → hmh ∈ θ(N ) 21.24 (a, b) ∈ A × B ∧ (e, β ) ∈ {e} × B → (a, b) ∈ A × B−1∧ (−e,1 β) ∈ {e} × B ∧ (a−1, b−1) ∈ A × B )(e, β )(a )(a , b ) ∈ A × B → (a, b)(e, → (a, b)(e, )(e, β )(a )(a−1 , b−1 ) = (ae, ae, bβ)( bβ)(a a−1 , b−1 ) )(e, β )(a )(a−1 , b−1 ) = (aea−1 ,bβb−1 ) → (a, b)(e, )(e, β )(a )(a−1 , b−1 ) = (e,bβb−1 ) → (a, b)(e, → (a, b)(e, )(e, β )(a )(a−1 , b−1 ) = (e,bβb−1 ) ∧ (e,bβb−1 ) ∈ {e} × B )(e, β )(a )(a−1 , b−1 ) ∈ {e} × B → (a, b)(e,
assumed def. of images, since G is onto H normalcy of N N θ is well-defined def of images 18.2(c) algebraic replacement assumed inverse property of A closure property of A algebra algebra algebra definition of e B algebraic replacement
×B ×B
{ }×
21.25 Theorem 15.1 tells us that this intersection is a subgroup. We need only verify that it is normal. assumed ∈ C ∧ g ∈ G → (∀C ∈ C )(c )(c ∈ C ∧ g ∈ G) definition of memb ership in C )(c ∈ C ∧ g ∈ G ∧ C G) defin definit itio ion n of C → (∀C ∈ C )(c −1 → (∀C −1∈ C )(gcg )(gcg ∈ C ) definition of normalcy definition definition of membership membership in C → gcg ∈ C 21.26 Recall the definitions of cosets: N g = {ng : n ∈ N }, gN = {gn : n ∈ N }. If N is normal, then these two c
i
i
i
i
i
i
i
cosets are subsets of each other: N G ( g ( g ( g ( g
↔ ↔ ↔ ↔
∀ ∀ ∀ ∀
N G ( g ( g ( g ( g
gng −1 )
N )(∃m ∈ N )( N )(m m= ∈ G, n ∈ N )( ∈ G, n ∈ N )( N )(∃m ∈ N )( N )(mg mg = gn) gn) N )(∃m ∈ n)(gn )(gn = mg)] mg)] ∈ G) [(∀n ∈ N )( )(gN ⊆ N g ) ∈ G)(gN −1
assumed defin definiti ition on of norm normal alcy cy algebra rearrange rearrangemen mentt of quantifier quantifierss
assumed conseq consequen uence ce of exerci exercise se 12. 12.20 20 algebra rearrange rearrangemen mentt of quantifier quantifierss
∈ G, n ∈ N )( N )(∃m ∈ N )( N )(m m = g ng) ng) N )(∃m ∈ N )( N )(gm gm = ng) ng) ∈ G, n ∈ N )( N )(∃m ∈ n)(ng )(ng = gm)] gm)] ∈ G) [(∀n ∈ N )( )(N g ⊆ gN ) gN ) ∈ G)(N 21.27 i) Because [G [G : N ] N ] = 2, we know that |N | (the number of elements in N ) is exactly one-half that of |G|. ↔ ↔ ↔ ↔
∀ ∀ ∀ ∀
ii) Let a be any element of G that is not in N . N . From the properties of cosets, we know that N a is disjoint from N e and also that aN is disjoint from eN . eN . But since N e = eN = N , N , this means that both N a and aN are disjoint from N . N . Another property of cosets tells us that N a = N = aN .
| | | | | |
35
iii) Because N a is disjoint from N and N a + N = G , we know that N a is the set of all elements of G that aren’t in N :i.e., N :i.e., N a = G N . N . For the same reasons, we know that aN = G N . N . Therefore N a = aN . aN .
−
| | | | | |
−
iv) From problem 21.26, we seen that N a = aN implies N G. 21.29
n
∈ H ∩ N ∧ g ∈ H → n ∈ H ∧ n ∈ N ∧ g ∈ H → n ∈ H ∧ n ∈ N ∧ g ∈ H ∧ g ∈ G → (n ∈ −N 1 ∧ g ∈ G) ∧ (n ∈ H ∧ g ∈ H ) → (gng −1 ∈ N ) N ) ∧ (n ∈ H ∧ g ∈ H ) N ) ∧ (n ∈ H ∧ g ∈ H ∧ g−1 ∈ H ) → (gng −1 ∈ N ) N ) ∧ (gng −1 ∈ H ) → (gng−1 ∈ N ) → gng ∈ H ∩ N
assumed definition of intersection property of subgroups rearrangement of terms from N G inve inverse rse propert property y of H of H closure property of H H definition of intersection
21.31 If every element of G can be expressed as some power of an , then every element of θ(G) can be expressed as θ (an ). By theorem 18.2(c) 18.2(c) this is equivalen equivalentt to saying saying that every element element of θ (G) can be expressed as θ (a)n , which means that the image of θ is equal to θ(a) .
21.32 From property 18.2(a), we know that θ must map the identity element of Z2 onto the identity element of Z3 : so θ([0]2 ) = [0]3 . In order to preserve homomorphism, it must be the case that θ([0]2 ) = θ ([1]2 [1]2 ) = θ ([1]2 )θ ([1]2 ) = [0]3 . This This mean meanss that that θ([1]2 ) must be its own inverse in Z3 , which which necessitate necessitatess that θ ([1]2 ) = [0]3 . (see exercise 11.17(b) for a full proof of this). So there is only one homomorphism, and that is defined by θ ([0]2 ) = θ ([1]2 ) = [0]3 .
⊕
21.33 θ (a + b) = eH = eH eH = θ (a)θ(b) 21.34 Note that the operation on Z Z is the direct product operation, and the operation on the group Z is addition instead of multiplication. So with these two groups, ( a, b)(c, )(c, d) = (a + c, b + d).
×
θ((a, ((a, b)(α, )(α, β )) = θ((a ((a + α, b + β )) = a + α + b + β = (a + b) + (α ( α + β ) = θ ((a, ((a, b)) + θ ((α, ((α, β )) 21.35 If θ is a homomorphic function Z (∆x) But this Z, then it must be the case that θ (x + ∆x) = θ (x) + θ (∆x indicates that θ must be a linear function (of the form θ (x) = ax + b), because:
→
θ(x + ∆x ∆x) ∆x
(∆x) − θ(x) = θ(∆x ∆x
which means that θ(a + ∆x) = θ(b + ∆x) for any possible values of a, b. But we also know that θ must be an odd function (as opposed to an even one), because: θ(0) = θ(x + ( x)) = θ (x) + θ ( x) = 0
−
−
which means that θ( x) = θ (x). So not only must θ be a linear function, it must also be an odd linear function. So the set of homomorphisms can be completely described by:
−
−
{θ : θ(x) = ax,a ∈ Z} 22 Quotient Groups 22.5 Let N a and N b be arbitrary elements of G/N , with a, b (N a)(N )(N b) = N ( N (ab) ab) = N ( N (ba) ba) = (N ( N b)(N )(N a)
∈ G.
defin definit itio ion n of quoti quotien entt grou group p multi multipl plic icat atio ion n abelianism of G G definition definition of quotient quotient group multiplicat multiplication ion
36
22.6
G is cyclic ( a G)(G )(G = a ) ( a G)( g G)( n N)(g )(g = an ) ( a G)( g G)( n N)(N )(N g = N ( N (an )) ( a G)( g G)( n N)(N )(N g = (N a)n ) ( N a G/N )( )( N g G/N )( )( n N)(N )(N g = (N ( N a)n ) ( N a G/N )( )( N g G/N )(N )(N g Na ) ( N a G/N )(G/N )(G/N = N a ) G/N is cyclic
→ → → → → → → →
∃ ∃ ∃ ∃ ∃ ∃ ∃
∈ ∈ ∈ ∈
∈ ∈ ∈
∀ ∈ ∀ ∈ ∀ ∈
∃ ∈ ∃ ∈ ∃ ∈ ∀ ∈ ∃ ∈ ∀ ∈ ∈
definition of cyclic definition of a N is well-defined definitio tion of quotient group multiplicatio tion chang changee of variabl ariablee definition of a definition of set equality definition of cyclic
22.7 G consists of n cosets of size n , or of m cosets of size m . So n n = m m , which is equivalent to | n = ||m = m/n . m n|
| |
| |
22.8
N a(NbNc) NbNc) = N a(N ( N (bc)) bc)) = N (a(bc)) bc)) = N (ab) ab)c) = N (ab) ab)N c = (NaNb) NaNb)N c
| |
| |
| |
definit definition ion of of the quotie quotient nt produ product ct operat operation ion defin definit itio ion n of the the quot quotie ient nt produ product ct opera operatio tion n asso assocciati iativi vitty of the the grou group pG definition of the quotient product operation definition of the quotient product operation
22.9a [G : N ] N ] represents the order of the group G/N . If it is prim prime, e, then then G/N G/N is isomorphic to Z[G:N ] by theorem 19.3. And each group Zn is cyclic. 22.9b Let G = Z8 and let N = [4] . In this this case case,, [G [ G : N ] N ] is not prime even though G/N is cyclic (G/N (G/N is generated by N [1] [1] ).
22.10 [G : N ] N ] = 12. In example 22.3, [3] [2] represent represented ed the direct direct product of two two subgroups: one of order 4 and one of order 2. But ([3], ([3], [2]) is a single subgroup of order 4.
×
22.11 This is a direct consequence consequence of problem problem 21.17. By theorem 22.2, we know that there is a homomorphis homomorphism m η:G G/N defined as η (a) = N a. We are told that that a G. Therefore Therefore,, by exercise exercise 21.17, o(η(a)) o(a), which means o(N a) o(a).
→
∈
|
|
22.12a Note that the normal group in this problem is Z, not N . So the coset containin containingg a is indicated by Za instead of N a. Za
∈ Q/Z → (∃q ∈ Q)(Za = Zq) → (∃r, s ∈ Z)(Za = Z ) → (∃r, s ∈ Z)((Za) = ( Z ) ) → (∃r, s ∈ Z)((Za) = Z( ) ) → (∃r, s ∈ Z)((Za) = {z + ( ) s
r s
s s
r s s r s s
r s s
:z
∈ Z})
assumed definition of membership in Q/Z definition of membership in Q coset operation is well-defined definition of coset operation defin definit itio ion n of Z sr
Remember that we are dealing with an additive group, so exponents are actually repeated addition: in an additive group, 23 = 2 + 2 + 2 = 6. Techni echnical cally ly,, I should should haved haved used used the notation notation r−s to represent division instead of a fraction, but it made the flow of the proof less clear. But, despite the sloppy notation, in an additive group it’s true that ( rs )s = r∗ss = r. So this last step implies: s
→ (∃r, s ∈ Z)((Za) = {z + r : z ∈ Z}) → (∃s ∈ Z)((Za) = Z) → o(Za)|s → o(Za) ≤ s → o(Za) is finite s
algebra closure of addition in Z Theorem 14.3 (Z is the identity element of Q/Z)
The difficult part of this problem is understanding exactly what “the order of an element of Q/Z” is even referring to. Q/Z is a group consisting of an infinite number of elements. Each of these elements is of the form Za = z + a : z Z, a Q , so each Za Q/Z also has an infinite number number of elements. elements. But Za is not generally a group, so it’s order is not the number number of elements elements it contains: contains: the order of Za is the
{
∈
∈ }
∈
37
number of distinct cosets in generated group Za . And the number of distinct distinct cosets depends on what a is.
Z = Z and the order of Z0 is sho If a = 0, for example, then Z0 = z + 0 : z shown to to be 1. If 1 2 2 a = 1/3, then then Za = z + 3 : z Z, a Q , (Za) = z1 + z2 + 3 : z1 , z2 Z, a Q , and 1 3 (Za) = z1 + z2 + z3 + 1 : z1 , z2 , z3 But no matte matterr what what a is chosen, the Z = Z: so o(Z 3 ) = 3. But order of Za is still finite.
{ ∈
{
{
∈ }
∈ } ∈ }
22.12b Let n be some arbitrary integer, and let q =
1 n
{
∈
∈ }
. For the reasons given above, o(N q) = n.
22.13 Let Za be an arbitrary element of R/Z. Remem Remember ber that the only only operati operation on defined defined on this group is 3 addition, addition, so a is equivalent to 3a 3a. o(Za) is ( p ( p ( p ( p ( p ( p ( p
→ → → → → → →
finite
∈ Z)(o )(o(Za) = p) ∈ Z)((Za) = e) ∈ Z)(Z(a ) = e) ∈ Z)(Z(a ) = Z) ∈ Z)({z + a : z ∈ Z, a ∈ R} = Z) ∈ Z)(a )(a ∈ Z) )(a = q ) ∈ Z, q ∈ Z)(a
∃ ∃ ∃ ∃ ∃ ∃ ∃
p
p p
p
p
p
assumed definition of finite theorem 14.3 theorem 22.2 and 18.2(c) Z is the identity of R/Z defin definit itio ion n of Za closure property of Z definition of Z
∈
Just Just as in the previo previous us proble problem, m, note note that that exponen exponents ts in additi additive ve groups groups are actually actually repeate repeated d p 3 addition: addition: in an additive additive group, 2 = 2 + 2 + 2 = 6. So the statem statemen entt a = q is equivalent to the statement a p = q. So this last step implies:
∗
)(a ∗ p = q) → (∃ p ∈ Z, q ∈ Z)(a → (∃ p ∈ Z, q ∈ Z)(a )(a = ) →a∈Q → Za ∈ Q/Z q p
algebra definition definition of Q definition definition of Q/Z
So we have shown that all the finite members of R/Z are members of Q/Z. Q is a subset of R. And problem 22.12 showed that all members of Q/Z are finite. Therefore, the set of all finite members of R/Z is Q/Z. 22.14
G/N is abelian ( a, b G)(N )(N ((a)N ( N (b) = N ( N (b)N ( N (a)) ( a, b G)(N )(N ((ab) ab) = N ( N (ba)) ba)) ( a, b G)(N )(N ((ab) ab)N ( N (ba) ba)−1 = N e) ( a, b G)(N )(N ((ab) ab)N (( N ((ba ba))−1 ) = N e) ( a, b G)(N )(N ((ab) ab)N ( N (a−1 b−1 ) = N e) ( a, b G)(N )(N ((aba−1 b−1 ) = N e) ( a, b G)(N )(N ((aba−1 b−1 ) = N ) N ) ( a, b G)(aba )(aba−1 b−1 N ) N )
↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔
∀ ∀ ∀ ∀ ∀ ∀ ∀ ∀
∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈
∈
defin definit itio ion n of abeli abelian an definition of quotient group oper peratio tion N e is the identity element of G/N Theo Theore rems ms 22 22.2 .2 and and 18 18.2 .2(c (c)) algeb lgebrra defin definit itio ion n of quot quotie ien nt grou group p oper operat atio ion n definition of N N e closure property of N N
22.15 Let n be an arbitrary element of N and let g be an arbitrary element of G. N (e) = N ( N (n) = N ( N (gg −1 ) = N ( N (g )N (g−1 ) = N ( N (ge) ge )N ( N (g −1 ) = N ( N (g )N (e)N ( N (g −1 ) = N ( N (g )N (n)N ( N (g−1 ) = N ( N (gng −1 )
n N iff N ( N (n) = N (e) identity property of G G def. def. of quot quotie ien nt grou group p ope operati ration on,, N is well-defined identity proper perty of G f G def. def. of quot quotie ient nt grou group p opera operati tion on,, N is well-defined n N iff N ( N (n) = N (e) def. of quotient group oper peration, N is well-defined
N (e) = N ( N (gng −1 ) → N ( −1 → gng ∈ N
summar summary y of the above above equali equalitie tiess n N iff N ( N (n) = N (e)
∈ ∈ ∈
38
23 Fundamental Homomorphism Theorem 23.7 (φ η)(a )(a) = φ(η (a)) = φ(N a) = θ (a)
◦
23.8 Let θ : G G be defined as the identity function, θ (g ) = g. The kernel kernel of this function is just K = e , the identity element of G. So, by the fundamental homomorphism theorem, G/K G.
→
{}
≈
23.11 i) Let G be a simple simple abelian group. By the definition definition of “simple”, the only normal subgroups subgroups of G are −1 e and G itself. itself. But all subgroups subgroups of an abelian group are normal (n (n = ngg = gng −1 ), so the only subgroups of G, normal or otherwise, are e and G.
{}
{}
ii) But this means that G cannot be isomorphic to any direct product Zm e G. If it were, then G would have subgroups of size m and n.
{ }×
×Z
n, m , n
> 1 other than
iii) And we care about this because the fundamental theorem of finite abelian groups (section 19), tells us that every abelian group G is isomorphic isomorphic to the direct product of cyclic groups of prime order: i.e., that G = Z p1 . . . Z pn .
× ×
iv) So we see that G is isomorphic to a direct product of cyclic groups of prime order, and that the only direct product isomorphic to G is G e . This implies that both G and e are cyclic groups of prime order. order. And since since G is obviously isomorphic to G, this means that G is isomorphic to a cyclic group of prime order, which is what we wanted to prove.
