Solutions to Modern Algebra (Durbin, 5E)

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

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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  R. So α(a) = b, and therefore β (a) = b. that α(a) ∈ R and 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.

◦

→

◦

→

→

→



















( 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

∈ × → ∈ × ∧y ∈ R → ∈ × ∧y =y → ∼

∼

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

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 )

assumed definition of transitivity of = definition of

→ → → → → →

∼

∼

∼

∼

∈ × R ∧ y1 = y2 ∈ × R ∧ y2 = y1

∧ ∼ ∈ × ∧ ∈ × ∧

∼ ∼

∧ y1 = y3

∼ ∼

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).

∈ 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}



















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 .

{

≤ }

∈ 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 n 2 relation. relation. One set of equivalence equivalence class representa representative tivess is the set 0 + a1 x + a2 x + . . . + an x : 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

∼





















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 n

n

n

n

n

n

n

m



















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 )



















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



11.10

⊕



⊕  ⊕  [a] + [ −a] = [a + (−a)]

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] [2]

[1] [1] [2]

[2] [2] [1]



















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] [3] [6] [9]

[0] [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 + 1 ))

∃ ∈ →∃ ∈ →∃ ∈

− ∧ −

⊕



















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

−

| ∧ |∧ | ∧ | → | ∃ ∈ ∧ ∧ ∃ ∈ ∧ ∧ | ∧ | ∧ | ∧ | → |

→

assumed theorem 12.2 the fractions are defined because d a, d b. corollary of theorem 12.2

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)

→ → → → → → → → → →

∃ ∃ ∃ ∃ ∃ ∃

∈ ∈ ∈ ∈ ∈

∈ ∈ ∈ | | ∧ | ∧ | | ∧ |

∧

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

→ → → → → → →

12.16

∧

→

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))

| ∧ | | ∧ | ∧ | ∧ | ∧

∨ ∨ | ∧ ∨ | ∧

| |

| ∧

∧

∧

|

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

→∃ →∃ →∃ →

∧ ∈ ∈ ∈

| ∧ | → |

∧∃ ∈

theo theore rem m 12 12.2 .2 multiply both equations algebra theorem 12 2





















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)



















7

×

7 = pe1 pe2 pek 7 = 20 30 50 71 110 . . . f 6 = pf pf 6 = 21 31 50 70 110 . . . 1 p2 k 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

13.15

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 ) t1

i

sk k t2 s 2

s1

t1 t2 t2 k

2

tk k )

s2

i

n is odd n = 20 pe2 pek 2n = 2 1 pe2 pek φ(2n (2n) = φ(21 pe2 pek ) φ(2n (2n) = φ(21 )φ( pe2 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

s1 s2

··· ···

k

k

2

2

··· 2

− 2

··· ··· ···

2

k

···

k

2

k

···

k

k

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 ) e e 2n = 2 2 ( p1 p2 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 ) e e φ(2 ) φ( e )

∃ ∈ → → → → → →

1

2

1

1

··· ··· ···

k

2

k

2

k

1

2

1

1

1

··· 2

− 2

k

··· k

k

2

···

definition of divisibility defin definit itio ions ns of m of m and n algebra standard form cannot have negative exponents

k

assumed definition of even FToA algebra p1 = 2 φ is well-defined theorem 13.3 theo theore rem m 13 13.1 .1 l b



















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 e e e n = ( p1 p2 pk ) ( 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 2 ( p1 p2 · · · p ) FToA ↔ n = ( p 2 2 2 ↔ n = ( p ( p1 p2 · · · p ) 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 → (∃q, r ∈ N)√n = → (∃q, r ∈ N)n2= 2 → (∃2q, r 2∈ N)r n2 = q )( p1 p2 · · · p ) = ( p21 p22 · · · p2 → ( p12 +p2 ·2· · p+ )( p → ( p1 p2 · · · p2 + ) = ( p21 p22 · · · p2 ) )(2r + n = 2q ) → (∀i ∈ N)(2r → (∀i ∈ N)(n )(n = 2(q 2(q − r )) → (∀i ∈ N)(2|n ) → (∀i ∈ N)(n )(n is even) → p1 p2 · · · p is a perfect square → n is a perfect square √ 13.20 A general proof for all n: √n is rational 13.19

q2 r2

r1 r1

r2

n1

q r

rk k n2

r2

i

n1 n2 rk k

i

i

i

i

i

i

n1 n2

i nk k

nk

nk k

q1

q1

q2

q2

qk k

qk k )

assumption definition of rational algebra algebra FToA, FToA, definit definition ion of square squaress algebra properties of exponents algebra definition of divisibility definition of even 13.18(b) definition of n

x

x

assumption

}



















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 ( )

( )



















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) −1 o(a ) 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 e = ba( ba(ab) ab)−1 −1 (ba) ba) e = (ba ( ba))−1 ba( ba(ab) ab)−1 (ba) ba)−1 = (ab) ab)−1 a−1 b−1 = b−1 a−1

↔ ↔ ↔ ↔ ↔

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)

assumed 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

  





















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 ) −1 ( a, b G)((ab )((ab)) (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)

