Dummit and Foote Soln

MAT7400 Assignment #3

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Dummit & Foote Text Exercise 3.4-2: Exhibit all 3 composition series for Q for Q 8 and all 7 composition series for D for D 8 . List the composition factors in each case. Solution :

In Exercise 3.1-32, we showed every subgroup of Q of Q 8 is normal. Of the normal subgroups, i , j , and k are maximal. In addition, these three only have one non-trivial subgroup. Every quotient in each series has order two: Q8 / i , i / 1 /, 1 /1, Q8 / j , j / 1 /, 1 /1, Q8 / k , k / 1 /, 1 /1, So the every quotient is isomorphic to the simple abelian group Z /2Z. This gives us the following three composition series for Q 8 :



 

      −  −      −  −      −  − 

1

−1  i  Q , 1  −1   j   Q , 1  −1  k   Q . 8

8

8

Exercise 3.1-33 informed us that there are three maximal subgroups of D of D 8 which are normal: s, r2 , r , and rs,r2 . Each have order four. s, r2 has three subgroups of order two: s , r2 , r2 s . Since each of these subgroups have index two, they are normal to s, r2 . r has one subgroup of order two: r2 and since it’s index is two, it is normal in r . rs,r2 has three subgroups of index two (and hence normal): r 2 , rs , r3 s . Again, every quotient is isomorphic to the simple abelian group Z /2Z. This gives us the following seven composition series for D 8 : 1. 1  s  s, r 2  D8 2. 1  r2 s  s, r2  D8 3. 1  r2  s, r2  D8 4. 1  r2  r  D8 5. 1  r2  rs,r2  D8 6. 1  rs  rs,r2  D8 7. 1  r3 s  rs,r2  D8

      



                   







  

     



 

     

 



 



Dummit & Foote Text Exercise 3.4-5: Prove Prove that subgroups and quotient quotient groups of a solvable solvable group are solvable. Solution :

A subnormal series of series of a group G is a chain of subgroups G = G = G s >

· · · G > G > G 2

1

0

such that G that G i  Gi+1 for all i all i..  Gs−1  Gs = G Let G Let G be a solvable group. So G has a subnormal series 1 = G = G0  G1  = G with with abelian factors: for each i = 1, . . . , s, s, G i+1 /Gi is an abelian group. Let H G. Consider H Consider H Gi and H and H Gi+1 . Each is a group. If g g H Gi and h and h H Gi+1 , then hg then hgh h−1 H since H since g g,, h H . Also, hg Also, hgh h−1 Gi . Since Gi  Gi+1 , H Gi  H Gi+1 . This also implies that H that H Gi+1 N G (H Gi ). Hence H Hence H is is solvable and the composition (subnormal) series is

∈ ∩ ∩ ∩

∈ ∈

∈ ∩

1  H G1 

∩ ∩

∈

∩ ∩

· · ·  H ∩ ∩ G

s−1

 H

≤ ≤ ≤

···

∩ ∩

∈

∩ ∩

∩ ∩ ∈

∩ ∩ G = H ∩ ∩ G = H. = H. s

We can also use the second (diamond) isomorphism theorem to show this as well. Again let H each i each i let H i = G = Gi H . Then for every i, G i1  Gi , and we have the subnormal series

≤ ≤ G and for

∩

1 = H 0  H 1 

· · ·  H

s−1

 H s = H.

By the second (diamond) isomorphism theorem, for every i, i , H i /H i−1 = H i /(Gi−1

∩ H ) =∼ (H ( H G i

i−1 )/Gi−1 ,


2

Dummit & Foote Text Exercise 3.4-2: Exhibit all 3 composition series for Q for Q 8 and all 7 composition series for D for D 8 . List the composition factors in each case. Solution :

In Exercise 3.1-32, we showed every subgroup of Q of Q 8 is normal. Of the normal subgroups, i , j , and k are maximal. In addition, these three only have one non-trivial subgroup. Every quotient in each series has order two: Q8 / i , i / 1 /, 1 /1, Q8 / j , j / 1 /, 1 /1, Q8 / k , k / 1 /, 1 /1, So the every quotient is isomorphic to the simple abelian group Z /2Z. This gives us the following three composition series for Q 8 :



 

      −  −      −  −      −  − 

1

−1  i  Q , 1  −1   j   Q , 1  −1  k   Q . 8

8

8

Exercise 3.1-33 informed us that there are three maximal subgroups of D of D 8 which are normal: s, r2 , r , and rs,r2 . Each have order four. s, r2 has three subgroups of order two: s , r2 , r2 s . Since each of these subgroups have index two, they are normal to s, r2 . r has one subgroup of order two: r2 and since it’s index is two, it is normal in r . rs,r2 has three subgroups of index two (and hence normal): r 2 , rs , r3 s . Again, every quotient is isomorphic to the simple abelian group Z /2Z. This gives us the following seven composition series for D 8 : 1. 1  s  s, r 2  D8 2. 1  r2 s  s, r2  D8 3. 1  r2  s, r2  D8 4. 1  r2  r  D8 5. 1  r2  rs,r2  D8 6. 1  rs  rs,r2  D8 7. 1  r3 s  rs,r2  D8

      



                   







  

     



 

     

 



 



Dummit & Foote Text Exercise 3.4-5: Prove Prove that subgroups and quotient quotient groups of a solvable solvable group are solvable. Solution :

A subnormal series of series of a group G is a chain of subgroups G = G = G s >

· · · G > G > G 2

1

0

such that G that G i  Gi+1 for all i all i..  Gs−1  Gs = G Let G Let G be a solvable group. So G has a subnormal series 1 = G = G0  G1  = G with with abelian factors: for each i = 1, . . . , s, s, G i+1 /Gi is an abelian group. Let H G. Consider H Consider H Gi and H and H Gi+1 . Each is a group. If g g H Gi and h and h H Gi+1 , then hg then hgh h−1 H since H since g g,, h H . Also, hg Also, hgh h−1 Gi . Since Gi  Gi+1 , H Gi  H Gi+1 . This also implies that H that H Gi+1 N G (H Gi ). Hence H Hence H is is solvable and the composition (subnormal) series is