×{ }
23.12 Let θ : Z18
([a]18 ) = [a]3 . → Z3 be defined as θ([a
θ is well defined [a]18 = [b]18 ( n Z)(a )(a = 18n 18 n + b) ( n Z)(a )(a = 3(6n 3(6n) + b) [a]3 = [b]3 θ([a ([a]18 ) = θ ([b ([b]18 )
assumed defin definit itio ion n of modul modular ar equi equiv valen alence ce alge algebr braa definition of modular equivalence definition of θ
θ is a homomorphism θ([a ([a]18 [b]18 ) = θ ([a ([a + b]18 ) = [a [ a + b ]3 = [a [ a]3 [b]3 = θ ([a ([a]18 ) θ ([b ([b]18 )
definition definition definition definition definition definition
→∃ ∈ →∃ ∈
⊕
⊕
⊕
of of θ of θ of of θ θ
⊕ ⊕
θ is onto H (trivial) kernel of θ = [3] θ([a ([a]18 ) = [0]3 [a]3 = [0]3 ( n Z)(a )(a = 3n 3 n + 0) Ker(θ Ker(θ ) = [3n [3n] : n Z Ker(θ Ker(θ ) = [3]
→ →∃ ∈ → →
{
∈ }
assumed definition definition of θ of θ defin definit itio ion n of modul modular ar equi equiv valen alence ce definition of kernel definition definition of x
This shows that there is a homomorphic function θ : Z18 the fundamental homomorphism theorem, Z18 / [3] Z3 .
≈
23.13 Let θ : Zn
→Z
k
be defined as θ([a ([a]n ) = [a [ a]k .
θ is well defined [a]n = [b]n ( r Z)(a )(a = nr + b) ( r, s Z)(a )(a = (ks) ks)r + b) ( r, s Z)(a )(a = k (sr) sr) + b) [a]k = [b]k θ([a ([a]n ) = θ ([b ([b]n )
→∃ ∈ →∃ ∈ →∃ ∈
→ Z3 with a kernel of [3].
assumed defin definit itio ion n of mo modu dula larr equi equiv valen alence ce from from the the fac factt tha thatt k n associa associativ tivee propert property y definition of modular equivalence definition of θ
|
39
Theref Therefore ore,, by
θ is a homomorphism θ([a ([a]n [b]n ) = θ ([a ([a + b]n ) = [a [ a + b ]k = [a [ a]k [b]k = θ ([a ([a]n ) θ([b ([b]n )
⊕
⊕
definition definition definition definition
⊕
of of θ of of θ θ
⊕ ⊕
θ is onto H (trivial) kernel of θ = [k] θ([a ([a]n ) = [0]k [a]k = [0]k ( r Z)(a )(a = rk + 0) Ker(θ Ker(θ ) = [rk] rk ] : r Z Ker(θ Ker(θ ) = [k ]
→ →∃ ∈ → →
{
∈ }
assumed definition definition of θ of θ defin definit itio ion n of modul modular ar equi equiv valen alence ce definition of kernel definition definition of x
This shows that there is a homomorphic function θ : Zn Zk . the fundamental homomorphism theorem, Zn / [k]
≈
→Z
k
with a kernel of [k ] . Theref Therefore ore,, by
23.16 We know that A is a subgroup from the proof in exercise 21.10. So we need only prove normalcy of A G: n
∈A∧g ∈ G → θ(n) ∈ B ∧ g ∈ G → θ(n) ∈ B ∧ θ(g) ∈ H → θ(n) ∈ B ∧ θ(g) ∈ H ∧ θ(g)−−11 ∈ H → θ(n) ∈ B ∧ θ−(1g) ∈ H ∧ θ(g ) ∈ H → θ(g)θ(−n1)θ(g ) ∈ B → θ(gng−1 ) ∈ B → gng ∈ A
assumed definition of membership in A range of θ is H closure property of H 18.2(c) normalcy of B H property property of homomorphi homomorphism sm of θ definition definition of A of A
23.17 Proof by contrapositiv contrapositive. e. Assume Assume that G/N is not simple simple.. This This mea means ns that there there exists exists some some normal normal subgroup M G/N such that M = G/N and M = N e (the identity element of G/N G/N ). ). Because Because M is a subgroup of G/N , it must contain the identity element of G/N : so N M . M .
⊂
∈
Because η : G G/N is a homomorphism and M G/N , G/N , exercise 23.16 tells us that we can construct a normal set A = g G : η (g ) M . This This tells tells us that A is a normal subgroup of G that contains every element from every coset in M . But we we know know that N is one of the cosets in M , so N A. And A is a normal normal subgroup subgroup of G, so we know that A G. So we know that N A G.
→ { ∈
∈ }
⊆
⊆ ⊆
⊆
In order to prove the contrapositive and show that A is a subgroup strictly between N and G, we must now rule out the possibility that A = N or that A = G. Because M = G/N , we know that there are some elements of G that aren’t in any of the cosets in M . M . And since A is just the set of all elements of all cosets of M , this means that A = G. For similar reasons, reasons, because M = N e, we know that A = N . N . Therefore Therefore N A G.
⊂ ⊂
So by assuming that G/N is not simple, we’ve shown that there is a normal subset strictly between N and G. This is the contrapositive of what we wanted to prove. 23.18 If G is simple, the only normal subgroups of G are G and eG . The kernel of θ(G) is a normal subgroup of G by theorem 21.1, so the kernel is either G or eG . If the kern kernel el is eG , then θ is one-to-one and onto θ (G) which means that G θ(G). If the kernel is G then θ(G) = 0 which means that o(θ (G)) = 1.
≈
{}
23.19 b) This is a direct consequence of theorems 21.2 and 23.1. c) A is an extension of R by R# because A contains a normal subgroup B such that R B and A/B (It can be shown that R B with the function θ : B R defined as θ (α1,b ) = b).
≈ ≈R → d) C contains all the same subgroups that R does, but C ≈ R (section 32). 23.20 a) Both H and K must contain the identity element of G, so e e ∈ H K . This is the identity element of H K , since (hk (hk)( )(ee e ) = hk = (e e )(hk )(hk)) for all hk ∈ H K . And each hk ∈ HK has an inverse: ≈
G G
G G
G G
40
hk
∈ H K ∧ h ∈ H ∧ k ∈ K assumed → h−−11 ∈ H ∧ k ∈ −K 1 existence of inverses in group H from normalcy of K → h−1 ∈ H ∧ hkh −1∈−K 1 → h−1 ∈ H ∧ (hkh ) ∈ K existence of inverses in group K −1 −1 h ∈ K theorem 18.2(e) → h−1 ∈ −H 1 ∧−hk definition of H K → h−1 hk−1 h 1 ∈ HK → k h−1 ∈ H K cancellation of inverses hk ) ∈ H K theorem 18.2(e) → (hk) Note that ab ∈ H K does not necessarily imply that a ∈ H or b ∈ K . The The most most we can assu assume me is that there exist h ∈ H, k ∈ K such that hk = ab. ab. For this reason both the preceeding preceeding and proceeding proceeding proofs explicitly state not only that hk ∈ H K , but also that h ∈ H and k ∈ K . Next, proof that H K is closed:
→ (h1k1 ∈ H K ) ∧ (h2k2 ∈ H K ) ∧ (h1, h2 ∈ H ) ∧ (k1, k2 ∈ K ) → (h1 ∈ H ) ∧ (h2 ∈ H ) ∧ (k1 ∈ K ) ∧ (k2 ∈ K ) → (h1h2 ∈ H ) ∧ (k1−1∈ K ) ∧ (k2 ∈ K ) → (h1h2 ∈ H ) ∧ (h−2 1k1h2 ∈ K ) ∧ (k2 ∈ K ) → (h1h2 −∈1H ) ∧ (h2 k1h2k2 ∈ K ) → h1h2h2 k1h2k2 ∈ K → h1k1h2k2 ∈ K →
assu assume med d closure property of H H normalcy of K , exercise 21.20 closure property of K K definition of H K cancellation of inverses
Finally, we show that K H K .
(hk1 H K ) (h H ) (k K ) ( j K )[(h )[(h H ) (k K ) ( j K )] )] −1 −1 ( j K )[(h )[(h H ) (k K ) ( j K )] )] −1 −1 −1 ( j K )[(h )[(h H ) ( j kj K )] −1 −1 −1 ( j K )(h )(h j kj H K ) ( j K )((h )((h−1 j −1 )kj −1 H K ) ( j K )(( jh )(( jh))kj −1 H K ) ( j K )( j )( j((hk) hk) j −1 H K ) K HK
→ → → → → → → →
∈ ∀ ∈ ∀ ∈ ∀ ∈ ∀ ∈ ∀ ∈ ∀ ∈ ∀ ∈
∧ ∈ ∧ ∈ ∈ ∧ ∈ ∧ ∈ ∈ ∧ ∈ ∧ ∈ ∈ ∧ ∈ ∈ ∈ ∈ ∈
assumed existe existence nce of inv invers erses es in in group groupss closure property of K K definition of H K associativity of operation on G theorem 18.2(e) associativity of operation on G defintion of normalcy
23.20 b) θ(ab) ab) = K (ab) ab) = K (a)K (b) = θ (a)θ (b) 23.20 c) Proof that ker θ
⊆ H ∩ K :
x
∈ ker θ assumed domain of θ is H , identity of codomain is K → x ∈ H ∧ θ(x) = K definition of θ → x ∈ H ∧ K x = K partial partial definition definition of set equality equality → x ∈ H ∧ K x ⊆ K )(kx = j ) quantification of ⊆ → x ∈ H ∧ (∀k ∈ K )()(∃ j ∈ K )(kx −1 −1 → x ∈ H ∧ (∀k ∈ K )()(∃ j ∈ K )(k )(k kx = k j) j ) algeb lgebra ra −1 )(x = k j) j ) cancellation of inverses → x ∈ H ∧ (∀k ∈ K )()(∃ j ∈ K )(x → x ∈ H ∧ x ∈ K closure property of K definition of intersection → x ∈ H ∩ K Proof that H ∩ K ⊆ ker θ: x ∈ H ∩ K assumed definition of intersection → x ∈ H ∧ x ∈ K x H x K θ ( x ) = Kx definition of θ → ∈ ∧ ∈ ∧ → x ∈ H ∧ θ(x) = K lemma 16.1, from fact that x ∈ K domain of θ is H , identity of the range is K . → x ∈ ker θ 41
Given the properties that were proven in steps (a-c), the isomorphism of the last two groups is a direct consequence of the fundamental homomorphism theorem.
25 Integral Domains and Subrings *Double-check 25.18. The first assumption doesn’t seem to be valid. 25.10 Let p be prime, and let [a [ a] and [b [b] be arbitrary elements of Z p such that [a [a] [b] = 0. We can prove that Z p is an integral domain by showing that one of these two elements must be the zero of Z p :
⊗
( a, b Z p )([a )([a] p [b] p = [0] p ) ( a, b Z p )([ab )([ab]] p = [0] p ) ( a,b,n Z)(ab )(ab = np + 0) ( a, b Z p )( p ab) ab) ( a, b Z p )( p a p b) ( a,b,m,n Z)( pn )( pn = a pm = b) ( a, b Z p )([0] p = [a] p [0] p = [b [ b] p )
∃ → → → → → →
∃ ∃ ∃ ∃ ∃ ∃
∈
⊗
∈ ∈ ∈ | ∈ | ∨ | ∈ ∨ ∈ ∨
assumed definition of the operation definition of modu odular equivalence definition of divisibility property of p being prime defin definit itio ion n of divi divisi sibi bili lity ty definit definition ion of modul modular ar equi equiv valence alence
⊗
√
√
25.11 Z[ 2] is a subset of R. If there were zero divisors in Z[ 2], these would also be zero divisors of R. And since we are given that R is an integral domain, there must not be any such zero divisors. Therefore Z[ 2] is an integral domain.
√
25.12 Let f and g be piecewise defined functions such that f (1) f (1) = 1, 1, f ( f (x) = 0 otherwise and g (2) = 1, 1, g (x) = 0 otherw otherwise ise.. Both Both of these function functionss are in M ( M (R), neither of them are the zero function 0(x 0( x), but their product is equivalent to the zero function. 25.13 Let R be a ring with zero element 0 R , and let S be a ring with zero element 0 S . case i) If S is not an integral domain, then there are nonzero elements s1 , s2 such that s1 s2 = 0S . This means that the nonzero elements (0R , s1 )(0R , s2 ) = (0R , 0S ) and therefore R S is not an integral domain. Similar results occur if R is not an integral domain.
×
case ii) If R If R is an integral domain and S contains some other element than 0 S , then the nonzero elements (0R , s1 )(r )(r1 , 0S ) = (0R , 0S ) and therefore R S is not an integral integral domain. domain. Similar Similar results occur if S is an integral domain and R contains a nonzero element.
×
case iii) If R is an integral domain and S contains only the zero element 0S , then (r (r1 , 0S )(r )(r2 , 0S ) = (r1 r2 , 0S ). (r1 r2 , 0S ) is the zero element of R S when and only when one of the terms on the left-hand side of the equation is the zero element of R S : this means that R S has no zero divisors. R S also has the nonzero unity element (1 R , 0S ) where 1R is the unity element of R. So in in this this case, R S is an integral domain. domain. Similar Similar results occur if S is an integral domain and R contains only the zero element 0 R .
×
×
×
×
25.14 Let D be a commutative ring with a,b,c b = c). Proof by contradiction: D has at least one zero divisor ( a, b D)(ab )(ab = 0 a = 0 b = 0) ( a, b D)(ab )(ab = a0 a = 0 b = 0) ( a, b D)(b )(b = 0 a = 0 b = 0)
→∃ ∈ →∃ ∈ →∃ ∈
×
∧ ∧ ∧ ∧ ∧ ∧
( ab = ac → ∈ D, a = 0, and the property of left cancellation (ab
defin definiti ition on of zero zero divi diviso sorr prope propert rty y ooff zer zeroo left left canc canceellat llatio ion n
And this last statemen statementt is contrad contradict ictory ory.. So if all of our assumpti assumptions ons are true, it cannot cannot be the case that D has zero divisors. Note that this does not necessarily prove that D is an integral domain, since we have not shown that it has a nonzero unity. 25.16 C (R) is nonempty nonempty becaus b ecausee it contains contains 0x 0x. For each continuous function f ( f (x), its negative function f ( f (x) is also continuous and therefore f ( f (x) C (R). The sum or product of two two continuous continuous functions functions is also continuous, but I haven’t taken an analysis course and therefore have no idea how to prove this.
−
−
∈
42
25.17 From theorem 15.1 we know that the intersection properties. nonemptiness eR R ( C i C i )(e )(eR eR C i
∈ → ∀ ∈ → ∈
i
necessary property of groups property of subgroups, lemma 7.1 definition of intersection
∈ C ) i
closed under its operations a C i b C i ( C i C i )(a )(a C i b C i ) ( C i C i )(ab )(ab C i (a + b) ab C i (a + b) C i
∈ ∧ ∈ → ∀ ∈ ∈ ∧ ∈ → ∀ ∈ ∈ ∧ → ∈ ∧ ∈
∈ C ) i
existence of negatives a R ( C i C i )(a )(a C i ) ( C i C i )( a C i ) a C i
∈ → ∀ ∈ → ∀ ∈ →− ∈
C is a subgroup of R, so we need only prove ring
assumed definition of intersection closur closuree propert property y of rings rings closure closure property property of rings
necessary property of groups property of subgroups, lemma 7.1 property of subgroups definition of intersection
∈
− ∈
25.18 Let S be any subset of a ring R, and let S denote the intersection of all of the subrings of R that contain S . Then S is the unique smallest subring of R that contains S in the sense that
(a) S contains S
(b) S is a subring
(c) if T is any subring of R that contains S , then T contains S . The proof of this follows follows directly from Theorem Theorem 15.2 and exercise exercise 25.17. Because Because rings are just additive additive groups with an extra operation defined on them, theorem 15.2 tells us that S is the smallest unique subgroup of R that contains S . Exercise 25.17 tells us that this subgroup is also a subring. Together, they tell us that S is the unique smallest subring.
25.19 Let C i be an intersection of integral domains of ring R. We proved that this intersection is a subring of R in exercise 25.17, so we need only prove that it is also an integral domain.
∈ ∧ ∈ → ∈ ∧ ∈ ∧ ∈ → ∀ ∈ ∈ ∧ ∈ ∧ ∈ C ) → ∀ ∈ ∈ ∧ ∈ → ∀ ∈ ∈ ∧ ∧ ∈ C ) → ∈ ∧ ∧ ∈ → lemma 2: no zero divisors (by contradiction) C has atleC ast has one zero divisor )(ab = 0 ∧ a = → (∃a, b ∈ C )(ab 0 ∧ b = 0) )(ab = 0 ∧ a = → (∀C ∈ C )(∃a, b ∈ C )(ab 0 ∧ b = 0) lemma 1: C i is commutative a C i b C i a C i b C i ab C i ( C i C i )(a )(a C i b C i ab ( C i C i )(ab )(ab C i ab C i ) ( C i C i )(ba )(ba C i ab = ba ab ba C i ab = ba ab C i C i is commutative
i
i
assumed closure of subrings definition of intersection closure of subrings commutativity of integral domains definition of intersection definition of commutative
i
i
i
i
i
→ (∀C ∈
i
C i )(C )(C i has at least one zero divisor)
i
assumed definition of zero divisor definition of intersection definition of zero divisor
∈
The second lemma shows, by contradiction, that if none of the elements C i C i contain a zero divisor, then C i itself has no zero divisors. So the fact of whether or not C i is an integral domain rests entirely entirely on whether or not it contains contains a unity element. element. It turns out that it does: each each element element C i , being an integral domain, has a unity element and the following proof shows that these unity elements must all be identical:
Let C 1 and C 2 be two elements of
C . Let e i
1
and e2 be their respective unity elements (which they must 43
have as a consequence of being integral domains): ( a C 1 , b C 2 )(a )(a = ae1 b = be2 ) ( a C 1 , b C 2 )(ab )(ab = ab a = ae1 ( a C 1 , b C 2 )(ae )(ae1 b = abe2 ) ( a C 1 , b C 2 )(abe )(abe1 = abe2 )
∀ → → →
∈ ∈ ∀ ∈ ∈ ∀ ∈ ∈ ∀ ∈ ∈
∧
∧
definition of unity
∧ b = be2)
algebraic replacement on ab = ab commutativity from lemma 2
Note that we can’t just use the left-cancel left-cancellation lation property property of groups here, here, since left-cancell left-cancellation ation depends on the existence existence of inverses inverses and integral integral domains don’t necessaril necessarily y have have multiplica multiplicative tive inverse inverses. s. But it turns out that left cancellation cancellation also works in integral integral domains (albeit for a different different reason): )((ab((e1 − e2 ) = 0) algebra → (∀a ∈ C 1, b ∈ C 2)((ab → (∀a ∈ C 1, b ∈ C 2)(ab )(ab = 0 ∨ e1 − e2 = 0) lemma 2 )(a = 0 ∨ b = 0 ∨ e1 − e2 = 0) lemm lemmaa 2 ag agai ain n → (∀a ∈ C 1, b ∈ C 2)(a algebra → a = 0 ∨ b = 0 ∨ e1 = e2 And it can’t be the case that (∀a ∈ C 1 )(a )(a = 0): if every element element were were zero, then it wouldn’t wouldn’t have a unity element and therefore wouldn’t be an integral domain. For the same reason, it can’t be the case that (∀b ∈ C 2 )(b )(b = 0). This means that this last statement implies: algebra → e1 = e2
Therefore each element of C i contains the same unity element e, which means that This shows that C i meets all criteria for an integral domain.