∈ → ∈ ∧ ∈ → ∈ ∧ ∈ → ∈ → → ∈

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

−

−





















14.31

o(a−1 ba) ba) = m assumed −1 m (a ba) ba) = e definition of o(x) −1 −1 m−1 a (baa ) 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 m−1 (b) 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 in reverse to justify the claim that (o ( o(b) = n) (o(a−1 ba) ba) o(b)). And, And, since since o(b) o(a−1 ba) ba) o(b), this mean that o(b) = o(a−1 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 m−1 (ba) ba) b = a−1 (ba) ba)m−1 ba = a−1 a (ba) ba)m = e o(ba) ba) m o(ba) ba) m (ba) ba) (ab) ab)

→ → → → → → → →

|

≤

assumed definition of o(x) associativity left multiplication right multiplication theorem 14.3 property of divisibility d fi f

≤

≤

≤

≤





















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

→ → → → → → →

∧ | ∧ | | ∧ |

(

) is a common multiple of (

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

) and it’s less than or equal to the least such multiple, it must





















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 −1 a 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 × {e }) B

assumed





















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

| ∈  ∈ 

15.23

↔ ↔ ↔ → → ↔

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

∃ ∈ ∃ ∈ ∃ ∈ | ∧ | ∃ ∈

x

∈ a, b → (∃m, n ∈ Z)(x )(x = a )(x = a → (∃m, n ∈ Z)(x

m n

b ) b da

m n

∧ | ∧ d|b)

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 ) m x=a

∃ ∈ ∃ ∈ ∃ ∈ ∃ ∈ n

assumed definition of x a, b definition of d and the gcd

∈ 

|























15.25

15.26

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

↔∃ ∈ ↔∃ ∈ ↔∃ ∈ ↔∃ ∈ ↔ ∈  ↔  x ∈ [a] )(x = [a [ a] ) → (∃r ∈ N)(x → (∃r ∈ N)(x )(x = [a [ a] ∧ d|a) )(x = [a] ∧ d → (∃r,s,t ∈ N)(x

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 Zn

r

r

s

definition of x [a] definition of d and gcd defin definit itio ion n of divi divisi sibi bili litty

∈ 

r

= a)























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 a A b A ab A

∈ × → ∈ ∧ ∈ → ∈

∈ →

∈ ×

assumed definition of cartesian product closure property of A























16.10 16.10

refle reflexi xiv ve a G a G a−1 a−1 a G a a

∈ → ∈ ∧ → ∈ → ∼

∈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 inverse property of groups closure closure property property of groups groups

assumed definition of

definition of

∼ ∼























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























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,

}

























both A and B . From the properties of the GCD, we know then that The only group of size one is , so the fact that ( ) 1 proves that

| |

| |

{}

|

|c| must also divide gcd(|A|, |B|) = 1. therefore that   { } and therefore

























θ((g, ((g, h)(i, )(i, j )) = θ((g ((g i, h# j)) j )) = (h ( h# j,g i)

∗

definit definition ion of of direct direct produ product ct operat operation ion d fi iti f θ

























θ is one-to-one θ([ ] ) θ([b ([b] )

d

























18.16 G is defined to be the set of all matrices of the form

a b 0

1

with a = 0. This is a subgroup of



























19.25 Let θ, λ be arbitrary isomorophisms from G

→ G. Let Aut

G

= f : f is an isomorphism G

{

→ G}.

































































































































































































































































































































































































































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:

|

|

|



















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

→ {0} × S to be

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 ) θ

S be a homomorphic function.



















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:



















( 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] n θ (a0 + a1 x + . . . + an x ) = (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



















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) mn (a + b) = 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

→∃ ∈ →∃ ∈

∧ | ∧ | | ∧ | ∧ |

|

∈ (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 I I +J homomorphis homomorphism m theorem theorem for rings, rings, then, I ∩J . 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, (gcd( f, p ) e p( p ( x ) gcd(f, gcd(f, p)|f ( f (x) definition of the gcd → | ∨ | gcd(f, g)) ∧ 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 )

assumed



















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

(b)

⊆ (a) → b ∈ (a) → (∃r ∈ D)(b )(b = ar) ar) → a|b

{ ∈ }

∈ ∈ |

⊆

| −

⊆

assumed (b) is an ideal containing b definition of (a) definition of divisibility

⊆

⊆

∈

∈

|



















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

|

|

|

| → g|e ∧ (∃x, y ∈ D)(g )(g = px + ay) ay) )(gr = e) ∧ (∃x, y ∈ D)(g )(g = px + ay) ay) → (∃r ∈ D)(gr → (∃r ∈ D)(gr )(gr = e) ∧ (∃x, y ∈ D)(gr )(gr = ( px + ay) ay )r)

| ∧ |

assumed lemma 2 of theorem 37.1 defin definit itio ion n of divi divisi sibi bili litty alge algebr braa

|



















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 to ns of alge al gebr bra a 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 3 (corollary (corollary 2 of theorem 42.3). 42.3). Proof by contradic contradiction tion that this is

√



















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) = 2 2 2 x 4x +2). But note that x 4x + 2 is equivalen equivalentt to (x ( x 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.

−





















√ √ − 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 of (a + bi) bi)2 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 ).

⊆

⊆

Solutions to Modern Algebra (Durbin, 5E)

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