∈ ∩ ∩ ∩

∈ ∈

∈ ∩

1  H G1 

∩ ∩

∈

∩ ∩

· · ·  H ∩ ∩ G

s−1

 H

≤ ≤ ≤

···

∩ ∩

∈

∩ ∩

∩ ∩ ∈

∩ ∩ G = H ∩ ∩ G = H. = H. s

We can also use the second (diamond) isomorphism theorem to show this as well. Again let H each i each i let H i = G = Gi H . Then for every i, G i1  Gi , and we have the subnormal series

≤ ≤ G and for

∩

1 = H 0  H 1 

· · ·  H

s−1

 H s = H.

By the second (diamond) isomorphism theorem, for every i, i , H i /H i−1 = H i /(Gi−1

∩ H ) =∼ (H ( H G i

i−1 )/Gi−1 ,


3

which is a subgroup of the abelian group G i /Gi−1 and therefore is abelian. Hence, H has H has a subnormal series with abelian factors and that implies H is H is solvable. (I had help with this part of problem.) Let K Let K be be a quotient group of G of G with K with K = G/N with N with N  G. Define π Define π : G : G K as as the projection homomorphism. and last define K i = π = π((H i ). Then for each i, i , K i−1  K i . We have now obtained the subnormal series:  K s−1  K s = K. 1 = K 0  K 1  = K.

→

···

Again by the second (diamond) isomorphism theorem we obtain

∼

K i /K i−1 = (Gi N ) N )/(Gi−1 N ) N ) = Gi /(Gi

∩G

N ) i−1 N )

which is a quotient of G of G i /Gi−1 by the third isomorphism theorem since G i−1  Gi quotient groups K groups K i /K i−1 are abelian and therefore K is K is solvable.//

∩G

N . i−1 N

Thus, the

Dummit & Foote Text Exercise 3.5-10: Find a composition series for A for A 4 . Deduce that A 4 is solvable. Solution :

It is quite fortuitous for me as Prof. Okoh had a very similar problem assigned as an exam

problem! Let H Let H = (), (), (1, (1, 2)(3, 2)(3, 4), 4), (1, (1, 3)(2, 3)(2, 4), 4), (1, (1, 4)(2, 4)(2, 3) . We know that

{

}

A4 = (), (), (1, (1, 2, 3), 3), (1, (1, 3, 2), 2), (2, (2, 3, 4), 4), (2, (2, 4, 3), 3), (1, (1, 2, 4), 4), (1, (1, 4, 2), 2), (1, (1, 3, 4), 4), (1, (1, 4, 3), 3), (1, (1, 2)(3, 2)(3, 4), 4), (1, (1, 3)(2, 3)(2, 4), 4), (1, (1, 4)(2, 4)(2, 3) .

{

}

Since H Since H is is a finite and nonempty nonempty subset subset of A of A 4 and closed under under the binary binary operation, it is a subgroup of A A 4 . −1 To show it is a normal subgroup. We need to show that aha H for H for all h all h H and a and a A4 . Since conjugation conjugation preserves preserves shapes (see Example Example 7.2.3 in Beachy Beachy and Blair’s text), aH text), aH a−1 preserve the cycle structure of H of H .. Since H Since H contains contains all of elements of A A 4 that have this cycle structure, aH a−1 = H . H . Hence H Hence H is is normal in A 4 . The definition of solvable is a chain of normal subgroups G = G = N N 0 N 1 N 2 N n = e such that

∈

∈

∈

⊇ ⊇ ⊇ · · · ⊇

{}

(1) N (1) N i  N i−1 for i for i = = 1, 2, 3, . . . n (2) N (2) N i−1 /N i is abelian for i for i = 1, 2, 3, . . . n (3) N (3) N n = e .

{}

The chain to verify is A 4 H e where G where G = = N N 0 , A 4 = N = N 1 , H = H = N 2 and N and N 3 = e . (1) H (1) H  A4 (just proven) and e  H since H since e is the trivial normal subgroups to all subgroups.

⊇ ⊇ ⊇ { } { }

{}

{ }

(2) By Lagrange, the factor group A 4 /H /H will will have three cosets. Since this is a group of prime order, A4 /H = Z/3Z. So it is cyclic and therefore abelian.

∼ H/{e} = {(1), (1), (12)(34), (12)(34), (13)(24), (13)(24), (14)(23)} = H . H . So H So H is is the identity in the factor group and the identity commutes with itself. So H/ H /{e} is abelian. (3) N (3) N = {e}. 3

Therefore A Therefore A 4 is solvable.// Dummit & Foote Text Exercise 3.5-11: Prove that Σ4 has no subgroup isomorphic to Q to Q 8 . Solution :

∼ −

Assume a subgroup H subgroup H Σ4 exists such that H that H = Q 8 . Q8 contains 6 elements of order 4 namely ( i, i,j, j, k, and k, and k). Σ4 also contains exactly 6 elements of order 4: the 4-cycles. So H contains all 4-cycles in Sigma4 . Since H Since H is is a subgroup, it is closed under the binary operation. That implies H implies H contains contains elements of order three (e.g., (1234)(1342) = (143) and (1234)(1423) = (243)). Q8 has no elements of order 8 (by Lagrange). A contradiction. And # H > 8 since it contains six elements of order four, one identity, and more than one element of order three. Another contradiction. Therefore, Σ4 has no subgroup isomorphic to Q to Q 8 .//

≤ ≤

− −


4

Dummit & Foote Text Exercise 7.1-1 Let R be a ring with 1. Show that ( 1)2 = 1 in R.