C contains e. i
25.20 From exercise 25.18, we see that the smallest subring containing n is n . The only such such subrings subrings that contains the multiplicative unity are 1 and 1 , so these are the only subrings that are integral domains.
−
25.21 This subring does not contain zero divisors in Z. I had to use Mathematica Mathematica to determine determine this. this. I’m sure there’s some elegant proof that treats the roots as isomorphic to Z3 , but I can’t find it. 25.22 Let C represent the center of R. existence of an identity element eR R ( a R)(e )(eR r = re R ) eR C
∈ →∀ ∈ → ∈
closure of multiplication a C b C ( c R)(ac )(ac = ca bc = cb) cb) ( c R)(acb )(acb = cab abc = acb) acb) ( c R)(abc )(abc = cab) cab) ab C
∈ ∧ ∈ →∀ ∈ →∀ ∈ →∀ ∈ → ∈
∧
∧
closure of addition a C b C ( c R)(ac )(ac = ca bc = cb) cb) ( c R)(ac )(ac + bc = ca + cb) cb) ( c R)((a )((a + b)c = c(a + b)) (a + b) C
∈ →∀ →∀ →∀ →
∧ ∈ ∈ ∈ ∈ ∈
∧
identity element of group R property of the identity element definition of the center assumed definition of the center left left and and righ rightt mult multipl iplica icatio tion n algebraic replacement definition of the center
assumed defin definit itio ion n of the the cen center ter algebra left left and righ rightt distri distribut butiv ivee propert property y definition of the center
44
existence of negatives a C assumed ( c R)(ac )(ac = ca) ca) definition of the center ( c R)(ac )(ac = ca c = c) ( c R)(ac )(ac = ca 0c = c0) property of the zero element ( c R)(ac )(ac = ca (a a)c = c(a a)) propert property y of negati negative vess ( c R)(ac )(ac = ca ac ac = ca ca) ca) dist distri ribu butiv tivee prope propert rty y of ring ringss ( c R)(ca )(ca ac = ca ca) ca) algebraic replacement ( c R)( (ac) ac) = (ca) ca)) left cancellation ( c R)(( a)c = c( a)) 24.2(b) a C definition of the center The center is a commutative ring R is R itself.
∈ →∀ →∀ →∀ →∀ →∀ →∀ →∀ →∀ →−
∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈
∧ ∧ ∧ − ∧ − − − − − − −
− −
25.23 The center is the subring consisting of all matrices of the form
x 0 0
y
.
25.24 M (R) is its own center because its multiplicative operation is commutative. 25.25 A subset S of integral domain R is itself an integral domain, by definition, if it is a commutative subring of R with no zero divisors and a unity element. But as long as S is a ring, it inherits commutativity and lack of zero divisors from R. So it is sufficient that (a) S is a subring of R (b) S contains a unity element
26 Fields 26.11 By definition, a field is a commutative ring whose nonzero elements form a group with respect to multiplication. tiplication. If R is an integral domain, then it is by definition a commutative ring with a nonzero unity element. Because R is a ring, multiplica multiplication tion is closed closed and associative. associative. The only way R could fail to be a field is if the multiplicative operation is not closed on the set of nonzero elements. But if we assume that each nonzero element has an inverse, then the operation is closed on this set. Proof by contradiction: R
− {0} is not closed closed under under multi multipli plicat cation ion )(ab = 0) → (∃a, b ∈−R1 − {0})(ab → (∃a,b,b−1 ∈ R − {0})(ab )(ab = 0 ∧ bb−1 = e) )(ab = 0 ∧ abb−1 = a) → (∃a,b,b−1 ∈ R − {0})(ab )(ab = 0 ∧ 0b−1 = a) → (∃a,b,b−1 ∈ R − {0})(ab → (∃a,b,b ∈ R − {0})(ab )(ab = 0 ∧ 0 = a) )(a = 0) → (∃a ∈ R − {0})(a
hypoth hypothesi esiss to be contra contradic dicted ted definition of closure given: given:eac each h nonz nonzero ero elemen elementt has has an inve inverse rse alge algebr braa alge algebr brai aicc repl replac acem emen entt property of zero
This This last last statem statemen entt is a contra contradic dictio tion, n, which which means that our assumpti assumption on is false: false: if R is an integral domain with an inverse for each nonzero element, then the nonzero elements in R are closed with respect to multiplication. And this was the final property needed for R to be a field. 26.12 As in the last problem, we need only prove that the multiplicative operation is closed on the set of nonzero elements. And if there is a unique solution for ax = b, a = 0 then the operation is closed on this set. Proof by contradiction:
R
− {0} is not closed closed under multiplic multiplication ation )(ab = 0) → (∃a, b ∈ R − {0})(ab )(ab = 0 ∧ a0 = 0) → (∃a, b ∈ R − {0})(ab → (∃b ∈ R − {0})(b )(b = 0)
hypothesis hypothesis to to be contradi contradicted cted definition of closure property of zero unique solution for ax = b
This This last last statem statemen entt is a contra contradic dictio tion, n, which which means that our assumpti assumption on is false: false: if R is an integral domain with a unique solution for ax = b, then the nonzero elements in R are closed with respect to multiplication. And this was the final property needed for R to be a field. 26.13 Z# n is a group with respect to
when n is prime. 45
26.15 The commutativity of the field along with the properties of zero and unity determine every product involving 0 and e. The only other products are aa, bb, and ab. ab. It cannot cannot be the case that that aa = 0, ba = 0, ab = 0, or bb = 0 (nonzero elements are closed w/r/t multiplication) or that aa = a,ba = a,ab = a,ba = b,ab = b, or bb = b (uniquenes (uniquenesss of the unity element). element). By elimination elimination it must be the case that ab = ba = e. This This means that it cannot be the case that aa = e or bb = e (uniqueness of inverses), so by elimination it must be the case that aa = b,bb = a. 26.16
⊕
(0,0) (0,1) (1,0) (1,1)
(0, (0,0) (0, (0,0) (0, (0,1) (1, (1,0) (1, (1,1)
(0,1) (0,1) (0,0) (1,1) (1,0)
(1,0) (1,0) (1,1) (0,0) (0,1)
(1,1) (1,1) (1,0) (0,1) (0,0)
+ 0 e a b
0 0 e a b
e e 0 b a
a a b 0 e
b b a e 0
These two functions are isomorphic under the function θ(0) = (0, (0, 0), 0), θ(e) = (0, (0, 1), 1), θ(a) = (1, (1, 0), 0), θ(b) = (1, (1, 1). 26.17 ([1], ([1], [0]) and ([0], ([0], [1]) are two nonzero elements of R R, but their their product product is zero. zero. This This means that that the ring R R contains contains a zero divisor. divisor. This isn’t in conflict conflict with 26.16 because we’ve we’ve merely merely created an isomorphism between the additive group of the ring R R and the additiv additivee group of the field. field. But the function θ is not an isomorphism between their multiplicative groups, so there is no reason to expect that R R should be a field.
× ×
×
×
26.18 Let F = F 1 F 2 F n be a direct product of fields. fields. Choose a F such that a = (0, (0, a2 , a3 , . . . , an ) where each element ai is an arbitrary element of F i . Choos Choosee b F such that b = (b1 , 0, 0, . . . , 0). These These are both nonzero elements, but their product is zero.
× ×···×
∈
∈
26.19 If a set K meets all of the properties listed in theorem 26.2, then it already meets the definition of a subring of F . F . This This mean meanss that that K inherits multiplicative associativity and commutativity from the multiplicative group of F and additive associativity from the additive group of F . F . It also also inheri inherits ts the distributive property and a lack of zero divisors from F . F . This meets all the requiremen requirements ts for K being a field.
26.20 Because each the field F is also a ring, we know from problem 25.17 that C i is a subring. As explained in problem 26.19, this means that it inherits most of the necessary field properties. In order to show that C i is a subfield, we need only to show that it contains the zero and unity of F , F , is closed under both operations, and that it contains negatives and inverses of each of its elements. Since every subfield in C i contains the unity and zero of F , F , so must C i itself. Proofs of the remaining properties:
additive and multiplicative closure a C i b C i ( C n C i )(a )(a C n b C n ) ( C n C i )(ab )(ab C n a + b C n ) (ab C i a + b C i )
assumed definitio tion of intersectio tion clos closur uree of of both both grou groups ps in the the fiel field d C n definition of intersection
negatives and inverses a C i ( C n )(a C n ) C i )(a ( C n C i )( a C n a−1 a C i a−1 C i
assumed definition of intersection comple com pleten teness ess of inv invers erses es and and nega negativ tives es in in the the field field C n definition of intersection
∈ ∧ ∈ → ∀ ∈ ∈ ∧ ∈ → ∀ ∈ ∈ ∧ ∈ → ∈ ∧ ∈ ∈ → ∀ ∈ ∈ → ∀ ∈ − ∈ ∧ →− ∈ ∧ ∈
∈ C ) n
26.21 Let S be a subset of a field F , F , and let S denote the intersection of all of the subfields of F that contain S . Then S is the unique smallest subfield of F that contains S , in that
(a) S contains S
(b) S is a subfield
(c) If T is any subfield of F that contains S , then T contains S
Property (a) is true, since each element of S contains S by definition. Property (b) was proven in exercise 26.20. Property (c) is true by the property of intersections that states ( S S i )( S i S ). ).
∀ ∈
46
⊆
∀ ∈ Z)( 1 ∈ F ). F ). And it must
26.22 Let F be a field containing Z . It must contain contain multiplicativ multiplicativee inverse inverses, s, so ( n m be closed under multiplication, so ( m, n Z)( n F ). F ). This field describes Q.
∀
∈
∈
n
26.23a We’re asked to show that the set of invertible elements is a subgroup of the multiplicative group of R. By theorem 7.1, we need to show show that the set of invertib invertible le elements elements is nonempty nonempty,, closed closed under multiplication, tiplication, and contains contains inverses inverses of each of its elements. elements. Let H represent the set of invertible elements: H = a R : ( b R)(ab )(ab = e) .
{ ∈
∃ ∈
nonempty e R ee = e e H
∈ → → ∈
}
given property of R property of unity definition of H
closed under multiplic multiplication ation a H b H ( r, s R)(ar )(ar = e bs = e) ( r, s R)(ar )(ar = e bs = e rs R) ( r, s R)(ar )(ar((bs) bs) = e rs R) ( r, s R)(ab )(ab((rs) rs) = e rs R) ( rs R)(ab )(ab((rs) rs) = e) ab H
∈ ∧ ∈ →∃ ∈ →∃ ∈ →∃ ∈ →∃ ∈ →∃ ∈ → ∈
∧ ∧
∧ ∈ ∧ ∈ ∧ ∈
assumed definition of H H multip multiplic licati ative ve closur closuree of R of R multip tiply both both equations commutativity of R R change of variable definition of H
invertibility of all elements a H ( r R)(ar )(ar = e a H ) ( r R)(ar )(ar = e a R) ( r R)(ar )(ar = e ar R a R) ( r R)(ar )(ar = e (ar) ar)−1 R a R) ( r R)(ar )(ar = e (ar) ar)−1 R arr −1 R) ( r R)(ar )(ar((ar) ar)−1 = e arr −1 R) ( r R)(arr )(arr −1 a−1 = e arr −1 R) ( r R)(a )(a−1 (arr −1 ) = e arr −1 R) −1 a H
∈ →∃ →∃ →∃ →∃ →∃ →∃ →∃ →∃ →
∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈
∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∧ ∈
assumed definition of H H H is a subset of R closure property of R R existen tence of inverses in R existe existence nce of inve inverse rsess in R definition of inverses theorem 14.1 commutativity of R R definition of H
26.24 Let a be an arbitrary element of commutative ring R Proof by contradiction: a is invertible and a is a zero zero divi diviso sorr ( b, c R 0 )(ac )(ac = e ab = 0) ( b, c R 0 )(bac )(bac = b ab = 0) ( b, c R 0 )(abc )(abc = b ab = 0) ( b, c R 0 )(0c )(0c = b) ( b R 0 )(b )(b = 0)
→ → → → →
∃ ∃ ∃ ∃ ∃
∈ ∈ ∈ ∈
−{ } −{ } −{ } −{ } ∈ −{ }
∧ ∧ ∧
assu assume med d definit definition ion of inve inverti rtible ble,, zero zero diviso divisorr alge algebr braa commu com mutat tativi ivity ty of R of R algebraic replacment property of zero
This last statement is a contradiction: it cannot be the case that b
∈ R − {0} and b = 0.
26.25a Ker(ρ Ker(ρr ) = 0 iff ( a R 0 )(ar )(ar = 0) iff (r (r = 0 or r is a zero divisor). The “iff”s here are true by consequence of the definitions of ρr , 0, and zero divisors.
{}
∃ ∈ −{ }
26.25b The field R, by definition, definition, has no zero divisors. divisors. If r = 0, then Ker(ρ Ker(ρr ) = 0 by the result of 26.25a. From theorem 21.1, this means that the function ρr is a one-to-one function. function. In order to show that it is an isomorphism, then, we need only show that the function is homomorphic:
{}
ρ(a + b) = (a + b)r = ar + br = ρ(a) + ρ(b)
27 Isomorphism and Characteristic 27.1 If θ is an isomorphism between two rings R and S , then by definition it is also an isomorphism between their multiplicative groups. So, by theorem 18.2, θ (eR ) = eS . Note that θ is also an isomorphism between the two additive groups, so the same theorem tells us that θ(0R ) = 0S .
47
27.2
θ:R S is an isomorphism and R is commutative ( s, t S )( )( a, b R)(θ )(θ (a) = s θ (b) = t ab = ba) ba) ( s, t S )( )( a, b R)(θ )(θ (a) = s θ (b) = t θ (ab) ab) = θ(ba)) ba)) ( s, t S )( )( a, b R)(θ )(θ (a) = s θ (b) = t θ (a)θ(b) = θ(b)θ(a)) ( s, t S )( )( a, b R)(θ )(θ (a) = s θ (b) = t st = ts) ts) ( s, t S )(st )(st = ts) ts) S is commutative
→ → → → → →
∀ ∀ ∀ ∀ ∀
→
∈ ∈ ∈ ∈ ∈
∃ ∃ ∃ ∃
∈ ∈ ∈ ∈
∧ ∧ ∧ ∧
∧ ∧ ∧ ∧
assumed definition of onto, commutativity θ is well-defined homo homomo morp rphi hism sm of θ of θ algebraic replacement removal of unnecessary quantifiers definition of commutative
27.3 We know that S must contain a unity from exercise 27.1 We need only show that S contains no zero divisors. Proof by contrapositive: The ring S contains a zero divisor ( s, t S 0S )(st )(st = 0S ) ( s, t S 0S , a , b R 0 R )(st )(st = 0S ( s, t S 0S , a , b R 0 R )(st )(st = 0S ( s, t S 0S , a , b R 0 R )(st )(st = 0S ( s, t S 0S , a , b R 0 R )(st )(st = 0S ( a, b R 0 R )(ab )(ab = 0R ) the ring R contains a zero divisor
→ → → → → → →
∃ ∃ ∃ ∃ ∃ ∃
∈ ∈ ∈ ∈ ∈ ∈
−{ } −{ } −{ } −{ } −{ } −{ }
∈ ∈ ∈ ∈
−{ } −{ } −{ } −{ }
∧ θ(a) = s ∧ θ(b) = t) ∧ θ(a)θ(b) = 0 ) ∧ θ(ab) ab) = 0 ) ∧ ab = 0 ) S
S
R
assumed definition of a zero divisor onto onto-n -nes esss of θ of θ algebraic replacement homomorphism of θ θ exercise 27.2 removal of unnecessary quantifiers definition of a zero divisor
By contrapositive we have shown that if R contains no zero divisors, then S contains no zero divisors. 27.4 S is a field if it is an integral domain with multiplicative inverses for each nonzero element. From exercise 27.3, θ guarantees that S is an integral domain. We need only show that it contains multiplicative inverses. s
27.9
assumed ∈ S )(θ (a) = s) onto-ness of θ → (∃a ∈ R)(θ −1 → (∃a ∈ R)(θ )(θ (a) = s ∧ a ∈ R) exis existe tenc ncee of inv inverse ersess in R −1 )(θ (a) = s ∧ θ (a ) ∈ S ) θ : R → S → (∃a ∈ R)(θ → (∃a ∈ R)(θ )(θ (a) = s ∧ θ (a)−1 ∈ S ) 18.2 18.2(b (b)) )(θ (a) = s ∧ s−1 ∈ S ) alge algebr brai aicc repl replac acem emen entt → (∃−a1 ∈ R)(θ removal of unnecessary quantifiers → s ∈ S a ∈ E assumed −1 existence of inverses in E → aa −=1 e → θ(aa )−=1 θ(e ) 0 θ is well-defined, a = θ homomorphic → θ(a)θ(a−1) = θ(e ) 18.2(a) → θ(a)θ−(1a ) = e−1 −1 → θ(a)−1 θ(a)θ(a−1 ) = θ(a) e algebra properties of inverses and unity → θ(a ) = θ(a) E
E
E
F
F
27.10 From theorem 25.2, we need only show that θ(R) is closed under both operations and that it contains the negative of each of its elements. multiplicative closure a θ(R) b θ (R) ( s, t R)(θ )(θ (s) = a θ (t) = b) ( s, t R)(θ )(θ (s) = a θ (t) = b st ( s, t R)(θ )(θ (s)θ(t) = ab st R) ( s, t R)(θ )(θ (st) st) = ab st R) ( st R)(θ )(θ(st) st) = ab) ab) ab θ(R)
∈ ∧ ∈ →∃ ∈ →∃ ∈ →∃ ∈ →∃ ∈ →∃ ∈ → ∈
∧ ∧
∧ ∈ R) ∧ ∈ ∧ ∈
assumed definition of θ(R) clos closur uree of R f R algebra homomorphism of θ θ change of variable definition of θ (R)
48
additive closure a θ(R) b θ (R) ( s, t R)(θ )(θ (s) = a θ (t) = b) ( s, t R)(θ )(θ (s) = a θ (t) = b s + t R) ( s, t R)(θ )(θ (s) + θ (t) = a + b s + t R) ( s, t R)(θ )(θ (s + t) = a + b s + t R) ( (s + t) R)(θ )(θ(s + t) = a + b) a + b θ ( R)
∈ →∃ →∃ →∃ →∃ →∃ →
∧ ∈ ∈ ∈ ∈ ∈ ∈ ∈
∧ ∧
∧
∧ ∧
∈
∈ ∈
existence of negatives a θ(R) ( s R)(θ )(θ(s) = a) ( s R)(θ )(θ(s) = a ss−1 = eR ) ( s R)(θ )(θ(s) = a θ(ss−1 ) = θ (eR )) ( s R)(θ )(θ(s) = a θ(s)θ (s−1 ) = θ (eR )) ( s R)(θ )(θ(s) = a θ(s)θ (s−1 ) = eS ) ( s R)(aθ )(aθ((s−1 ) = eS ) ( s R)(θ )(θ(s−1 ) = a−1 ) a−1 θ (R)
∈ →∃ →∃ →∃ →∃ →∃ →∃ →∃ →
∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈
∧ ∧ ∧ ∧
assumed definition of θ (R) closu losurre of R of R algeb lgebrra homomorphism of θ f θ change of variable definition of θ (R) assumed definition of θ (R) closure of R θ is well-defined homo homomo morp rphi hism sm of θ of θ 18.2(a) algebraic replacement uniqueness of inverses definition of θ (R)
27.11 Let the function θ : Z p D be defined as θ ([a ([a]) = ae. ae. We wish to show that the image of θ is a subring of D isomorphic to Z p . Exercise 27.10 showed that the image of a ring homomorphism is itself a subring, so we need only the isomorphism Z p θ (Z p ). In order order to do this, this, we must must show show that θ is well-defined, one-to-one and onto the image of θ, and preserves the operations of D.
→
≈
θ is well-defined and one-to-one [a] p = [b] p ( r Z)(a )(a = pr + b) ( r Z)(a )(a b = pr) pr) ( r Z)((a )((a b)e = pre) pre) (a b)e = 0 ae be = 0 ae = be θ ([a ([a]) = θ([b ([b])
↔ ↔ ↔ ↔ ↔ ↔ ↔
∃ ∈ ∃ ∈ ∃ ∈ − −
− −
assumed definition of modular equivalence algebra property of unity D has characteristic p distributive property of D D algebra definition of θ
Note that the right arrow of each step (the “if” direction) shows that θ is wellwell-defi defined ned,, and the left left arrow (the “only if” direction) shows that θ is onto. θ preserves addition θ([a ([a] [b]) = θ ([a ([a + b]) = (a ( a + b)e = ae + be = θ ([a ([a]) + θ([b ([b])
⊕
definition of definition of θ distributive property of D
⊕
θ preserves multiplication θ([a ([a] [b]) = θ ([ab ([ab]]) definition of = (ab ( ab))e definition definition of θ of θ = (ae ( ae)( )(be be)) exercise 27.6 θ([a ([a])θ ])θ([b ([b]) definition of θ And since θ is onto its image by definition, this is all that is necessary to prove that θ (Z p ) is a subring of D isomorphic to Z.
27.12 Theorem 19.2 can be applied to both the additive and multiplicative groups of rings. 27.13 Commutativ Commutativity ity,, lack lack of zero divisors, divisors, existence existence of a unity unity element, element, being an integral integral domain, having characteristic 0
49
27.14 It must have either characteristic 1 (if it is isomorphic to Z1 ) or 2, since 2x 2 x = (x + x) = (x 27.15 Let θ : R
→ S be an isomorphism between two rings.
R has characteristic m ( r R)(mr )(mr = 0) ( r R)(mr )(mr = 0) ( s ( r R)(mr )(mr = 0) ( s ( r R)(mr )(mr = 0) ( s ( s S )(θ )(θ(0) = ms) ms) ( s S )(ms )(ms = 0)
→ → → → → →
∀ ∀ ∀ ∀ ∀ ∀
− x) = 0.
∈ ∈ ∈ ∈ ∈ ∈
∧ ∀ ∈ S )()(∃r ∈ R)(θ )(θ(r ) = s) )(mθ((r) = ms) ms) ∧ ∀ ∈ S )()(∃r ∈ R)(mθ )(θ(mr) mr) = ms) ms) ∧ ∀ ∈ S )()(∃r ∈ R)(θ
assumed definition of characteristic θ is onto alge algebr braa 18.2(c 18. 2(c)) for for additi additive ve groups groups algebraic replacment of mr m r =0 18.2(a)
Becaus Becausee isomor isomorphi phism sm is reflexi reflexive ve,, the same steps steps can be used used to show show that that if S has characteristic n, then ( r R)(nr )(nr = 0). This This shows shows that m : ( r R)(mr )(mr = 0) = n : ( s S )(ns )(ns = 0) . The leas leastt such shared element is the characteristic of both S and R.
∀ ∈
27.16 Z2
{
∀ ∈
} {
∀ ∈
}
× Z2 has characteristic 2, while Z4 does not.
27.17 θ ([a ([a]
([ab]) ]) = ([ab ([ab]]2 , [ab] ab]3 ) = ([a ([a]2 , [a]3 ) ⊗ ([b ([b]2 , [b]3 ) = θ([a ([a]) ⊗ θ ([b ([b]) [b]) = θ([ab 27.18 In Z4 : (2[2] (2[2] = [2] ⊕ [2] = [2 + 2] = [0]) but (2[1] = [1] ⊕ [1] = [1 + 1] = [2]). Or, more trivially, let a = 0. 27.19 The subring {[0], [0], [2], [2], [4]} of Z6 is a ring of characteristic 3, but it is not an integral domain and its nonzero elements are not closed under the multiplicative operation.
27.20a Let m, n be integers and let z represent represent their least common multiple. multiple. Define the function θ : Zz ([a]z ) = ([a ([a]m , [a]n ). Zm Zn as θ ([a
×
([a ([a]m , [a]n ) = ([b ([b]m , [b]n ) [a]m = [b]m [a]n = [b]n m (a b) n (a b) z (a b) [a]z = [b [ b ]z
↔ ∧ ↔ | − ∧ | − ↔ | − ↔
→
assumed definition of ordered pair equivalence defin definit itio ion n of mo modu dula larr equi equiv valen alence ce z is the lcm: theorem 12.3 definition of modular equivalence
The “if” portion of this proof shows that the function is well-defined, and the “only if” portion shows that it is one-to-one one-to-one.. We showed showed that θ preserves both the additive and multiplicative operations in exercise ercise 27.17. Because Because θ is one-to-one, it is onto if and only if Zz = Zm Zn . Sinc Sincee mn = gcd(m, gcd(m, n) lcm(m, lcm(m, n) (exercise 13.22), this means that θ is onto if gcd(m, gcd(m, n) = 1. This proves that θ is an isomorphism if m and n are relatively prime.
| | | × |
×
27.20b The proof from 27.20a shows that Zm Zn Zz , where z is the least common multiple of m and n. If the gcd(m, gcd(m, n) = 1, then the lcm(m, lcm( m, n) = mn (because mn = gcd(m, gcd(m, n)lcm(m, )lcm(m, n)). )). But But if z = mn, mn, obviously Zz Zmn. And since Zm Zn Zz and isomorphism is an equivalence relation, we know that Zm
27.2 27.21 1
≈
× Z ≈ Z n
mn
lemm lemma: a: x = (-x (-x)) 2 x=x = ( x)2 = x
− −
× ≈ × ≈
definition of boolean ring theorem 24.2(c) definition of boolean ring
commutativity a+b= a+b (a + b)2 = a + b (a2 + ab + ba + b2 ) = a + b (a + ab + ba + b) = a + b (ab + ba) ba) = 0 (ab ba) ba) = 0 ab = ba
→ → → → → →
−
definition of boolean ring left and right distributive properties definition of boolean ring left and right additive cancellation lemma
50
2x = 0 2x =x+x = x + ( x) =0
−
lemma additive inverses
27.22 Let R be a finite ring of order p and let m be its characteristic. ( ( ( ( (
)( pa + mb = gcd( p,m gcd( p,m))) ∃a, b ∈ Z)( pa ∀r ∈ R)(∃a, b ∈ Z)(( pa )(( pa + mb) mb)r = (gcd( p,m)) p,m))rr) )( p((ar) ar) + m(br) br) = (gcd( p,m)) p,m))rr ) ∀r ∈ R)(∃a, b ∈ Z)( p ∀r ∈ R)(∃a ∈ Z)( p )( p((ar) ar) + 0 = (gcd( p,m (gcd( p,m)) ))rr) p,m))rr) ∀r ∈ R)(0 = (gcd( p,m))
theorem 12.2 algebra algeb lgebra ra R is characteristic m p(ar) ar) = 0 by 14.2 and Lagrange’s theorem: o(a) p p
|
So gcd( p,m gcd( p,m)) is a zero divisor for all elements of R. Beca Becaus usee it is a diviso divisorr of m, we know that 1 gcd( p,m gcd( p,m)) m. And since m, by the definition of characteristic, is the least integer that is a zero divisor for all elements of R, it must be the case that m = gcd( p,m gcd( p,m). ). This of course means that m divides p = R , which is what we wanted to prove.
≤ | |
≤
27.23
√
√
√
√
√
θ(a + b 2)θ 2)θ (c + d 2) = (a (a + b 3)(c 3)(c + d 3) = (ac (ac + 3bd 3bd + (bc (bc + ad) ad) 3)
√
√
√
√
θ((a ((a + b 2)(c 2)(c + d 2)) = θ(ac + 2bd 2bd + (bc (bc + ad) ad) 2) = (ac (ac + 2bd 2bd + (bc (bc + ad) ad) 3) x2 = eR + eR θ(x2 ) = θ(eR + eR ) θ(x)θ (x) = θ (eR ) + θ (eR ) [θ(x)]2 = θ(eR ) + θ(eR ) [θ(x)]2 = eS + eS ( s S )(s )(s2 = eS + eS )
27.24a
→ → → → →∃ ∈ √ 27.24b (0 + 1 2)2 = 2
assumed θ is well-defined additive additive and multiplica multiplicative tive homomorphi homomorphisms sms algebra theorem 18.2a or exercise 27.1 fro from the the onto onto-n -nes esss of θ of θ
27.24c Proof by contradiction: Assume that there exists an isomorphism θ : R
√ (x + y√3)) ∃x, y ∈ Q)(θ )(θ(1 + 1 2)√= (x √3)) )(θ(1) + θ (1 √2) = (x (x + y √3)) ∃x, y ∈ Q)(θ )(θ(1)√ + θ (1) 2 =√ (x ( x + y 3)) ∃x, y ∈ Q)(θ ∃x, y ∈ Q)(1 + 2 =√(x ( x + y√ 3)) ∃x, y ∈ Q)(1 − x = y 3 − 2)
( ( ( ( (
→ S :
θ is well-defined homo homomo morp rphi hism sm of θ of θ 18.2(c 18. 2(c)) for for additi additive ve groups groups 18.2(a) or exercise 27.1 algebra
The left-hand side of this last equality is rational and the right-hand side contains the irrational square root of 2, so there is no way way for this equality equality to ever be valid. By contradicti contradiction, on, our assumption assumption that an isomorphism exists must be false. 27.26 Modify the proof in exercise 27.25, but use Z p instead instead of Z.
28 Ordered Integral Domains 28.1
x
∈ D2 assumed 2 theorem 28.2(f ) → x2 ∈ D ∨ x = 20 → x2 + e ∈ D ∨ x + e = e closure of D trichotomy → x + e = 0 a >b ∧c< 0 assumed → (a − b) ∈ D ∧ (0 − c) ∈ D definition of < → (a − b)(0 − c) ∈ D multiplicative closure of D bc) ∈ D distributivity of rings → (−ac + bc) → (bc − ac) ac) ∈ D additive commutativity of rings definition of < → ac < bc p
p
28.2
p
p
p
p
p
p
p
51
28.3 28.3
lemm lemma: a: a D p ab case i) a D p ( b) D p a( b) D p (ab) ab) D p ab D p
p
p
∈ ∧ ∈D →b∈D ∈ ∧− ∈ → − ∈ →− ∈ → ∈
case ii) a D p b = 0 ab = 0 ab D p
assumed property of zero trichotomy
∈ ∧ → → ∈
case iii) a D p b ab D p
p
∈ ∧ ∈D → ∈
proof: (ac (a (a 28.4
assumed closure of D p
ac > bc c > 0 bc) bc) D p c D p b)c D p c D p b) D p
∧ → − ∈ ∧ ∈ → − ∈ ∧ ∈ → − ∈ a < 0∧b< 0 → (0 − a) ∈ D ∧ (0 − b) ∈ D → (0 − a)(0 − b) ∈ D → (−a)(−b) ∈ D → ab ∈ D → ab > 0 p
p
p
p
p
28.5
assumed closure of D p 14.1 trichotomy
a >b (a b) D p (a + ( b)) D p ( ( a) + ( b)) D p ( b ( a)) D p b> a
assumed definition of > distributivity of rings lemma assumed definition of < multiplicative closure of D p property of zero theorem 14.1 definition of >
assumed
definition definition of > of > → − ∈ → − ∈ → −− − ∈ →−−− ∈ →− − definition definition of > of > 28.6 a > b ∧ a ∈ D ∧ b ∈ D assumed definition of > of > → (a − b) ∈ D ∧ a ∈ D ∧ b ∈ D definition additive closure of D → (a − b) ∈ D ∧ a + b ∈ D )(a + b) ∈ D multiplicative closure of D → (a2 − b)(a 2 → a2 + ab − ba − b2 ∈ D left and right distributivity of rings commutativity of integral domains → a2 + ab2 − ab − b ∈ D → a2 − b 2 ∈ D definition definition of > of > →a >b 28.8 Let E = E ∩ D . p
p
p
p
p
p
p
p
p
p
p p
p
p
p
52
additive and multiplicative closure of E p a E p b E p a E b E a D p b D p ab E ab D p (a + b) E (a + b) D p ab E D p (a + b) E D p ab E p (a + b) E p
∈ → → → →
∧ ∈ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∈ ∧ ∈ ∧ ∈ ∧ ∈ ∩ ∧ ∈ ∩ ∈ ∧ ∈
trichotomy of E p a E a E a D a E (a D p a = 0 ( (a E a D p ) (a E (a E p ) (a = 0) (a E (a E p ) (a = 0) (( a) (a E p ) (a = 0) (( a)
∈ → → → → → →
∈ ∧ ∈ ∈ ∧ ∈ ∈ ∧ ∈ ∈ ∨ ∈ ∨ ∈ ∨
∨ ∨ ∨ ∨ ∨
assumed definition of E p closures of E and D p definition of intersection definition of E p
∈
p
∨ −a) ∈ D ) ∈ ∧ a = 0) ∨ (a ∈ E ∧ (−a) ∈ D ) ∈ ∧ (−a) ∈ D ) − ∈ E ∧ (−a) ∈ D ) − ∈ E )
assumed property of subrings trichotomy of D
p
p
p
p
definition of E p E existence of negatives in ring E definition of E p
28.9 The only subsets of D that are integral domains are Z p where p is prime (example 25.2, exercise 25.10) or D itself. itself. But for each nonzero nonzero a Z p , pa = 0 and p( a) = 0 (theorem 14.2 and Lagrange’s theorem). p So D is not closed under multiplication for these finite groups. Therefore the only subset of D that is an ordered ordered integral integral domain is the one with characte characteristic ristic 0: that is, the set of integers. integers. And if D = Z, then D p = N.
∈
−
28.10b R is not ordered because R is not an integral domain: it is not an integral domain because it has zero divisors. divisors. For each f M ( M (R), define a function f 0 for which f 0 (x) = 0 iff f iff f ((x) = 0. If f ( f (x) = 0 for any value of x, then f 0 is a zero divisor of f . f .