−

Solution :

( 1)2 + ( 1) = = = =

−

and

−

( 1) + ( 1)2

−

= = = =

−

( 1)( 1) + ( 1) ( 1)(( 1) + 1) ( 1)(0) 0 ( 1)2 = 1.

− − − − − − ⇒− (−1) + (−1)(−1) (1 + (−1))(−1) (0)(−1) 0 ⇒ (−1) = 1. 2

Dummit & Foote Text Exercise 7.1-2: Prove that if u is a unit in R then so is Solution :

−u.

Since u is a unit, there exists some v R such that uv = vu = 1. From Exercise 7.1.1, we know that ( 1)2 = 1. So 1 = uv = u 1 v = u( 1)2 v = u( 1)( 1)v = ( u)( v) and 1 = vu = v 1 u = v( 1)2 u = v( 1)( 1)u = ( v)( u). Hence u is a unit.//

−

· ·

−

∈ · · − − − − − − − − − −

Dummit & Foote Text Exercise 7.1-4: Prove that the intersection of any nonempty collection of subrings of a ring is also a subring. Solution :

Let R be a ring and let X be a nonempty set of subrings of R. We know that X R is a subgroup. To show that X is closed under multiplication, let a, b X . Then a, b S for all S X , and xy S for all S X . Hence xy X , and by definition X R is a subring. Note that the distributive properties did not have to be checked as those are inherited if multiplicative closure holds.//



∈

⊆

∈

∈

∈



∈

  ∈ ⊆



Dummit & Foote Text Exercise 7-1-5: Decide which of the following (a)- (f) are subrings of Q : (a) the set of all rational numbers with odd denominators (when written in lowest terms) (b) the set of all rational numbers with even denominators (when written in lowest terms) (c) the set of nonnegative rational numbers (d) the set of squares of rational numbers (e) the set of all rational numbers with odd numerators (when written in lowest terms) (f ) the set of all rational numbers with even numerators (when written in lowest terms). Solution :

(a) Let X be the given set. Associativtity in X for addition is inherited from Q . +bc X is closed under addition. If ab , dc X , then b and d are odd. Therefore bd is odd. Next ab + dc = adbd and is equal to some other fraction pq in lowest terms, such that q bd. Because bd is odd, q must also be odd. So X is closed under addition. 0 is the additive identity element for X since 01 + ab = ab = ab + 01 for all ab X . 1 For every ab X , −ba X . Because ab + −ba = 0b = 01 and −ba + ab = 0b = 01 , every element of X has an additive inverse. So X is a group under addition. if fracab and dc are written in lowest terms and b and d are odd, then when ac/bd is written in lowest terms, its denominator must divide bd and therefore is odd. Hence X is closed under multiplication and X is a subring.//

∈

∈

|

∈

∈

(b) Let X be the given set. If a is odd, then 2a is even. So closed under addition, X cannot be a subring of Q .

1 2a

∈ X but

1 2a

+

1 2a

=

1

a

∈ X . Since X is not

(c) This set does not contain additive inverses for all of its elements and therefore it is not a group under addition. Hence, it cannot be a subring of Q . (d) This set X , is not closed under addition. ( 13 )2 therefore cannot be in X .

∈ X but (

1 2 3

) + ( 13 )2 =

2 9

∈ X since √ 2 ∈ Q and


5

(e) This set X is not closed under addition. 15 + 15 = 25 X but 15 X . −bc (f ) Let Y be the given set. If ab and dc are in lowest terms and a and c are even, then ab + −dc = adbd . We ad−bc must have b and d be odd because when reduced to lowest terms, the numerator of bd is even since 2 divides a and b. Since Y contains 2 = 21 , Y Q is a subgroup. Now if ab and dc Y and in lowest terms, then b and d are odd. Thus when expressed in lowest terms, 2 must divide ac. Thus Y is closed under multiplication. Now to check for the multiplicative identity. Assume Y has an identity element dc . Then for all ab we have ca = ab , so that cab = dab. Thus c = d. Since c must be even and d odd, we have a contradiction. So Y is a db not ring with a multiplicative identity. Y is a ring but not a unital ring. The definition of subring requires the identity element of the group operation (addition) and Y meets that definition. Y can be considered a subring of Q .

∈

∈

≤

∈

Dummit & Foote Text Exercise 7.1-7: The center of a ring R is z R zr = rz for all r R (i.e., is the set of all elements which commute with every element of R). Prove that the center of a ring is a subring that contains the identity. Prove that the center of a division ring is a field.

{ ∈ |

Solution :

∈ }

Z (R) contains the zero element (additive identity). 0 Z (R) because 0 r = 0 = r 0 for all r R Hence, Z (R) is nonempty. If a, b Z (R) and r R, then (a b)r = ar br = ra rb = r(a b). By the subgroup criterion, Z (R) R. Next abr = arb = rab, so that ab Z (R). Hence by definition of a subring, Z (R) is a subring. If R has a 1 (multiplicative identity), then, 1 a = a 1˙ = a for all a R. Hence 1 Z (R). Next let R be a division ring, and consider its center Z (R). If a Z (R), then by the cancellation law, the inverse of a is unique. Let the inverse of a be a −1 . We know that (r−1 )−1 = r. Since (ab)(b−1 a−1 ) = 1, we have (ab)−1 = b −1 a−1 . Now let r R. a−1 r−1 = (ra)−1 = (ar)−1 = r −1 a−1 . Since r −1 R, a −1 Z (R). Since Z (R) is a commutative division ring, it is a field.

∈

∈

≤

∈

∈

·

∈

− ∈

∈

·

−

−

·

−

∈

∈

∈

∈

Dummit & Foote Text Exercise 7.1-13: An element X in R is called nilpotent if xm = 0 for some m Z+ . (a) Show that if n = a k b for some integers a and b then ab is a nilpotent element of Z /nZ. (b) If a Z is an integer, show that the element a Z/nZ is nilpotent if and only if every prime divisor of n is also a divisor of a. In particular, determine the nilpotent elements of Z /72Z explicitly. (c) Let R be the ring of functions from a nonempty set X to a field F . Prove that R contains no nonzero nilpotent elements.