∈
∈ D , then a > −a but a2 > (−a)2. 28.13 If b = −a, then (b (b + a) = 0 and the following proof holds: p
28.12 Disproof: if a
a> b (a b) D p (a b) D p (a + b)2 D p (a b)2 D p (a b) D p (a2 + b2 + 2ab 2ab)) D p (a2 2ab + b2 ) (a b) D p (a2 + b2 ) > 2ab (a2 + b2 ) > 2ab (a b) D p (a2 + b2 ) > 2ab (a b) D p (a2 + b2 ) > 2ab > ab (a b) D p (a2 + b2 ) > ab (a b) D p (a2 + b2 ) > ab (a b) D p (a2 + b2 + ab) ab) D p (a b)(a )(a2 + b2 + ab) ab) D p (a3 b3 ) D p a3 > b 3
→ → → → → → → → → → → →
− − − − − − − − − − −
∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
∈
∈
∈
∧ − ∈ ∈ ∧ − − ∧ | | | | | | | | − ∈
assumed definition of > 28.2(f) algebra definition of > definition of absolute value
p
∈D
transitivity of > partial definition of absolute value definition of > closure of D p algebra, commutativity of integral domains definition of >
If b = a, then a > b implies a > a, which means that a D p . It must must then then be the case case that that a3 > ( a)a2 , from which algebraic replacement shows that a3 > b3 .
− −
−
∈
29 The Integers 29.1 Assume that there is some least positive element q Q. Then Then q D p and 12 D p since they are both positive elements of Q. But then multipl multiplica icativ tivee closur closuree of D p requires 2q D p , which contradicts our p assumption that q is the least element of D .
∈
∈
∈
∈
29.2 Theorem 29.1 tells us that if D were an ordered integral domain, then it would be isomorphic to Z. Proof by contradiction that there can exist no isomorphism θ : Z[ 2] Z:
√ →
53
√ → √
Z There exists an isomorphism θ : Z[ 2] ( a, b Z)( c Z)(θ )(θ (a + b 2) = c) Choose a, b such that a = 0 and b = 1. ( c Z)(θ )(θ (1 2) = c) ( c Z)(θ )(θ (1) 2 = c) Z ( c )(1 2 = c)
→ → → →
assumed onto-ness of θ θ
∀ ∈ ∃ ∈ √ onto-ness of θ ∃ ∈ √ 18.2(d) for additive groups ∃ ∈ √ ∃ ∈ 18.2(a) √ But 2 is irrational and c is rational, rational, so this last statement statement must be false. By contradicti contradiction, on, then, there can exist no isomorphism between the two groups. 29.3
a a
∈D ∈D∧e∈D → a ∈ D∧0+e ∈ D → a + (−a) + e ∈ D → a + (−a) + (−(−e)) ∈ D → a − (a + (−e)) ∈ D →a > a−e p
p
p
p
p
assumed lemma 28.1 existence of additive identity in rings existence of additive inverses in rings 14.1 14.1 definition definition of D of D p
29.4 Proof by contradiction: a > a2 a a2 D p a(e a) D p (a D p e a D p ) ( a (a D p e a D p ) ( a (a D p e a D p ) ( a (a D p a < e) e ) ( a D p
→ → → → → →
− − ∈ ∈ ∈ ∈
∈
∈ ∧ − ∈ ∨− ∧ − ∈ ∨− ∧ − ∈ ∨− ∧ ∨− ∈
p
p
∈ D ∧ −(e − a) ∈ D ) ∈ D ∧ −(e − a) + e + e ∈ D ) ∈ D ∧ e − (−a) ∈ D ) e) ∧ (−a) < e) p
p
p
p
assumed definition of > left distributivity distributivity of rings lemma from exercise 28.3 closu losurre of D of D p algebra definition of >
This last statement is false: neither of the conditions can be true, since they both imply the existence of a member of D p smaller than e, which is the smallest member of D p by definition. definition. So, by contradic contradiction, tion, 2 2 2 it cannot be the case that a > a . Either a = a (when a = 0 or a = 1) or a < a (all other cases). This proof is only valid valid for well-orde well-ordered red integral integral domains. domains. For example, Q is an ordered integral domain and it is not the case that ( 12 )2 > 12 . 29.5 For every nonempty subset S of Dn , define a complementary set T = s : s S . Beca Becaus usee T is a nonempty nonempty subset of D of D p , it must have a least element. Let a represent this least element of T . T . Proof by contradiction that a is the greatest element of S :
−
( a, b S )(b )(b > a) a) ( a, b S )( )( b < a) ( a, b T )( T )( b <
∃ ∈ →∃ ∈ − − → ∃ − − ∈ − −a)
{−
∈ }
hypothesis of contradiction exercise 28.5 defin definit itio ion n of T of T
But this last statement cannot be true, since we defined a as the least element of T . T . So our our initi initial al n n assumption that there was some element in D greater than a is false: a is the greatest element of D .
−
29.6 Proof by contradiction that there is no integer between n and n + 1: ( m Z)(n )(n + 1 > m > n) n) ( m Z)(n )(n + 1 m D p m n D p ( m Z)(1 (m n) D p m n D p ( m Z)(1 > m n m n D p
∃ → → →
∈ ∃ ∈ ∃ ∈ ∃ ∈
− ∈ ∧ − ∈ − − ∈ ∧ − ∈ − ∧ − ∈
hypothesis of contradiction definition of > commutativity of addition definition of >
This last statement cannot be true, or 1 would not be the smallest element of D p . So by contradic contradiction tion there cannot be any integer between n and n + 1.
54
29.8
θ must be the identity mapping θ(x) = θ(xe) xe) = xθ( xθ(e) = xe =x
property of the identity element 18.2(d) for additive groups 18.2(a)
θ(x) = 2x is an additive isomorphism θ(x + y ) = 2(x 2(x + y ) = 2x + 2y 2y = θ (x) + θ(y )
30 Field of Rational Numbers 30.1 ([2], ([2], [1]) ([0], ([0], [2]) and ([0], ([0], [2]) ([0], ([0], [1]). If were an equivalence relation on Z4 should guarantee guarantee that ([2], ([2], [1]) ([0], ([0], [1]) but this is not the case.
∼
30.2 Let F = R 30.5 30.5
∼ ∼
refle reflexi xive ve (a, b) D D a D b D ab = ba (a, b) (a, b)
assumed definition of cartesian product and D commutativity of integral domain D definition of
symmetric (a, b) (c, d) ad = bc cb = da (c, d) (a, b)
assumed definition of commutativity of integral domain D definition of
→ → →
∼ ∼
∼
transitive (a, b) (m, n) (m, n) an = bm my = nx an( an(my) my) = bm( bm(nx) nx) ay( ay (nm) nm) = bx( bx(nm) nm) ay = bx
→ → → →
∼
∼
∼
∧
∧
∼ (x, y)
→ (a, b) ∼ (x, y)
assu assume med d definition of algebra commutativity of integral domain D theorem 25.1 note: this is not right-can right-cancella cellation, tion, since right-cancell right-cancellation ation requires the existence of inverse elements and multiplication is not necessarily a group. This step would not be legal for rings that were not integral domains.) definition of
∼
∼
∼
[a1 , b1 ] = [a2 , b2 ] [c1 , d1 ] = [c [ c2 , d 2 ] (a1 , b1 ) (a2 , b2 ) (c1 , d1 ) (c2 , d2 ) (a1 b2 = b1 a2 ) (c1 d2 = d1 c2 ) (a1 b2 )(c )(c1 d2 ) = (b1 a2 )(d )(d1 c2 ) (a1 c1 )(b )(b2 d2 ) = (b1 d1 )(a )(a2 c2 ) (a1 c1 , b1 d1 ) (a2 c2 , b2 d2 ) [a1 c1 , b1 d1 ] [a2 c2 , b2 d2 ]
→ → → → → →
× Z#4 then transitivity
(5, 1) ∼ (15, (15, 3), but (5, (5, 1) = (15, 3). × R. (5, (15,
∈ × → ∈ ∧ ∈ → → ∼
30.6
∼
∧ ∧ ∧
∼ ∼
∼
assumed definit definition ion of = definition of algebra associativity definition of definition of =
∼
equi equiv valence alence equivalence
∼ equivalence
55
equivalence
30.7 30.7
zero ero of F of F D is [0, [0, eD ] [0, [0, e] + [a, [ a, b] = [0b [0b + ea, ea, eb] eb] = [ea,eb] ea,eb] = [a, b] = [ae,be] ae,be] = [ae + b0, be] be] = [a, b] + [0, [0, e]
definition of addition in F D proper perty of zero in multiplicative groups property of the unity element of D D property of the unity element of D D prop propeerty rty of zero zero in multi ultipl pliicati cativ ve group roupss definition of addition in F D
negative of [a, b] is [ a, b] [a, b] + [ a, b] = [ab [ ab + ( a)b,bb] b,bb] = [ab [ ab ab, ab, bb] bb] = [0, [0 , bb] bb] = [0, [0 , e]
−
− −
−
addition is associative [a, b] + ([m, ([m, n] + [[x, [[x, y]) = [a, [ a, b] + [my [ my + nx,ny] nx,ny] = [any [ any + b(my + nx) nx),bny] ,bny ] = [any [ any + bmy + bnx, bnx, bny] bny] = [(an [( an + bm) bm)y + bnx, bnx, bny] bny] = [an [ an + bm, bm, bn] bn] + [x, [ x, y] = ([ ( [a, b] + [m, [ m, n]) + [x, [x, y] addition is abelian [a, b] + [x, [ x, y] = [ax [ ax + by,by] by,by] = [xa [ xa + yb,yb] yb,yb] = [x, [ x, y] + [a, [ a, b]
definition of addition in F D 14.1 from (0, bb) bb)
∼ (0, (0, e)
defin definit itio ion n of addi additi tion on in F D definit definition ion of additi addition on in F D left left distri distributi butivit vity y of the ring ring D right right distri distribut butivi ivity ty of the ring ring D defin definit itio ion n of addi additi tion on in F D defin definit itio ion n of addi additi tion on in F D
definition of addition in F D com omm mutat utativ ivit ity y of inte integr gral al doma domain in D definition of addition in F D
The previous four properties show that the additive operation on F D is an abelian group. multiplication is associative [a, b]([m, ]([m, n][x, ][x, y]) = [a, [ a, b]([mx,ny ]([mx,ny]]) = [a [ a(mx) mx), b(ny) ny)] = [(am [( am))x, (bn) bn)y ] = [am [ am,, bn][ bn][x, x, y] = ([ ( [a, b][m, ][m, n])[x, ])[x, y] multiplication is commutative [a, b][x, ][x, y] = [ax [ ax,, by] by] = [xa,yb [ xa,yb]] = [x, [ x, y][a, ][a, b] left distributive property [a, b]([m, ]([m, n] + [x, [ x, y]) = [a, [ a, b][my ][my + nx,ny] nx,ny] = [a [ a(my + nx) nx),bny] ,bny ] = [amy [ amy + anx, anx, bny] bny] = [b [ b(amy + anx) anx), b(bny) bny)] = [bamy [ bamy + banx, banx, bbny] bbny] = [am [ am((by) by) + bn( bn(ax) ax), bn( bn(by)] by )] = [am [ am,, bn] bn] + [ax,by [ ax,by]] = [a, [ a, b][m, ][m, n] + [a, [ a, b][x, ][x, y]
definition of multiplication in F D definition of multiplication in F D associativity of multiplication in ring D definition of multiplication in F D definition of multiplication in F D
definition of multiplication in F D commutativity of multiplication in integral domain D definition of multiplication in F D
definition of addition in F D definition of multiplication in F D distributive property of ring D from (x, y) (nx,ny) nx,ny) distri tributiv tive proper perty of ring D commu com mutat tativ ivee propert property y of integ integral ral domain domain D definition of addition in F D defin definit itio ion n of multi ultipl plic icat atio ion n in F D
∼
56
right distributive property ([m, ([m, n] + [x, [ x, y])[a, ])[a, b] = [my [ my + nx,ny][ nx,ny][a, a, b] = [(my [( my + nx) nx)a,nyb] a,nyb] = [mya [ mya + nxa, nxa, nyb] nyb] = [b [ b(mya + nxa) nxa), b(nyb) nyb)] = [bmya [ bmya + bnxa, bnxa, bnyb] bnyb] = [ma [ ma((yb) yb) + nb( nb(xa) xa), nb( nb(yb)] yb )] = [ma [ ma,, nb] nb] + [xa,yb [ xa,yb]] = [m, [ m, n][a, ][a, b] + [x, [ x, y][a, ][a, b]
definition of addition in F D definition of multiplication in F D distributive property of ring D from (x, y) (nx, nx, ny) ny) distributive proper perty of ring D comm commuta utativ tivee prope propert rty y of inte integr gral al doma domain in D definition of addition in F D definition of multiplication in F D
∼
The previous four properties show that the F D is a commutative ring unity of [a, b] is [eD , eD ] [a, b][e, ][e, e] = [ae [ ae,, be] be] = [a, [ a, b] = [ea [ ea,, eb] eb] = [e, [ e, e][a, ][a, b]
definition of multiplication in F D property of the unity element of D D property of the unity element of D D definition of multiplication in F D
F D contains no zero divisors [a, b][c, ][c, d] = [0, [0, e] [ac,bd] ac,bd] = [0, [0, e] (ac,bd) ac,bd) (0, (0, e) ace = bd0 bd0 ac = 0 a=0 c=0 a = b0 c = d0 ae = b0 ce = d0 (a, b) (0, (0, e) (c, d) (0, (0, e) [a, b] = [0, [0, e] [c, d] = [0, [0, e]
→ → → → → → → → →
∼
∨ ∨ ∨ ∼
∨ ∨
∼
assumed definition of multiplication in F D definition of = equivalence in F D definition of equivalence in F D properties of unity and zero integral domain D has no zero divisors property of zero property of unity defini efiniti tion on of equivalence in F D defin definit itio ion n of = equi equiv valen alence ce in F D
∼ ∼
The previous two properties show that F D is an integral domain inverse of [a, b] is [b, a] [a, b][b, ][b, a] = [ab [ ab,, ba] ba] = [ab [ ab,, ab] ab] = [e, [ e, e] = [ba [ ba,, ba] ba] = [ba [ ba,, ab] ab] = [b, [ b, a][a, ][a, b]
definition of multiplication in F D commutativity of integral domain D from (ab,ab) ab,ab) (e, e) from (e, e) (ba,ba) ba,ba) commutativity of integral domain D definition of multiplication in F D
∼
∼
This shows that F D is a field. 30.8 A subring of a field cannot cannot have any zero divisors: divisors: if it did, they would also be zero divisors of the field. Z6 has zero divisors, and the existence of zero divisors is something that is preserved by isomorphism (section 27). 30.9 Theorem 30.1 tell us that the field of quotients F D is the smallest field into which the integral domain D can be embedded. If D is already a field, then the smallest field into which it can be embedded is D itself, in which case F D D.
≈
)([n, e][m, ][m, n] = [m, e] ∀ ∈ D, n ∈ D)([n,
30.10 ( m
30.11 If K is any field of prime characteristic p, then K contains a subfield isomorphic to Q p (the quotient field of Z p ). Proof: Theorem 27.3 tells us that if K is a field of prime characteristic p, then K contains a ring isomorphic to Z p . So by theorem 30.1, K must also contain a subfield isomorphic to the equivalence classes of Z p Z p . And by definition, this set of equivalence classes is Q p .
×
57
30.12 [1, [1, 2] = [2, [2, 4] and [0, [0, 1] = [0, [0, 1], but ([1, ([1, 2] + [0, [0, 1]) = [1, [1, 3] = [2, [2, 5] = ([2, ([2, 4] + [0, [0, 1]).
34 Polynomia Polynomials: ls: Definition Definition and Element Elementary ary Properties 34.8 In order for R[x] to be a field, it would have to have multiplicative inverses for each of its nonzero elements. Let f ( f (x) = 0 + r1 x. In order for some other polynomial to be its multiplicative inverse, there would have to be some (a (a0 + a1 x + . . . an xn ) such that ((a ((a0 0)+(a 0)+(a1 r1 )x + . . . (an 0)x 0)xn ) = e. But there can be no a0 R such that a0 0 = e, so there can be no multiplicative inverse for f ( f (x).
∈
34.9a False. Consider the sum of two nth degree polynomials an xn + an xn .
−
34.9b False. Consider Consider two polynomials polynomials in Z4 [x]. ([2] ([2]x5 )([2]x )([2]x5 ) is equal to [0], which certainly does not have degree 10. 34.10 Let R be a commutative ring and let R[x] be an integr integral al domain. domain. If R is not an integral domain, then there is some nonzero a, b R such that ab = 0. 0. But But ab is also represents the product of two nonzero polynomials of degree zero in R[x]: any zero divisor of R is also a zero divisor of R[x].
∈
This means that (R (R is not an integral domain R[x] is not an integral integral domain). By contrapositiv contrapositive, e, If R[x] is an integral domain, then R is an integral domain.
→
34.11 If p is a zero divisor for every element of R, then it is a zero divisor for every element of R[x]. The The converse is also true. Proof: p is a zero divisor of R ( r R)( pr = 0) ( f R[x])( ri f )( f )( pr0 + pr1 x1 + . . . + prn xn = 0) ( f R[x])( ri f )( f )( p( p(r0 + r1 x1 + . . . + rn xn ) = 0) ( f R[x])( pf = 0) p is a zero divisor of R[x]
↔ ↔ ↔ ↔ ↔
∀ ∀ ∀ ∀
∈ ∈ ∈ ∈
∀ ∈ ∀ ∈
assumed definition of zero divisor definit definition ion of a polynom polynomial ial in R[x] distri distribut butiv ivee proper property ty of rings rings definition of a polynomial in R[x]
Since R and R[x] share all zero divisors, divisors, they also share the least such divisor. divisor. This least zero divisor divisor is the characteristic by definition. 34.12 Let θ : R S represent the isomorphism between R and S . Defined Defined a second function function φ : R[x] S [x] n n 1 1 as φ(a0 + a1 x + . . . + an x ) = θ (a0 ) + θ (a1 )x + . . . + θ (an )x . The The fact fact that that φ is a bijection follows directly from the bijection of θ. Proof of homomorphism:
→
→
additive homomorphism φ((a ((a0 + a1 x1 + . . . + an xn ) + (b ( b0 + b1 x1 + . . . + bn xn )) = φ((a ((a0 + b0 ) + (a ( a1 + b1 )x1 + . . . + (a (an + bn )xn )) The previous step required the associativity and commutativity of addition on R and the distributive property of R. = θ (a0 + b0 ) + θ (a1 + b1 )x1 + . . . + θ (an + bn )xn ) = θ (a0 ) + θ (b0 ) + θ (a1 )x1 + θ (b1 )x1 + . . . + θ(an )xn + θ (bn )xn = θ (a0 ) + θ (a1 )x1 + . . . + θ (an )xn + θ(b0 ) + θ (b1 )x1 + . . . + θ(bn )xn φ(a0 + a1 x1 + . . . + an xn ) + φ(b0 + b1 x1 + . . . + bn xn ))
→
definition of φ additive homomorphism of θ additive associativity of S definition of φ
multiplicative homomorphism φ((a ((a0 + . . . am xm )(b )(b0 + . . . bn xn )) = φ(a0 b0 + . . . (horrifying binomial)x binomial)x? . . . + am bn xm+n ) = θ (a0 b0 ) + . . . θ(horrifying θ (horrifying binomial) . . . + θ (am bn )xm+n = φ(a0 + . . . am xm )φ(b0 + . . . bn xn ) 34.14a D[x] p is closed under the operations operations of addition addition and multiplica multiplication: tion: If p, q are two elements of D[x] p p then they are of finite order with highest terms pm , qn D (true by definition). So their sum is of degree max(m, max(m, n) with a highest term of either pm , qn , or pm + qn : all of which which are are D p (true by the additive closure property of integral domain D). Their product is of degree m + n with highest term pm qn which is
∈
58
∈
in D p (true by the multiplicative closure property of integral domain D). Because a polynomial is member of D[x] p iff its highest coefficient is a member of D p , the trichotomy of D[x] p follows directly from the trichotomy of D. 34.14b Proof by contradiction: there exists a pos positive pol polynomial less than 1 p ( f Z[x] )(f )(f < 1) p ( f Z[x] )(1 f Z[x] p ) ( a0 , . . . , an−1 Z, an Z p )(1 (a0 + a1 x1 . . . + an xn ) Z[x] p ) ( a0 , . . . , an−1 Z, an Z p )(1 a0 a1 x1 . . . an xn ) Z[x] p )
→ → → →
∃ ∈ ∃ ∈ ∃ ∃
− ∈ ∈ ∈ ∈ ∈
− − −
−
∈ ∈
hypot pothesis of contradiction formalization of hypothesis definition of > defin definit itio ion n of f of f Z[x] p dist distri ribu bute te prope propert rty y
∈
If the degree n of f is greater than 0, then the highest term of (1 f ) f ) is an for some an Z p : this means p that (1 f ) f ) Z[x] . If the degree n of f is 0, then the highest term is (1 a0 ). This can only be positive if a0 < 1, which is not possible since a0 Z p (it has to be to make f a member of Z[x] p ) and 1 is the least member of Z p by definition. definition. So no matter what polynomial we choose choose for f , f , (1 f ) f ) Z[x] p , which contradicts our initial assumption that 1 > f . f .
−
− ∈
∈
− −
∈
− ∈
34.14c If Z[x] were a well-ordered domain, then every nonempty subset of Z[x] p would have a least element. Let S be the set of all polynomials of degree 2 in Z[x]. There There can be no least least elemen elementt of S : If a0 + a1 x were the least positive element, then there is an immediate contradiction since a0 + a1 x > (a0 1) + a1 x and (a (a0 1) + a1 x Z[x] p . In other other words, words, S has no least element because a0 Z and Z has no least element.
−
∈
∈
−
34.16 Define a function θ : R[x] R[X ] as θ (a0 + a1 x1 + . . . + an xn ) = (a0 , a1 , . . . , an ). The The proof proof that that θ is a bijection bijection follows follows directly directly from the definitions definitions of equality equality for polynomials and sequences. sequences. Proof of homomorphism:
→
additive homomorphism θ[(a [(a0 + a1 x1 + . . . + an xn ) + (b ( b0 + b1 x1 + . . . + bn xn )] = θ ((a ((a0 + b0 ) + (a ( a1 + b1 )x1 + . . . + (a (an + bn )xn ) The previous step required the associativity and commutativity of addition on R and the distributive property of R. = (a ( a0 + b0 , a1 + b1 , . . . , an + bn ) = (a ( a0 , a1 , . . . , an ) + (b ( b0 , b1 , . . . , bn ) 1 θ(a0 + a1 x + . . . + an xn ) + θ(b0 + b1 x1 + . . . + bn xn ) multiplicative homomorphism θ((a ((a0 + . . . am xm )(b )(b0 + . . . bn xn )) = θ (a0 b0 + . . . (horrifying binomial)x binomial)x? . . . + am bn xm+n ) = (a ( a0 b0 , . . . , horrifying binomial coefficients, coefficients, . . . , am bn ) = (a ( a0 , . . . , am )(b )(b0 , . . . , bn ) = θ (a0 + . . . am xm )θ(b0 + . . . bn xn )
definition of θ definition of sequence addition defin definit itio ion n of θ of θ
dist distri ribu buti tivi vity ty of R of R defin definit itio ion n of θ of θ definition of sequence multiplication definition of θ
35 Division Algorithm 1 m−n 35.11 Because the highest term of g (x) is bn xn , the highest term of the product g (x)am b− is am xm . n x 1 m−n Since this is also the highest term of f ( f (x), we know that the degree of f ( f (x) g (x)am b− is less than n x or equal to the degree of f ( f (x).
−
35.13 f ([2]) f ([2]) = [30], which is equal to [0] for p
∈ {3, 5}. 59
35.14 Let F D represent the set of equivalence classes of Z p [x] that F D and Z p share all zero divisors:
× Z [x]. p
By lemma lemma 30.3 30.3,, F D is a field. field. Proof Proof
m is a zero divisor of every element in F D ( (f ( )(m(f ( (0, e)) f (x), g (x)) F D )(m f (x), g (x)) (0, ( f ( f (x) F D )(m )(m(f ( f (x)) = 0)) ( f ( f (x) F D )( [ai ] f ( f (x))(m ))(m[a0 ] + m[a1 ]x + . . . + m[an ]xn = 0)
↔∀ ↔∀ ↔∀
∈ ∀ ∈
∈ ∈
≈
assumed lemma 30.3: (0,e) is the zero of F F D lemma 30.1: definition of definiti definition on of zero zero polynom polynomial ial
∼
This last statement tells us that ma = 0 every coefficient a for every polynomial f ( f (x) in F D . But the set of every possible coefficient is just Z p , so we know that: )(ma = 0) ↔ (∀a ∈ Z )(ma ↔ m is a zero divisor of every element in Z p
definition of zero divisor
p
Because F D and Z p share all elements with the “characteristic property”, they also share a least such element: element: by definition, definition, this least zero divisor is their commo common n characte characteristi risticc which which we know to be p (theorem 27.3). 35.15 Let f ( f (x) = 3, g (x) = 2. In order for the division algorithm to be true, there would have to be some a Z such that 3 = 2a 2a + r, where the deg(r deg(r ) < deg(2). The only integer with degree less than deg(2) is 0, and there is no a Z such that 3 = 2a 2 a + 0.
∈
∈
35.16 Let f = a0 and g = b0 be two arbitrary nonzero polynomials of degree 0 in D[x]. In order for the division algorithm to be true, it must be the case that: ( a0 , b0
∀
)(a0 = b0 r + s), deg(s deg(s) < deg(a deg(a0 ) ∈ D)(∃r, s ∈ D)(a
But since the degree of a0 is zero, this means that s = 0. If we then choose a 1 as a specific value for a0 , the previous statement reduces to: ( b0
∀ ∈ D)(∃r ∈ D)(1 = b0r)
This tells us that every nonzero element b0 in the integral domain D has an inverse, which makes D a field by definition. 35.17 From Lagrange’s theorem, we know that the multiplicative group of Z5 is cyclic with order 5 and that, for any [a [a] Z5 , [a]5 = [a]. So the equation f ([ f ([a a]) = [a [a]6 [a] will be zero for any [a [ a], which means that 6 every [a [a] is a root of f ([ f ([a a]) = [a [a] [a].
∈
−
−
35.18 See exercise 35.17.
37 Unique Factorization Domains 37.7a N ( N (a + bi) bi) = a + bi 2 = a2 + b2 , and the last term must be
|
37.7b
z = 0 + 0i 0i N (z ) = 0 + 0i 0i 2 N (z ) = 0 2 + 02 N (z ) = 0
↔ ↔ ↔
37.7c
|
|
|
assumed definition definition of N of N definition of complex norm
N (( N ((a a + bi)( bi)(cc + di)) di)) = N (( N ((ac ac bd) bd) + (ad ( ad + bc) bc)i) 2 = (ac bd) bd) + (ad (ad + bc) bc)2 = (ac) ac)2 + (ad (ad))2 + (bc (bc))2 + (bd (bd))2 = (a2 + b2 )(c )(c2 + d2 ) = N ( N (a + bi) bi)N ( N (c + di) di)
−
−
≥ 0 from the lemma of theorem 28.1.
dist distri ribu buti tivi vity ty and and comm commut utat ativ ivit ity y of C 2 definition of N = (ac ( ac)) 2(abcd 2(abcd)) + (bd ( bd))2 + (ad (ad))2 + 2(abcd 2(abcd)) + (bc ( bc))2 algebra algebra algebra definition of N N
−
37.8 Proof by contradiction:
60
√−5] ± √−5 is reducible Z in [ √ 5])(zw = 2 ± −5) √ → (∃z, w ∈ Z[√√−5])(zw → (∃z, w ∈ Z[√−5])(N 5])(N ((zw) zw ) = N (2 (2 ± −5)) 5])(N ((zw) zw ) = 22 + 5 = 9) → (∃z, w ∈ Z[√−5])(N → (∃z, w ∈ Z[ −5])(N 5])(N ((z )N ( N (w) = 9) N (z ), N ( N (w) ∈ Z)(N )(N ((z )N ( N (w) = 9) → (∃N ( 2
hypothesis of contradiction definition of reducibility N is well-defined definition of N f N N is a multiplicative function The range of N N is Z The only factors of 9 in Z are 1,3,and 9. In order for N ( N (zw) zw ) = 9 to be reducible, it must have a divisor that is neither a unit of Z nor an associate of N ( unit and and 9 is N (zw) zw ) = 9. 1 is a un an associate, so in order to be reducible, N ( N (zw) zw ) must have a factor of 3.
→ (∃N ( N (z ), N ( N (w) ∈ Z)(N )(N ((z ) = 3) 2 2 )(a + b = 3) → (∃a, b ∈ Z)(a
definition of N N
And this last statement is false: there are no such integers. So, by contradiction, 2 37.10
± √−5 is irreducible.
No matter how the distance function d is defined, d(a) = d(b) for all a, b in the field F . F . Proof: d(a) d(b)
≤ d(aa−−11b) → d(a) ≤ d(b) ≤ d(bb a) → d(b) ≤ d(a)
This is odd, but it doesn’t prove that F isn’t isn’t a Euclid Euclidean ean domain: domain: it just just shows shows that d(x) is a constant function. function. Howeve However, r, the second second property property of Euclidean Euclidean domains is that all nonzero nonzero elements elements a, b F can be expressed as a = bq + r where d(r) < d(b). But we know know that that d(r) = d(b) for any choice of b, r. This This still doesn’t prove that F isn’t a Euclidean Euclidean domain: it might be the case that we can always always find a = bq with r = 0. And in a field, we can always do this: just choose q = b−1 a.
∈
37.11a The unity of Z[x] is 1, which is divisible by
±1.
37.11b Requirement (b) is not met because of the lack of multiplicative inverses in Z. Consider Consider the polynomials polynomials a(x) = 3x and b(x) = 2x 2 x. There are no polynomials in Z[x] such that 2x 2x = 3xg( xg (x) + h(x), deg(h deg(h) < 1. 37.12
a is a unit ae ( b D)(ab )(ab = e) ( b D)(ab )(ab = e d(a) d(ab) ab) d(ab) ab) ( b D)(d )(d(a) d(e) d(e) d(ae) ae)) ( b D)(d )(d(a) d(e) d(e) d(a)) d(a) = d(e)
→ → → → → →
| ∃ ∃ ∃ ∃
∈ ∈ ∈ ∈
≤ ≤
∧
∧ ∧
≤
≤ ≤
∧
aab)) ≤ d(aab))
37.13 ae = a, so a a. a0 = 0, so a 0. And And for for any any x conditions are met for a to be the gcd(a, gcd(a, 0).
|
|
61
assumed definition of a unit definition of divisibility propert property y of of Eucl Euclide idean an domain domainss algebraic replacement property of unity
∈ D, if x|a ∧ x|0, then obviously x|a.
And And so bot both h
37.14
a and b are associates ab ba ( r, s D)(ar )(ar = b bs = a) ( r, s D)(bsr )(bsr = b ars = a) ( r, s D)(b )(b(sr e) = 0 a(sr
→ → → →
|∧| ∃ ∈ ∃ ∈ ∃ ∈
−
∧ ∧
∧
assumed definition definition of associate associate definition of divisiblity algebraic replacement from last step distributiv distributivity ity,, negativ negatives. es. commutativit commutativity y
− e) = 0)
We can’t do any type of cancellation to simplify this equation because we can’t assume the existence of multiplicative inverses. Instead, note that D is an integral domain: by definition, there are no zero divisors. In order for b(sr e) = a(sr e) = 0, it must be the case that:
−
−
)([a = b = 0] ∨ [sr − e = 0]) → (∃r, s ∈ D)([a )([a = be] be] ∨ [sr = e]) → (∃r, s ∈ D)([a )([a = be] are unit units) s))) → (∃r, s ∈ D)([a be]∨(s and r are
algebra defin definit itio ion n of unit unit
So either a = be or a = rs. rs. But both e and r are units of D. So in either case, the lemma is true. 37.15 The GCD exists in Euclidean domains by virtue of the fact that the division algorithm applies to Euclidean domains. See page 66 for the proof. 37.16
d1 = gcd(a, gcd(a, b) d1 a d1 b d1 gcd(a, gcd(a, b) d1 d2
→ | ∧ | → | → |
definition of gcd second second part part of the definit definition ion of gcd gcd d2 = gcd(a, gcd(a, b)
The same proof with the variables exchanged shows d2 d1 . Beca Becaus usee d1 and d2 divide each other, they are associates by definition. 37.17
|
b is a unit ( c D)(bc )(bc = e) ( c D)(bc )(bc = e d(ab) ab)
↔∃ ∈ ↔∃ ∈
∧
abc) ∧ d(a) ≤ d(ab)) ab)) ≤ d(abc)
definition of unit propert property y of of Eucl Euclide idean an domain domainss
The “only if” ( note that d(ab) ab)
→) direction of this step is justified by algebraic algebraic replacement replacement.. To justify the “if” (←) step, abc) for some c. ≤ d(a) is only true if d(a) = d(abc) ab) ≤ d(a) ∧ d(a) ≤ d(ab) ab) algebraic replacement ↔ d(ab) ↔ d(ab) ab) = d(a)
37.18
gcd(a, gcd(a, b) = e a bc ( r, s D)(e )(e = ar + bs a bc) bc) ( r,s,t D)(e )(e = ar + bs at = cb) cb) ( r,s,t D)(ce )(ce = car + cbs) cbs) at = cb) cb) ( r,s,t D)(c )(c = car + ats) ats) ( r,s,t D)(c )(c = a(cr + ts) ts)) ac
→ → → → → →
∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ |
∧ |
∧ | ∧
∧
assumed lemma 37.2 defin definiti ition on of divi divisi sibi bili lity ty,c ,com ommu muta tati tivi vity ty algebraic replacement commutativity, distributivity definition of divisibility
37.19 Because p is irreducible, all of its divisors are either units or associates of p. x p p
)(xr = e) ∨ p|x] | → [(∃r ∈ D)(xr
Let x = gcd( p,a gcd( p,a). ). Because the gcd is a divisor of p, it also must either be a unit or an associate of p. case i) If x is an associate of p, then p x and x a (the former from x being an associate of p, the latter from x being the gcd) which means p a. case ii) If x is a unit:
|
|
|
62
x is a unit ( r D)(rx )(rx = e) ( r D)(rx )(rx = e) ( m, n D)(x )(x = pm + an) an) ( r D)(rx )(rx = e) ( m, n D)(rx )(rx = rpm + ran) ran) ( r,m,n D)(e )(e = rpm + ran) ran) ( r,m,n D)(be )(be = brpm + bran) bran) ( r,m,n D)(b )(b = p(brm) brm) + ab( ab(rn) rn))
→ → → → → →
∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈
∧∃ ∧∃
assumed definition of unit property of gcd dist distri ribu buti tivi vity ty algebraic replacement distributivity commutativity,property of unity
∈ ∈
We see that p divides both terms on the right-hand side of this last equation: p p( p(brm), brm), for obvious reasons, and we are told that p ab in the description of the problem. This means that p b. We’ve shown that either p a or p b, which is what we set out to prove.
|
37.21
|
|
|
u is a unit ( r D)(ur )(ur = e) ( r D)(ura )(ura = a) ua
|
assumed defini efiniti tion on of unit unit alge algebr braa definition definition of divides divides
→∃ ∈ →∃ ∈ → |
37.22 Let U represent the set of all units of D. Each Each element element of U has an inverse inverse:: ( r D)(ru )(ru = e) is both the necessary necessary condition for being a unit and the necessary necessary condition condition for having having an inverse inverse.. The unity of D is the unity of U , U , since ee = e. And U is closed under multiplication:
∃ ∈
closure a U b U a and b are units ( r, s D)(ar )(ar = e bs = e) ( r, s D)(ar )(ar((e) = e bs = e) ( r, s D)(ar )(ar((bs) bs) = e) ( r, s D)(ab )(ab((rs) rs) = e) ab is a unit ab U
∈ ∧ ∈ → →∃ ∈ →∃ ∈ →∃ ∈ →∃ ∈ → → ∈
∧
∧
assumed definition of membership in U definition of unit prope propert rty y of unit unity y algebraic replacement commutativity of D D definition of unit definition definition of mem membership bership in U
Therefore U meets all the requirements of being a multiplicative group.