∈

∈

Solution :

∈

(a) Let n = a k b, where k

≥ 1. Then (ab)

k

= a k bk = (ak b)bk−1 = nbk−1

≡ 0 mod n.

(b) ( ) Assume a Z/(n) is nilpotent. Then a m = nk for some m and k. If p is a prime that divides n, then p divides a m , so that it divides a. Hence every prime that divides n divides a. ( ) Let n = pb11 pbk and a = pc11 pck m, where 1 bi , ci for all i and m is an integer. Let t = max bi . Then a t = ( pc11 pck m)t = pc1 t pck t mt , where c i t bi for each i. Hence a t = nd for some integer d, and we have a t 0 mod n. Since 72 has prime factorization 2 3 32 and since every prime that divides n must divide a the only nilpotent elements in Z /nZ are those a such that a is a multiple of 6 (for any representative choice of a): 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 50, 66 .

⇒

∈

⇐

≡

·· · ·· ·

k

k

i

·· ·

· ·· i

k

≤ ≥

{ }

{

} (c) Proof by contradiction. Assume f ∈ R is nilpotent. If f  = 0, then there exists x ∈ X such that f (x)  = 0. Let m be the least positive integer such that f (x) = 0. Then f (x)f (x) = 0 and f (x)  =0  0. We have our desired contradiction as F contains zero divisors. Therefore no nonzero and f (x) = m

m−1

element of R is nilpotent.//

m−1


6

Dummit & Foote Text Exercise 7.1-14: Let x be a nilpotent element of the commutative ring R (cf. the preceding exercise). (a) Prove that x is either zero or a zero divisor. (b) Prove that rx is nilpotent for all r R. (c) Prove that 1 + x is a unit in R. (d) Deduce that the sum of a nilpotent element and a unit is a unit.

∈

(a) Let n be the least positive integer such that x n = 0. If n = 1, then x = 0. If n > 1, then x = 0, x n−1 = 0 and x xn−1 = 0, then x is a zero divisor by definition of a zero divisor.// (b) Let x be defined as in part (a). Because R is commutative, (rx)n = r n xn = r n 0 = 0.// (c) Let x be defined as in part (a). If x = 0, then clearly 1 = 1 + x is a unit in R. So assume x R and x = 0. We know that 1 = (1 xn ) = (1 + x)(1 x + x2 + ( 1)n−1 xn−1 ). We also know from the closure n−1 2 properties of a ring, (1 x + x + x ) R. Thus 1 + x is a unit in R.// (d) Let x be defined as in part (a). If u is any unit, then u + x = u(1 + u−1 x) is the product of two units (using the results of part (b). Hence (u + x) is a unit.// Solution :





·



·

−

−

· ··

∈

−

∈

· ·· −

Dummit & Foote Text Exercise 7.1-19: Let I be any nonempty index set and let R i be a ring for each i I . Prove that the direct product i∈I Ri is a ring under componentwise addition and multiplication.



∈

Solution :

In Dummit & Foote Text Exercise 5.1-15 we proved that the direct product of groups is a group and it certainly holds for abelian groups. We need to show associativity of multiplication and the left and right distributive properties.

              

                      

If ( ai ), ( bi ), ( ci ) I R i , then ( ai )(( bi )( ci )) = ( ai )( (bi ci )) = so multiplication is associative. Next, ( ( ai )(

ai (bi ci ) =

(ai bi )ci = (

ai bi )(

ci ) = ((

ai )(

ai )(( bi ) + ( ci )) = ( ai )( (bi + ci )) = ai (bi + ci ) = (ai bi + ai ci ) = ( bi ) + ( ai )( ci ). So, multiplication distributes over addition on the left.

(( ai ) + bi )( ( ai )( ci ) + ( Thus

∈

I R i is

bi ))(

ai bi ) + (

ci ),

ai ci ) =

ci ) = ( (ai + bi ))( ci )) = (ai + bi )(ci ) = (ai ci + bi ci ) = ( ai ci ) + ( bi ci ) = bi )( ci ). So, multiplication distributes over addition on the right as well.

a ring under componentwise addition and multiplication.//

Dummit & Foote Text Exercise 7.1-26: Let K be a field. A discrete valuation on K is a function v : K × Z satisfying (i) v (ab) = v(a) + v(b) (i.e., vis a homomorphism from the multiplicative group of nonzero elements of K to Z , (ii) v is surjective, and (ii) v (x + y) min v(x), v(y) for all x, y K × with x + y = 0. The set R = x K × v(x) 0 0 is called the valuation ring of v .

→

≥

{

}

∈



{ ∈

|

≥ }∪{ }

(a) Prove that R is a subring of K which contains the identity. (In general, a ring R is called a discrete valuation ring if there is some field K and some discrete valuation v on K such that R is the valuation ring of v .) (b) Prove that for each nonzero element x K either x or x −1 is in R. (c) Prove that an element x is a unit of R if and only if v (x) = 0.

∈

Solution :

v is an epimorphism (surjective homomorphism) since we are given that v is is surjective and it has the homomorphic property for addition. So if x R, then v (x) = v(1 x) = v(1) + v(x). This implies v(1) = 0 by cancellation. We can use this to show v( x) = v(x) as follows: 0 = v(1) = v(( 1)( 1)) = v( 1) + v( 1) implies v ( 1) = v( 1) and that in turn implies v( 1) = 0. So if x R, then v ( x) = v(( 1)x) = v( 1) + v(x) = v(x). Hence, v (x) = v(x). This fact is necessary for part (a).