38 Homomorphisms of Rings 38.8b π1 (a, b) = 0 iff a = 0, b S . So the kern kernel el of π1 is 0 θ (s) = (0, (0, s). Proof that this function is isomorphic:
{ } × S .
∈
θ is well-defined and one-to-one s1 = s2 s1 = s2 0 = 0 (0, (0, s1 ) = (0, (0, s2 ) (0, s1 ) = θ (0, (0, s2 ) θ (0,
↔ ↔ ↔
∧
Define Define a functio function n θ : S
definition of ordered pair equality definition of θ
θ is onto (0, (0, s1 ) 0 S s1 S θ (s1 ) = (0, (0, s1 ) ( s S )(θ )(θ (s) = (0, (0, s1 )
assumed definition definition of cartesian cartesian product definition of θ
∈{ }× → ∈ → →∃ ∈
θ is homomorphic θ(s1 + s2 ) = (0, (0, s1 + s2 ) = (0, (0, s1 ) + (0, (0, s2 ) = θ(s1 ) + θ(s2 ) θ(s1 s2 ) = (0, (0, s1 s2 ) = (0, (0, s1 )(0, )(0, s2 ) = θ (s1 )θ(s2 ) 38.9 Let R be a commutative ring and let θ : R
→ {0} × S to be
→ S be a homomorphic function. 63
a
∈ θ(R) ∧ b ∈ θ(R) assumed )(θ (r) = a ∧ θ(s) = b) defin definit itio ion n of imag imagee → (∃r, s ∈ R)(θ → (∃r, s ∈ R)(θ )(θ (r)θ (s) = ab) ab) )(θ (rs) rs) = ab) ab) homomorphism of θ θ → (∃r, s ∈ R)(θ → (∃r, s ∈ R)(θ )(θ (sr) sr) = ab) ab) commutativity of R R )(θ (s)θ(r) = ab) ab) homomorphism of θθ → (∃r, s ∈ R)(θ ab) algebraic replacement from θ (s) = b, θ(r ) = a → (ba = ab) 38.10 Let θ : R → S be a homomorphic function. (∀s ∈ θ (R))(∃r ∈ R)(θ )(θ (r) = s) definition of image )(θ(r) = s ∧ re = r = er) er) unity of R → (∀s ∈ θ(R))(∃r ∈ R)(θ )(θ(r) = s ∧ θ (re) re) = θ (r) = θ (er) er) θ is well-defined → (∀s ∈ θ(R))(∃r ∈ R)(θ → (∀s ∈ θ(R))(∃r ∈ R)(θ )(θ(r) = s ∧ θ (r)θ (e) = θ (r) = θ (e)θ(r) homo homomo morp rphi hism sm of θ of θ ))(sθ((e) = s = θ (e)s algebraic replacement → (∀s ∈ θ(R))(sθ → θ(e) is a unity of θ(R) definition of unity The existence of a unity for θ (R) is also an immediate consequence of theorem 18.2 n
{ 2 : n ∈ Z} b) {2 : n ∈ Z} c) {k 2 : n, k ∈ Z}
38.11 a)
n
n
d) Q (see example 38.4)
e) Q. Because Because of additive closure, closure, the smallest subfield subfield must contain every every integer. integer. And because a field −1 contains multiplicative inverses for each element, it must contain a for every integer a. And because of multiplicative closure, it must contain ab−1 for every integer a, b Z. And this is the definition of Q from section 30.
∈
38.12 Let S be a subring of Z. s
∈ S → (∀n ∈ Z)( s ∈ S ) 1 → (∀n ∈ Z)(ns )(ns ∈ S ) → S is an ideal of Z n
assumed addi additiv tivee clos closur uree defin definit itio ion n of inte intege gerr mult multip ipli lica cati tion on definition definition of ideal
38.13 The fact that the constant polynomials of Z[x] form a ring follows directly from the fact that Z forms a ring. That the constant polynomials aren’t an ideal of Z[x] can be shown by the fact that 3 is a constant polynomial, x Z[x], but 3x 3x is not a constant polynomial.
∈
38.14 We can show that (a ( a) is the smallest ideal containing a by showing that (a (a) must be a subset of any other ideal containing a. Let I be any ideal of R containing a. Proof that (a (a) is a subset of I : b
assumed ∈ (a) → (∃r ∈ R)(b )(b = ar) ar) definition of (a) )(b = ar ∧ ar ∈ I ) defin definit itio ion n of idea ideall I → (∃r ∈ R)(b → b ∈ I algebraic replacement 38.15 For each nonzero element a ∈ R, (a) is an ideal containing a. Since a ∈ (a), we know that (a (a) = {0}, so
it must be the case that (a (a) = R. We will prove that R is a field by proving that (a ( a) is a field. First, proof that each element in (a (a) contains a multiplicative inverse:
64
( a R 0 )((a )((a) = R) true for reasons outlined above ( a R 0 )(R )(R (a)) partial definition of set equality ( a R 0 )(x )(x R x (a)) defin definiti ition on of subs subset et ( a, x R 0 )(x )(x (a)) quantification Because this last statement is true for all x R, it is still true when we choose any particular value of x. Choose x to be the unity element of R (which is guaranteed to exist from the text of the exercise): ( a R 0 )(e )(e (a)) quantification ( a R 0 )( r R)(e )(e = ra) ra) defin definiti ition on of me memb mber ersh ship ip in (a)
∀ → → →
∈ ∀ ∀ ∀
−{ } ∈ −{ } ⊆ ∈ −{ } ∈ → ∈ ∈ −{ } ∈
∈
→ ∀ ∈ −{ } ∈ → ∀ ∈ −{ } ∃ ∈
And this last statement is just the quantification of “every nonzero a R has an inverse in R”. So we know that every element of R has an inverse, that R has a unity, and that multiplication on R is associative associative (guarante (guaranteed ed from the properties properties of rings): rings): this means that multiplicat multiplication ion is a group on R. In order for R to be a field, though, we would have to show that the set of nonzero elements forms a group: i.e., we need to show that R has no zero divisors. But this is true by virtue of the fact that each nonzero element a has an inverse: ab = 0 a = 0 a−1 ab = a−1 0 b=0
∧
→ →
∈
assumed each each nonz nonzer eroo a has an inverse properties of unity and zero
We’ve shown that R is a commutative ring in which the set of nonzero elements form a multiplicative group: this is the definition of a field. 38.17 If I = (0), then it must contain some nonzero element a. Since Since the ideal is a ring, ring, it must must also contai contain n two must be positive positive (from the trichotom trichotomy y property of ordered integral integral domains). domains). So the a. One of these two set of positive ideals is nonempty, which means that it must contain a least element (from the well-ordered property of Z, section 29). Let n represen representt this least positive positive integer integer in I .
−
Since ideals are subrings by definition, we know that I is closed closed under additio addition. n. This This tells tells us that that all multiples of n are in I : (n) I . But it’s also true that I (n):
⊆
⊆
x
assumed ∈ I → (∃r, s ∈ Z)(x )(x = rn + s, 0 ≤ s < n) n ) divisi division on alg algori orithm thm )(x = rn + s, s = 0) n is the least positive element → (∃r, s ∈ Z)(x → (∃r ∈ Z)(x )(x = rn) rn) property of zero definition of (n) → x ∈ (n) And since (n (n) ⊆ I ∧ I ⊆ (n), we know that (n (n) = I . 38.18a f ∈ I )(f = 2a 2 a0 + a1 x + a2 x2 + . . .) definition of membership in I → (∃a ∈ Z)(f → (∀g ∈ Z[x])(∃a , b ∈ Z)(f )(f g = 2a0 b0 + horrifying binomial expansion) ])(f g ∈ I ) definition of membership in I → (∀g ∈ Z[x])(f We’ve shown that (∀f ∈ I , g ∈ Z[x])(f ])(f g ıI ) which is the definition of I being an ideal. i
i
i
38.18b (x + 2) and (x (x + 4) are both members of I and are irreducible in Z[x]. Assume that I is equal to some principal ideal (a ( a). By the definition of principal ideal, both (x (x + 2) and and (x (x +4) would have to be multiples of (a). But But since since (x (x + 2) and (x (x + 4) are irreducible, their only common divisor is 1, so 1 (a). But But 1 clearly doesn’t have an even number as the constant term, so 1 I . So we have have shown that no principal principal ideal can contain all the members of I without also containing some members no in I : I cannot be a principal ideal.
∈
∈
38.19 We will disprov Z[x] as disprovee this by constructing constructing a homomorphi homomorphism sm that is not an ideal. Define θ : Z[x] θ (a0 + a1 x + . . . + an xn ) = (a0 ). The domain domain of this function function is all p olynomials, olynomials, and its codomain codomain is the set of constant polynomials. That θ is a homomorphism is trivially verified, and exercise 38.13 shows that the codomain is a subring that is not an ideal.
→
38.20 From theorem 38.1(b), we know that the kernel of θ is an ideal of F . F . From example 38.4, we know that the only ideals of F are 0 and F itself. If the kernel is 0 , then θ is one-to-one (theorem (theorem 28.1(c)). And if the kernel is F , F , then by definition of kernel θ(a) = 0 for all a F . F .
{}
{}
65
∈
38.21 From the definitions of Q given in section 30, we know that each element in Q can be expressed as ab−1 , a Z, b Z 0 . Let ab−1 be an arbitrary element of Q:
∈
∈ −{ }
α(ab−1 ) = α(a)α(b−1 ) = α(a)α(b)−1 = β (a)β (b)−1 = β (a)β (b−1 ) = β (ab−1 )
homo homomo morp rphi hism sm of α of α theorem 18.2(b) algebraic replacement with α(x) = β (x) theo theore rem m 18 18.2 .2(b (b)) homomorphism of β f β
38.22 Let I represent the set of all nilpotent elements in R. We need to prove that I is a subring and also that I is an ideal. I is nonempty 02 = 0 0 I
→ ∈
I is closed under both operations a I b I ( m, n Z)(a )(am = 0 bn = 0) (a + b)mn = 0 (ab) ab)mn = 0
∈ ∧ ∈ →∃ ∈ →
∧
assumed
∧
This previous step is justified because every term in the binomial expansion of (a ( a + b)mn contains either am or bn as a factor, which means that every term is zero. And the sole term of the multiplicative expansion of (ab) ab)nm also obviously contains a factor of am . Note that both of these facts rely on the commutativity of R. ab) ∈ I → (a + b) ∈ I ∧ (ab)
definition of I
I contains negatives for each nonzero element a I ( m N)(a )(am = 0) ( m N)(a )(a2m = 0) ( m N)((a )((a2 )m = 0) ( m N)((( a)2 )m = 0) ( m N)(( a)2m = 0) a I
∈ →∃ →∃ →∃ →∃ →∃ →−
∈ ∈ ∈ ∈ ∈ ∈
− −
assumed definition of memb ership in I
theorem 28.1 definition definition of membership membership in I
39 Quotient Rings 39.6
(I + a)(I )(I + b) = (I + b)(I )(I + a) (I + ab) ab) = (I ( I + ba) ba) (I + ab) ab) (I + ba) ba) = (I + i I ) (I + (ab (ab ba)) (I + i I ) ba)) = (I ab ba I
↔ ↔ − ↔ − ↔ − ∈
∈
∈
defin definit itio ion n of comm commut utaativi tivitty definition of addition in R/I zero zero elem elemen entt of R/I of R/I defin definit itio ion n of subt subtra racctio tion in R/I
So the set of elements a, b for which (I (I + a) and (I (I + b) commute is identical to the set of elements a, b for which ab ba I . So R/I is commutative iff ab ba I for every pair of elements in I .
− ∈ 39.7 Lemma Lemma: x ∈ (n) iff n|x:
− ∈
By the definition of principal ideal, (n ( n) is the set of all multiples of (n ( n). So x And this is also the definition of divisibility, so x (n) iff n x.
∈
|
i) Proof by contrapositive that n is prime if (n (n) is a prime ideal: n is not prime ( a, b Z)(n )(n = ab n a n b) definition of prime ( a, b Z)(n )(n ab n a n b) x x for all x ( a, b Z)(ab )(ab (n) a (n) b (n)) lemma (n) is not a prime ideal definition of prime ideal
→∃ ∈ →∃ ∈ →∃ ∈ →
∧ | ∧ | | ∧ | ∧ | ∈ ∧ ∈ ∧ ∈
|
66
∈ (n) iff rn = x for some r.
ii) Proof by contrapositive that (n ( n) is a prime ideal if n is prime: (n) is not a prime ideal ( a, b Z)(ab )(ab (n) a (n) b (n)) definit definition ion of prime prime ideal ideal ( a, b Z)(n )(n ab n a n b) lemma ( a Z)(gcd(n, )(gcd(n, a) = 1 n a) contrapos positive of theorem 13.1 n is not prime definition of prime
→∃ ∈ →∃ ∈ →∃ ∈ →
∈
∧ ∈ ∧ ∈ | ∧ | ∧ | ∧ |
39.8 The proof is an immediate consequence of the defintions of prime ideal and integral domain. Remembering that the zeroes of R/P of R/P are all elements of the from (P ( P + + p P ): P ): R/P is an integral domain (P + a)(P )(P + b) = (P ( P + p P ) P ) [(P [(P + a) = (P ( P + p
∈
↔ ∈ → ab) = (P ( P + p ∈ P ) P ) → [(P [(P + a) = (P ( P + p ∈ P ) P ) ↔ (P + ab) P ) → [a ∈ P ∧ b ∈ P ] P ] ↔ (ab ∈ P ) ↔ P is a prime ideal
J 39.9 To show that I ∩I J I + , define a function θ : I (I + I + J )/J as θ(i) = J i. The domain of this function is J I and the function is equal to 0 whenever i J . So the kernel of this function is I J . J . By the fundamental J homomorphis homomorphism m theorem theorem for rings, rings, then, I ∩I J I + . J
≈
∈
→
∩
≈
40 Quotient Rings of F[X] p(x) is irreducible in F [ F [x] divisors of p(x) must be units or associates ( g(x) F [ ])(g (x) p( F [x])(g p(x) g (x) e p( p(x) g (x))
40.2
→ →∀
∈
|
→
|∨
|
assumed definition of irreducible quantification of previous step
Choose g (x) such that g (x) = gcd(f, gcd(f, p). Since Since the gcd obviou obviously sly divides divides p(x), the right-hand side of the previous conditional gives us: (gcd(f, p)|e ∨ p( p(x)| gcd(f, gcd(f, g )) → (gcd(f, (gcd(f, p)|e ∨ p( p(x)| gcd(f, gcd(f, g )) ∧ gcd(f, gcd(f, p)|f ( f (x) definition of the gcd → (gcd(f, → (gcd(f, (gcd(f, p)|e ∧ gcd(f, gcd(f, p)|f ( f (x)) ∨ ( p( p(x)| gcd(f, gcd(f, g ) ∧ gcd(f, gcd(f, p)|f ( f (x)) log logica icall distri distribut butivi ivity ty (gcd(f, p)|e ∧ gcd(f, gcd(f, p)|f ( f (x)) ∨ ( p( p(x)|f ( f (x)) transitivity of divisibility → (gcd(f, → (gcd(f, (gcd(f, p)|e) ∨ ( p( p(x)|f ( f (x)) p∧q →p gcd(f, p)|e) we are told that p(x) |f ( f (x) → gcd(f, → gcd(f, gcd(f, p) = e) definition of gcd from theorem 36.1 The last step (moving from gcd(f, gcd(f, p)|e to gcd(f, gcd(f, p) = e) is justified by the definition of gcd given in
theorem theorem 36.1. By definition definition of gcd for polynomials, polynomials, the greatest greatest common divisor must must be monic: and by definition of monic, the highest coefficient must be e. 40.11
(f ) f ) = (g ) (f ) f ) (g) (g ) (f ) f ) ( a (f ))( f ))(a a (g )) ( b
assumed definition of set equality defini efiniti tion on of subs subset et
→ ⊆ ∧ ⊆ ))(b ∈ (f )) f )) →∀ ∈ ∈ ∧ ∀ ∈ (g))(b f ∈ (f ) f ) and g ∈ (g), so we can choose a to be f and choose b to be g. → (f ∈ (g)) ∧ (g ∈ (f )) f )) F [x])(f ])(f = gq) gq ) ∧ (∃r ∈ F [ F [x])(g ])(g = f r ) defin definit itio ion n of (f ) f ), (g) → (∃q ∈ F [ → g|f ∧ f |g definition of divisibility definition of associate → g and f are associates
This proves that (f (f )) = (g ( g) only if g if g and f are associates. For the “if” portion of the proof, the same steps can be used in reverse reverse order. order. The same justificatio justifications ns can be used for all of the steps except except for the step
67
from f (g ) to (f (f )) (g ). This step is justified because if f means that (f (f )) (g).
∈
⊆
⊆
∈ (g), then all multiples of f are in (g (g ), which
40.12 Theorem 40.3 tells us that there is some m F [ F [x] such that (m (m) = I , but the theorem does not guarantee this this polynomial is monic or that it is unique.
∈
To show that there is at least one monic polynomial a such that (a (a) = I , note that theorem 36.1 tells us that gcd(m, gcd(m, m) exists and is monic. From the properties properties of the gcd, we know that (gcd( (gcd(m, m) (m) (because gcd(m, gcd(m, m) m) and (m (m) (gcd(m, (gcd(m, m) (because m gcd(m, gcd(m, m)). )). So (gcd (gcd((m, m)) = (m (m) = I and gcd(m, gcd(m, m) is monic.
|
⊆
⊆
|
To show that this monic polynomial is unique, assume that there a and b are two monic polynomials and (a) = I = (b). Exercise Exercise 40.11 showed showed that a and b must must be associa associates tes:: ( r, s F [ F [x])(ar ])(ar = b a = bs). bs). Because a and b are of the same degree, both r and s must be constant polynomials of degree 0 (note: this fact depends on F being an integral integral domain). And because the highest highest term of a and b are both 1, this means that r = s = e. Therefore a = b.
∃ ∈
∧
40.13a If we aren’t supposed to assume that these are polynomials in a field F [ F [x], then this is false. Let R[x] 2 2 be a subring of Z4 [x] consisting of [0], [0], [2]x, [2]x, [2]x [2]x , [2]x [2]x + [2]x [2]x . [2]x [2]x is not the same degree as ([2]x ([2]x2 ), but ([2]x ([2]x) = ([2]x ([2]x2 ): ([2]x ([2]x) = [4]x [4]x2 , [4]x [4]x3 , [4]x [4]x3 + [4]x [4]x2 = [0]
{
}
{ } { } ([2]x ([2]x2 ) = {[4]x [4]x3 , [4]x [4]x4 , [4]x [4]x4 + [4]x [4]x3 } = {[0]}
This would not work in a field, since there would be no zero divisors. 40.13b False. In the commutative ring of integers, 1 and 2 both have degree 0, but 1 40.14 True for the same reason as above: 2
∈ (2) so (1) = (2).
∈ (1), deg(2) = deg(1), but (2) = (1).
41 Factorization and Ideals 41.1 Let I represent any ideal in Euclidean domain D. Because Because the division division algorithm algorithm holds for Euclidean Euclidean domains, for any two elements a, b I we find that gcd(a, gcd(a, b) I . Let x = gcd( gcd ( i I ), the greatest element that divides every element of I . We want want to show show that that (x (x) = I . First, First, assume assume that that a (x). Then a = rx for some r D (from the definition of (x (x)). But sinc sincee x I , rx I (from the definition of an ideal). ideal). And rx = a, so a I . Thus (x (x) I . Next, Assume Assume that that a I . By the division algorith algorithm, m, a = xq + r, d(r) < d(x) or, alternatively, a xq = r, d(r) < d(x). But But x a (from being the gcd of every element of I ) and x xq (from the definition of division) so x a xq = r. But it can’t be the case that x r and d(r) < d(x) unless r = 0 (otherwise it would be the case that d(r = xm) xm) < d(x), which would make D not a Euclidean domain). So the fact that we could write a as a = xq + r means that r = 0 and a = xq. xq. And this means that a (x). Thus I (x). We’ve shown that I (x) I , so I = (x).
∈
∈
∈
∈
−
|
∈
41.2a
41.2b
(b)
a and b are are assoc associa iate tess ab ba (b) (a) (a) (b) (a) = (b)
→ |∧| → ⊆ ∧ →
⊆
∈ ∈ |
⊆
| −
⊆
⊆ (a) → b ∈ (a) → (∃r ∈ D)(b )(b = ar) ar) → a|b a|b → (∃r ∈ D)(b )(b = ar) ar) )(bs = ars) ars) → (∀s ∈ D)(∃r ∈ D)(bs )(bs = a(rs)) rs)) → (∀s ∈ D)(∃r ∈ D)(bs → (∀s ∈ D)(bs )(bs ∈ (a)) → (b) ⊆ (a)
{ ∈ }
⊆
assumed (b) is an ideal containing b definition of (a) definition of divisibility assumed definition of divisibility algebra associa associativ tivity ity of rings rings definition of (a) definition of (b)
assu assume med d definition definition of associate associate exer exerci cise se 41 41.2 .2aa definition of set equality
68
⊆
∈
∈
|
41.3a Principal ideal domains are commutative. So, for all r
∈ D, i ∈ I :
ri = r(ax + by) by) = rax + rby = a(rx) rx) + b(ry) ry ) So ri I as long as rx and ry are members of D. And because x,y, and r are members of D, closure of D tells us that rx and ry are in D.
∈
41.3b) This is true by definition, because D is a principal ideal domain. 41.3c) Proof that d is a divisor of both a and b: I
⊆ (d) )(i = rd) rd) → (∀i ∈ I )()(∃r ∈ D)(i → (∀x, y ∈ D)(∃r ∈ D)(ax )(ax + by = rd) rd)
true from part (b) of this exercise definition of subset and of (d) defin definit itio ion n of me mem mbers bershi hip p in I We could choose x = 0, y = e to give us b = rd. rd. Or we could choose x = e, y = 0 to give us x = rd. rd. And because we could make either choice, this implies: ( r D)(b )(b = rd) rd) ( s D)(a )(a = sd) sd) db da definition of divisibility
→∃ ∈ → |∧ |
∧∃ ∈
Proof that d is the greatest such divisor: ca
assumed | ∧ c |b true from part (b) of this exercise → c|a ∧ c|b ∧ (d) ⊆ I )(dr = i) definition of (d) → c|a ∧ c|b ∧ (∀r ∈ D)(∃i ∈ I )(dr → c|a ∧ c|b ∧ (∀r ∈ D)(∃x, y ∈ D)(dr )(dr = ax + by) by) definition of memb ership in I )(cm = a ∧ cn = b) ∧ (∀r ∈ D)(∃x, y ∈ D)(dr )(dr = ax + by) by) definit definition ion of divisi divisibil bilit ity y → (∃m, n ∈ D)(cm → (∀r ∈ D)(∃m,n,x,y ∈ D)(dr )(dr = cmx + cny) cny ) algebraic replacement Choose r to be e: )(d = cmx + cny) cny ) → (∃m,n,x,y ∈ D)(d → (∃m,n,x,y ∈ D)(d )(d = c(mx + ny) ny)) distributivity definition of divisibility → c |d 41.4 From problem 41.3, we know that the set I = {ar + bs : r, s ∈ D} = (gcd(a, (gcd(a, b)). So we can prove prove that that e = ar + bs if we can show that e ∈ gcd(a, gcd(a, b). Althou Although gh we don’t don’t know for certain certain that e = gcd(a, gcd(a, b), we know that e|a and e|b, so e| gcd(a, gcd(a, b). But this this means means that er = gcd(a, gcd(a, b) for some r ∈ D which, by definition, definition, means that e ∈ (gcd(a, (gcd(a, b)). 41.5 If p is irreducible, irreducible, then its only divisors are units and associates. associates. Let g = gcd( p,a gcd( p,a). ). Since Since g is a divisor of p, then g must either be a unit (g ( g e) or an associate of p (g p p p g). Assume Assume g is an associate. associate. Then p g (from being an associate) and g a (from being the gcd( p,a gcd( p,a)). )). So, because divisibility divisibility is transitive transitive,, p a. Assume that g is a unit. Then g e, which implies:
|
|
|
|
| ∧ |
|
ge
assumed | → g|e ∧ (∃x, y ∈ D)(g )(g = px + ay) ay) lemma 2 of theorem 37.1 )(gr = e) ∧ (∃x, y ∈ D)(g )(g = px + ay) ay) defin definit itio ion n of divi divisi sibi bili litty → (∃r ∈ D)(gr → (∃r ∈ D)(gr )(gr = e) ∧ (∃x, y ∈ D)(gr )(gr = ( px + ay) ay )r) alge algebr braa )(e = pxr + ayr) ayr ) algebraic replacement, distributivity → (∃r,x,y ∈ D)(e )(be = bpxr + bayr) bayr) algebra → (∃r,x,y ∈ D)(be )(b = pbxr + abyr) commutativity → (∃r,x,y ∈ D)(b abyr) We are told p|ab, ab, so (∃S ∈ D)( ps )( ps = ab). ab). Using this replacement: → (∃r,s,x,y ∈ D)(b )(b = pbxr + psyr) psyr) algebraic replacement )(b = p(bxr + syr) syr )) distributivity of rings → (∃r,s,x,y ∈ D)(b → p|b definition of divisibility So the fact that g is either a unit or an associate implies that either p|a or p|b. 41.6 Proof by by induction. induction. Let p be irreducible and let S = {n : p|a1 a2 . . . a → (∃i)( p )( p|a )}. 1 ∈ S for trivial reasons and exercise 41.5 showed that 2 ∈ S . Now, Now, assume that n ∈ S . Then p|a1 a2 . . . (a a +1) implies that p either divides a for some i < n, n , or p|a a +1 . And if the latter is true, then 2 ∈ S tells us that p|a o p|a +1 . So n + 1 ∈ S . Therefore, by induction, S = N. n
i
n n
i
n n
n
n
69
41.8 Exercise 41.7 shows that all nonzero, non-unit elements of a principal ideal domain can be written as a product of irreducible irreducible elements. elements. Exercise Exercise 41.6 can be used to prove uniqueness uniqueness via the same steps as the proof of the fundamental theorem of arithmetic on pp70-71. 41.9 a,b,c) Every field is a Euclidean domain (example 38.4) d,g,h) F [ F [x] is a Euclidean domain for any field F . F . (p175, below definition) e) This is a Euclidean domain (example 37.2) f ) This is an integral domain but not a unique factorization domain (example 37.1) 41.10 1,3,4,5,9
{
}
42 Simple Extensions 42.a This This is an expand expanded ed proof of theore theorem m 42. 42.1. 1. Define Define a functi function on θ : F [ F [x] F [ F [a] by θ ( p( p(x)) = p(a). This This function can easily be shown to be a homomorphism, so by the fundamental homomorphism theory for quotient rings, we know that: F [ F [x]/Ker( /Ker(θ) θ (F [ F [x])
→
≈
θ (F [ F [x]), the range of θ , is the set of all linear combinations of elements of F and the element a. This range is, by definition, the field extension F ( F (a). So we see that: F [ F [x]/Ker( /Ker(θ )
F (a) ≈ F (
The kernel of θ is the set of all polynomials in F [ F [x] for which p(a) = 0. And we know know that kernel kernelss are always always normal subgroups subgroups and ideal subrings. subrings. And since F [ F [x] is a principal ideal domain if F is a field, we know that the kernel is an ideal domain: it is equal to ( p ( p((x)) for some p(x) F [ F [x]. So:
∈
F [ F [x]/( p( p(x))
≈ F ( F (a)
And since F ( F (a) is a field, that means that F [ F [x]/( p( p(x)) is a field and therefore, by theorem 40.1, p(x) is irreducible over F . F . 42.2
bi) C R(a + bi) x R(a + bi) bi) ( r R)(x )(x = r(a + bi) bi)) ( r R)(x )(x = ra + rbi) rbi) ( ra,rb R)(x )(x = (ra) ra) + (rb ( rb))i) x C
⊆
∈ →∃ ∈ →∃ ∈ →∃ ∈ → ∈ bi) : x ∈ C C ⊆ R(a + bi) → (∃m, n ∈ R)(x )(x = m + ni) ni) )(x = (a + bi) bi) + → (∃m, n ∈ R)(x bi) → x ∈ R(a + bi)
assumed unique representation from theorem 42.3 corollary 2 distributive property of R multiplicative closure of R definition of C
assumed definition of C n mb−na 2 (a + bi) bi) ) tons tons of alge algebr braa b b(a −b ) unique representation from theorem 42.3 corollary 2 Note that the requirement that a + bi is imaginary forces b to be nonzero, so the fractions in the penultimate step are all defined.
√
√
2
2
√
√
42.3 If Q( 2) = Q( 3), then every element of Q( 2) (including (including 2 itself) would have to be expressible as a p 2nd-degree linear combination of r 3 (corollary (corollary 2 of theorem 42.3). 42.3). Proof by contradic contradiction tion that this is not possible: possible:
√
√ √ assumed p artial definition of set√ equality → √ √ ⊆√ √ √ → ∈ 2 is an element of Q( 2) √ √ →∃ ∈ theorem 42.3 corollary 2 √ algebr alg ebra: a: squar squaree both sides sides →∃ ∈ √ →∃ ∈ − − √ algeb lgebra ra 2−3 − → (∃r, s ∈ Q)( 2 = 3) algebra The left-hand√side of this equation is also a rational number, so: )(t = 3) → √(∃t ∈ Q)(t → 3 is rational Q( 2) = Q( 3) Q( 2) Q( 3) 2 Q( 3) ( r, s Q)( 2 = r 3 + s) ( r, s Q)(2 = 3r 3r2 + 2rs 2rs 3 + s2 ) ( r, s Q)(2 3r2 s2 = 2rs 3) r 2 s2 rs
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And we know, from proving it a thousand times in every class since discrete math, that
√3 is not rational.
42.4 a) 0 + 7a 7a b)
−5 − a
c) 8 + 2a 2a
− 16 + 16 a 42.5 a) x2 − 2 b) x4 − 4 c) x4 − 2x2 − 9 √ d) x2 − 2 2x + 3 √ 42.6 f ( f (x) ∈ Q 5 √ √ → (∃a ∈ Q)(f )(f ((x) = a0 + a1 (√ 5) + . . . + a ( 5)√) √ )(f ((x) = ( a3 ( 5)3 ) + ( a3 +1 ( √5)3 +1 ) + ( a3 +2 (√ 5)3 +2 )) → (∃a ∈ Q)(f √ ) + (√ a3 +1(5) ) 5 + ( a3 +2(5) ) 25) )(f ((x) = ( a3 (5) → (∃a ∈ Q)(f → (∃a,b,c ∈ Q)(f )(f ((x) = a + b 5 + c 25) d)
3
3
n
3
i
n
k
3
i
k
i
k
k
3
k
3
k
k
k
3
k
3
k
k
k
3
3
42.7 If a2 is algebraic over F , F , then there exists some function f ( f (x) such that f ( f (a2 ) = 0. Define a new function g (x) = f ( f (x2 ). Then g (a) = f ( f (a2 ) = 0, so a is algebraic over F . F . 42.8 The field of quotients of F [ F [x] is defined in section 30 as the unique smallest field that contains a solution in F [ F [x] for the equation f ( f (x)λ(x) = g(x) for all nonzero elements f ( f (x), g(x) in F [ F [x]. This field is clearly clearly the field containing all elements of the form f ( f (x)/g( /g (x). 42.12 By theorem 42.5, we can prove that F ( F (α) F ( F (β ) just by showing that there exists an isomorphism F F . F . And this is trivially done by the function θ (a) = a.
≈
≈
42.12 (alternate proof) From theorem 42.1, we know that Q(α) Q[x]/( p( p(x) = x2 2) and Q(β ) Q[x]/(q (x) = x2 4x +2). But note that x2 4x + 2 is equivalen equivalentt to (x ( x 2)2 2. So ( p ( p((x)) (q(x)) under the mapping θ (f ( f (x)) = f ( f (x 2). And so, by the definition of extension at the start of the chapter, Q(α) Q(β ).
−
≈ − −
−
− √5)(x + x √5 + √25), which has two complex roots. 42.13 x3 − 5 = (x − 5)(x 3
3
− ≈
≈
≈
3
44 Splitting Fields 44.1 1,5,7, and 11 are roots. Z12 is not a field (it has zero divisors), so theorem 44.1 does not apply. [1]) = ([1] ([1]x x3 + [−7]x 7]x2 + [15]x [15]x + [−9]) 9]) = ([1] ([1]x x3 + [3]x [3]x + [1]) − [3])2(x − [1]) 44.3 (x − i)2 (x + 1) = (x (x3 − ix2 + x − i) = (x ( x3 + x) − (x2 + 1)i 1)i 44.2 (x
44.5 Use theorem 44.7 to generate the set of all possible rational roots, and then prove by exhaustion that none of these possible rational roots are roots of 12x 12 x3 3x + 2.
− 44.6 a) f ( f (x) = (x − 32 )(x )(x + 12 )(x )(x + 1), roots are 32 , − 12 , −1. b) roots are 2, 2, 12 , 13 (found using theorem theorem 44.7).
)(x2 + 4), 12 is the only root in Q. − 12 )(x 44.7 a) f ( f (x) = (x − 1)(x 1)(x2 − 5), 1 is the root in Q b) f ( f (x) = 3(x 3(x − 23 )(x )(x2 + 1), 23 is the root in Q c) f ( f (x) = x(x2 − 2x + 2), 0 is the root in Q √ √ 44.8 a) f ( f (x) = (x − 1)(x 1)(x2 − 5), 1, 1, 5, − 5 are the roots in R b) f ( f (x) = 3(x 3(x − 23 )(x )(x2 + 1), 23 is the root in R c) f ( f (x) = x(x2 − 2x + 2), 0 is the root in R c) f ( f (x) = 2(x 2(x
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√ √ − 1)(x 1)(x2 − 5), 1, 1, 5, − 5 are the roots in C b) f ( f (x) = 3(x 3(x − 23 )(x )(x2 + 1), 23 are the roots in C c) f ( f (x) = x(x2 − 2x + 2), 0 is the only root in C
44.9 a) f ( f (x) = (x
44.11 If f If f ((x) has an imaginary root of degree two, then ( a + bi) bi)2 is a root for some a, b R. From exercise 42.7, it is also the case that (a ( a + bi) bi) is a root. And from the corollary to theorem 44.5, the complex conjugates 2 of (a + bi) bi) and (a (a + bi) bi) are also roots. So the degree of f ( f (x) is at least 4.
∈
44.12 x2 + (2 + 2i 2i)x + 2i 2i = (x + (1 + i))2 , so
−1 − i is a root of multiplicity 2.
44.13 [1]x [1]x3 + [1]x [1]x2 + [1]x [1]x + [1] 44.14 F ( F (c1 , . . . , cn ) is by definition the smallest extension of F that contains all of the elements (c ( c1 , . . . , cn ). Since F ( F (c1 , . . . , cn ) is the smallest field containing those roots, we know that F ( F (c1 , . . . , cn ) H . So if it is also true that H F ( F (c1 , . . . , cn ), then H = F ( F (c1 , . . . , cn ).
⊆
⊆
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