− − −

∈

−

−

∈ − −

− −

− − −

·

−

−

(a) By definition, 0 R and that implies R is nonempty. Apply the subgroup criterion. Let x, y consider x y. If x = 0 and y = 0, then v (x y) = v(x) 0. If x = 0 and y = 0, then

−

∈ 

−

≥



∈ R and


7

v(x y) = v( y) = v(y) 0. If x = y = 0, then x y = 0 R. If x, y = 0, then either x x y = 0, and then v (x y) min(v(x), v( y)) = min(v(x), v(y)) 0.

− − 

−

≥ − ≥

−

−

∈

≥



− y = 0 ∈ R or

Now to check for multiplicative closure. If x, y R and xy = 0, then xy R. If xy = 0, then v(xy) = v(x) + v(y) 0, so that xy R. Thus R is a subring of K . We showed earlier that v (1) = 0, so that 1 R.

≥

∈

∈

∈

∈



(b) Assume x K is nonzero. So, 0 = v(1) = v(xx−1 ) = v(x) + v(x−1 ). Hence, v(x) = v(x−1 ), and either v(x) or v (x−1 ) is nonnegative and therefore either x orx−1 R. (c) ( ) If u R is a unit then u −1 R .

∈

From part (b), v (u) =

−1

−1

∈

⇒

−

∈

−1

−v(u ), and both v (u) and v(u ) are nonnegative. Hence v (u) = v(u (⇐) If v (u) = 0, then −0 = 0 = −v(u) = v(u ), so that u ∈ R and u ∈ R is a unit.// −1

∈

) = 0.

−1

Dummit & Foote Text Exercise 7.1.27: A specific example of a discrete valuation ring (cf. the preceding exercise) is obtained when p is a prime, K = Q and v p : Q×

→ Z by v

p

a a c = α where = pα , p c and p d. b b d



|

|

Prove that the corresponding valuation ring R is the ring of all rational numbers whose denominators are relatively prime to p. Describe the units of this valuation ring. Solution :

We are asked to prove that R = a/b

∈ Q | a = 0 or gcd(b, p) = 1}. (R ⊆ {a/b ∈ Q | a = 0 or gcd(b, p) = 1}). Assume that v ( ) ≥ 0, and let a = p c and b = p d, where p does not divide c or d. By definition, v ( ) = m − n ≥ 0. Take the case that is in lowest terms then  0, then m = 0, and v ( ) < 0, a contradiction. Hence n = 0 and gcd(b, p) = 1. either m or n is 0. If n = ({a/b ∈ Q | a = 0 or gcd(b, p) = 1} ⊆ R). If ∈ Q and p does not divide b, then = p ) for some α ≥ 0. Hence v (fracab) ≥ 0, and we have ∈ R.// {

a p b

m

n

a b

a p b

a b

a b

p

a p b

a b

α c d

Dummit & Foote Text Exercise 7.2-2 Let p(x) = a n xn + an−l xn−l + + a1 x + a0 be an element of the polynomial ring R[x]. Prove that p(x) is a zero divisor in R[x] if and only if there is a nonzero b R such that bp(x) = 0. [Let g(x) = b m xm + bm−1 xm−l + + b0 be a nonzero polynomial of minimal degree such that g(x) p(x) = 0. Show that b m an = 0 and so a n g(x) is a polynomial of degree less than m that also gives 0 when multiplied by p(x). Conclude that a n g(x) = 0. Apply a similar argument to show by induction on i that a n−i g(x) = 0 for i = 0, 1, . . . , n and show that this implies b m p(x) = 0.]

···

∈

·· ·

Solution :

⇐) If bp(x) = 0 for some nonzero b ∈ R, then letting b(x) = b states that p(x) is a zero divisor. m

( ) Assume p(x) is a zero divisor. For some q (x) = i=0 bi xi , p(x)q (x) = 0. We may choose q (x) to have minimal degree among the nonzero polynomials with this property.



⇒

Now for the induction argument to show that a i q (x) = 0 for all 0 of the problem.

 

≤ i ≤ n. I obtained help for this part



n+m

k n+m The base case: p(x)q (x) = k=0 in this product is a n bm i+j =k ai bj x = 0. The coefficient of x on the left side and 0 on the right side. Thus a n bm = 0. Now a n q (x) p(x) = 0, and the coefficient of x m in q is a n bm = 0. Thus the degree of a n q (x) is strictly less than that of q (x). Since q (x) has minimal degree among the nonzero polynomials which multiply p(x) to 0, a n q (x) = 0. So, a n bi = 0 for all 0 i m.

≤ ≤

For the inductive step, suppose that for some 0 s < 4, we have a r q (x) = 0 for all s < r n. Now n+m k m+s p(x)q (x) = k=0 is i+j =m+s ai bj on side side of the i+j =k ai bj x = 0. The coefficient of x equation and it is 0 on the other. Thus i+j =m+s ai bj = 0. By the induction hypothesis, if i s, then ai bj = 0. Hence all terms such that i s are zero. If i < s, then we must have j > m, a contradiction. Thus we have a s bm = 0. As in the base case, a s q (x) p(x) = 0 and a s q (x) has degree strictly less than that of q (x), so that by minimality, a s q (x) = 0.

 

≤



By induction, a i q (x) = 0 for all 0

≥





≤ i ≤ n. In particular, a b

i m =

0. Hence b m p(x) = 0.

≤

≥


8

Dummit & Foote Text Exercise 7.2-3: Define the set R[[x]] of formal power series in the indeterminate x with coefficients from R to be all formal infinite sums ∞



an xn = a 0 + a1 x + a2 x2 + a3 x3 +

n=0

·· · .

Define addition and multiplication of power series in the same way as for power series with real or complex coefficients i.e., extend p olynomial addition and multiplication to power series as though they were ”polynomials of infinite degree”: ∞

∞

∞

    ×    an xn +

n=0

bn xn =

n=0

∞

(an + bn )xn

n=0

∞

n

∞

an xn

bn xn =

n=0

n=0

n=0

ak bn−k

k =0



xn .

(The term ”formal” is used here to indicate that convergence is not considered, so that formal power series need not represent functions on R.) (a) Prove that R[[x]] is a commutative ring with 1. (b) Show that 1 x is a unit in R[[x]] with inverse 1 + x + x2 + . n (c) Prove that ∞ n=0 an x is a unit in R[[x]] if and only if a 0 is a unit in R.

−

· ··

n (a) Let r = ∞ n=0 an x , s = Addition is Associative:

Solution :



(r + s) + v

= (( = ( = =

∞ n=0

bn xn , and v =

∞ n=0

an xn ) + (

      

∞ n=0 (an +

∞ n=0 ((an + ∞ n=0 (an +



∞ n=0



∞ n=0

bn )xn ) + (

cn xn .

bn xn )) + ( ∞ n=0





cn xn )

(bn + cn ))xn

∞ n=0

an xn ) + (

= (

∞ n=0

an xn ) + ((

= r + (s + v).

∞ n=0 (bn +

 

∞ n=0

cn )xn )

bn xn ) + (

∞

∞

            n

an x

bn xn

+

n=0

n=0

∞

(an + bn )xn

n=0 ∞

=

(bn + an )xn

n=0

∞

=

bn x

n=0

= s + r



∞ n=0 0

n

·x

= 0 is the additive identity.

∞

n

an xn

+

n=0

∞ n=0



Addition is Commutative:

=

cn xn )

bn ) + cn )xn

= (

r + s =

∞ n=0

cn xn ))


9

∞ n=0

    

r+0 = (

an xn ) + (

∞ n=0 (an +

=

∞ n=0

= (

∞ n=0 0

= (

r + r

  

= (

∞ n=0

n

·x

)+(

∞ n=0

∞ n=0 (

an xn ) + (

∞ n=0 0

=



∈ R[[x]] as r =

∞ n=0 (an

=

n

·x

)

+ an )xn

= 0+r R[[x]] contains left and right inverses. Define r

∞ n=0 0

an xn ) = α

∞ n=0 (0

=

0)xn





an xn )



∞ n=0 (

−a

n n )x .

−a

Then we have

n n )x )

n n )x

−a n

·x

= 0

∞ n=0 0

  

=

∞ n=0

=

= (

n

·x (−a + a )x (−a )x ) + ( n

∞ n=0

n

n

n

n

= r + r.

∞ n=0



an xn )

Hence the left and right inverses coincide. r is the additive inverse for r. I had a lot of help sorting this one out. The definition of the product is identical to that of what electrical engineers know as convolution. Multiplication is Associative:

                   ∞

(rs)v

=

∞

n

an x

n=0

∞

n

n=0

ai b j

n=0

i+j =n

∞

=

n=0

t+k =n

n=0

=

n=0

t+k =n i+j =t

∞

n

x

cn xn

n=0

ai bj

ck

ai bj ck

xn

i+j =t

∞

cn xn

bn x

∞

=

          xn


10

                           ∞

=

xn

ai bj ck

n=0

i+j +k=n

∞

=

xn

ai bj ck

n=0

i+s=n j +k =s

∞

=

ai

n=0

i+s=n

j +k=s

∞

=

∞

n

an x

n=0

xn

b j ck

n=0

∞

=

xn

b j ck

j +k =n

∞

∞

n

n

an x

cn xn

bn x

n=0

n=0

n=0

= r(sv). Distributivity: ∞

r(s + v) =

∞

                       an xn

n=0

n=0

∞

=

∞

bn xn

(bn + cn )xn

an x

n=0

n=0

∞

=

n=0

n=0

xn

ai bj + ai cj

xn

i+j =n

i+j =n

∞

=

ai bj

n=0

+

i+j =n

ai bj

n=0

i+j =n

=

∞

n

x

∞

n

an x

n=0

= rs + rv.

ai cj

∞

n

+

cn xn

an x

n=0

xn

i+j =n

∞

n

bn x

n=0

+

n=0

i+j =n

∞

xn

ai c j

∞

=

                       

ai (bj + cj )

∞

=

cn xn

+

n=0

∞

n

   

n=0


11

∞

(r + s)v

=

∞

n

an x

n

+

cn xn

bn x

n=0

n=0

∞

=

∞

                                                  n=0

∞

n

cn xn

(an + bn )x

n=0

n=0

∞

=

xn

(aj + bj )ci

n=0

i+j =n

∞

=

xn

ai cj + bi cj

n=0

i+j =n

∞

=

ai cj

n=0

+

i+j =n

i+j =n

∞

=

∞

ai cj

n=0

n

x

+

b i cj

n=0

i+j =n

∞

=

xn

bi cj

∞

an xn

n=0

i+j =n

∞

cn xn

∞

bn xn

+

n=0

n=0

= rv + sv. Hence multiplication distributes over addition on the left and right. If R is commutative for multiplication, then so is R[[x]]:

              ∞

rs

=

∞

n

bn xn

an x

n=0

n=0

∞

=

n=0

ai b j

xn

bj ai

xn

i+j =n

∞

=

n=0

j +i=n

∞

=

∞

n

an xn

bn x

n=0

= sr.

n=0

xn

cn xn

n=0


12

         

If R has a multiplicative identity, then so does R[[x]]. Define 1 = That gives us ∞

·

en xn by e 0 = 1 and e i+1 = 0.

∞

an xn

r 1 =

∞ n=0

en xn

n=0

n=0

∞

=

xn

ai e j

n=0

i+j =n

∞

an e0 xn

=

n=0 ∞

an xn

=

n=0

= r ∞

1 r

en xn

=

·

∞

         an xn

n=0

n=0

∞

=

xn

ej ai

n=0

i+j =n

∞

e0 an xn

=

n=0 ∞

an xn

=

n=0

= r Hence R[[x]] has the multiplicative identity 1. (b). Let 1 x = product gives

−

∞ n=0



bn xn where b 0 = 1, b1 =

i

      −        ∞

(1 So if n (c) (

≥ 1, then



i+j =n

− x)r =

∞

bn xn

n=0

∞

xn

=

n=0

n=0

di = 0, and d 0 = 1. Hence (1

x)r = 1.

⇒) Suppose u ∈ R[[x]] is a unit, with inverse u ∞

1 = uu −1 =

−1

∞ n=0

=

∞

an xn

n=0

∞ n=0



−1, and b = 0 for i ≥ 2. Let r = bi

i+j =n

 

xn .

an xn . So, ∞

an xn

=

n=0

xn . Taking the

n=0

ai aj

i+j =n

 

xn .

The coefficient of x 0 in this power series is 1 on the left hand side and a 0 a0 on the right hand side. Thus a0 a0 = 1 R. We also have

∈

       ∞

1 = u

−1

u =

∞

n

an x

n=0

∞

n

an x

n=0

=

ai aj

n=0

i+j =n

xn .


13

Again, the coefficient of x 0 in this power series is 1 on the left hand side and a 0 a0 on the right hand side. Hence a 0 a0 = 1 R and a 0 is a unit in R.

∈

I had help on the following part (could not keep the indices on the sums consistent). n ( ) Suppose a 0 R is a unit, with a 0 a0 = a 0 a0 = 1. Define u = ∞ n=0 bn x with b 0 = a 0 and bk+1 = a0 i+j =k+1,j ≤k ai bj . Then

⇐

 ∈

−

        ∞

uu =

∞

n

an x

bn x

n=0

If n = 0,

 

i+j =n



ai bj = a 0 bn +

=

n=0

ai bj = a 0 a0 = 1. If n

i+j =n

∞

n

ai bj

n=0

xn

i+j =n

≥ 1, then

ai bj = a 0

i+j =n,j
  − a0

ai bj

i+j =n,j
ck+1 =

∞ n=0

+

i+j =n,j
cn xn where c 0 = a 0 and

   

Thus uu = 1. Similarly, we can see that if v =

    

ai cj

i+j =n,j ≤k

ai bj

 

= 0.

 −

( a0 )

then vu = 1. Since u has both a left and a right inverse, they are equal (see part (a)) and u is a unit in R[[x]].// Dummit & Foote Text Exercise 7.2-10: Consider the following elements of the integral group ring ZS 3 : α = 3(12) 5(23) + 14(123) and β = 6(1) + 2(23) 7(132). (where (1) is the identity of S 3 ). Compute the following elements: (a) α + β , (b) 2α 3β , (c) αβ , (d) β α, (e) α 2 .

−

−

−

Solution :

(a) α + β = 6(1) + 3(1 2) 3(2 3) + 14(1 2 3) 7(1 3 2) (b) 2α 3β = 18(1) + 6(1 2) 16(2 3) + 28(1 2 3) + 21(1 3 2) (c) αβ = 108(1) + 81(1 2) 21(1 3) 30(2 3) + 90(1 2 3) (d) β α = 108(1) + 18(1 2) + 63(1 3) 51(2 3) + 84(1 2 3) + 6(1 3 2) (e) α 2 = 34(1) 70(1 2) 28(1 3) + 42(2 3) 15(1 2 3) + 181(1 3 2)

−

− −

−

−

−

−

−

− − −

−

−

Dummit & Foote Text Exercise 7.2-11: Repeat the preceding exercise under the assumption that the coefficients of α and β are in Z /3Z (i.e., α, β Z/3ZS 3 ).

∈

Solution :

Evidently, (a) α + β = 2(1 2 3) + 2(1 3 2) (b) 2α 3β = 2α = 2(2 3) + (1 2 3) (c) αβ = 2(1 2) + 1(1 2 3) (d) β α = 1(1 3) + 2(1 3 2) (e) α 2 = 1(1) + 2(1 2) + 2(1 3) + 1(1 3 2)

−

Dummit & Foote Text Exercise 7.2-12: Let G = g1 , . . . , gn be a finite group. Prove that the element N = g1 + g2 + + gn is in the center of the group ring RG (cf. Exercise 7, Section 1).

{

·· ·

Solution :

}

The center of a ring R is z R r R, z r = rz . That is, the elements of R that commute with every element of r. In the group ring RG, for every g G we have gN = g(g1 + g2 + + gn ) = gg1 + gg 2 + + gg n , but left multiplication by g is a permutation of the element of G, so G = gg 1 , gg2 , . . . , g gn , hence gN = N . Let r 1 g1 + + rn gn RG be arbitrary. Then applying the definition of multiplication in a group product we obtain N (r1 g1 + + rn gn ) = r1 N g1 + + rn N gn .

· ··

···

{ ∈ | ∀ ∈ ·· · }

{

· ··

∈

}

· ··

∈


14

r1 N g1 + + rn N gn = r1 N + + rn N since N g = N . Then r 1 N + + rn N = r 1 g1 N + + rn gn N since gN = N . Then by the distributive property, we have r 1 g1 N + + rn gn N = (r1 g1 + + rn gn )N . So N (r1 g1 + + rn gn ) = (r1 g1 + + rn gn )N implies N is in the center of RG. //

·· ·

·· · ···

· ··

·· · ·· ·

· ··

Prove that the rings Z [x] and Q [x] are not isomorphic.

Dummit & Foote Text Exercise 7.3-2: Solution :

Suppose φ : Q[x] homomorphism,

·· ·

→ Z[x] is a ring homomorphism. Then since φ(1) = 1 by definition of a ring φ(2) = φ(1 + 1) = φ(1) + φ(1) = 1 + 1 = 2.

We also have 1 = φ(1) = φ(2 1/2) = 2 φ(1/2).

·

·

Thus φ(1/2) Z[x]× . By by Proposition 7.2.4(2) Z[x]× = Z× = +1, 1 . Since 2 we have a contradiction. Hence no such homomorphism φ exists.//

∈

{ −}

· −1 = 1 and 2 · 1 = 1,

Dummit & Foote Text Exercise 7.3-4: Find all ring homomorphisms from Z to Z /30Z. In each case describe the kernel and the image. Solution :

A homomorphism φ : Z Z/30Z requires that φ(x)φ(x 1) = x φ(1). We need to determine how many images of 1 are possible since each of them defines the map. We know that φ(x + y) = (x + y)φ(1) = xφ(1) + yφ(1) = φ(x) + φ(y). Also, φ(xy) = xyφ(1) while φ(x)φ(y) = xφ(1)yφ(1) = xyφ(1)2 This implies that we have a homomorphism if and only if φ(1) = φ(1)2 in Z /30Z. The eight elements of Z /30Z which meet this requirement are 0, 1, 6, 10, 15, 16, 21, 25 .

→

·

·

{

Map

Kernel

Image

φ(1)

→ 0

Z

{0}

φ(1)

→ 1

30

1

φ(1)

→ 6

5

6

φ(1)

→ 10

3

10

φ(1)

→ 15

2

15

φ(1)

→ 16 15 16 = 2

φ(1)

→ 21 10 21 = 3

φ(1)

→ 25

6

25 = 5

}


15

Dummit & Foote Text Exercise 7.3-5: Describe all ring homomorphisms from the ring Z Solution :

× Z to Z .

Z Z has two generators (0, 1) and (1, 0). So any homomorphism has to send generators to generators. If φ : mathbbZ Z Z is a ring homomorphism then φ is determined by where it sends the two generators, say φ(1, 0) = a and φ(0, 1) = b. So φ(1, 1) = φ(1, 0) + φ(0, 1) = a + b and φ(1, 1) = φ((1, 1)(1, 1)) = φ(1, 1)φ(1, 1) = (a + b)2 . This implies that a + b = 0 or a + b = 1. Next we have for all (s, t) Z Z, φ(s, t) = φ(s 1, t 1) = sa + yb and φ(s, t) = φ((1, 1)(s, t)) = (a + b)(sa + sb). So φ(s, t) = sa + tb = (a + b)(sa + tb). If a + b = 0, φ(s, t) = 0 (s, t) Z Z. this implies φ = 0. The remaining case is a + b = 1. Since φ is a ring homomorphism, φ((s, t)(u, v)) = φ(s, t)φ(u, v) = (as + bt)(au + bv) and φ((s, t)(u, v)) = φ(su, tv) = asu + btv. This implies asu + btv = a 2 su + (sv + tu)ab + b2 tu for all integers s, t, v and v . If t = v = 0 and s, u = 0, then asu = a 2 su. Hence a 2 = a. Since a is an integer, we have a 0, 1 . That implies (a, b) (1, 0), (0, 1) . So we have three ring homomorphisms (maps): 1. (1, 0) 0 and (0, 1) 0. 2. (1, 0) 1 and (0, 1)0. 3. (1, 0) 0 and (0, 1)1.

×

× →

∈ × ∈ ×



∈{

 → → →

·

·

∈{ }

}

→

Dummit & Foote Text Exercise 7.3-6: Decide which of the following are ring homomorphisms from M 2 (Z) to Z:

  →   →   →

a c a (b) c a (c) c (a)

b d b d b d

a (projection onto the 1,1 entry.) a + d (the trace of the matrix) ad

− bc (the determinant of the matrix).

 

1 1 Solution : (a) If A = , then φ(AA) = φ 1 0 this mapping is not a ring homomorphism.

   

 

1 1 2 1 (b) If A = , then φ(AA) = φ 1 0 1 0 mapping is not a ring homomorphism.

  2 1 1 0

= 2 and φ(A)φ(A) = 1, Since they are not equal,

= 3 and φ(A)φ(A) = 1.Since they are not equal, this

 

1 0 0 0 (c) If A = and B = , then φ(A + B) = 1 and φ(A) + φ(B) = 0. Since they are not equal, 0 0 0 1 this mapping is not a ring homomorphism.

Dummit & Foote Text Exercise 7.3-7: Let R = triangular matrices. Prove that the map φ : R

{  | ∈   → a b 0 d

→ Z × Z defined by

a,b,d

a b 0 d

is a surjective ring homomorphism and describe its kernel.

 

a Solution : Let A, B R with A = 1 0 φ is a ring homomorphism since

∈

 

b1 a and B = 2 d1 0

b2 . d2

Z be the subring of all upper

}

(a, d)


16

φ(A + B) = φ

    a1 0

b1 a + 2 d1 0

b2 d2

= (a1 + a2 , d1 + d2 ) = (a1 + a2 , d1 + d2 ) = φ

    a1 0

b1 d1

a2 0

+φ

b2 d2

= φ(A) + φ(B) and φ(AB) = φ

   a1 0

b1 d1

a2 0

b2 d2

= (a1 a2 , d1 d2 ) = (a1 , d1 )(a2 , d2 ) = φ

  ·   a1 0

b1 d1

φ

a2 0

b2 d2

= φ(A) φ(B). φ is surjective since if (a, d)

2

∈Z

, then φ

 ∈     ∈  

a b If 0 d

ker φ, then φ

0 b 0 0

kerφ =

a b 0 d

b

 · a 0 0 d

= (a, d).

= (a, d) = (0, 0) which implies a = d = 0. Since φ

  0 b 0 0

= (0, 0),

Z .

Dummit & Foote Text Exercise 7.3-8: Decide which of the following are ideals of the ring Z

× Z.

{(a, a) | a ∈ Z} {(2a, 2b) | a, b ∈ Z} {(2a, 0) | a ∈ Z} {(a, −a) | a ∈ Z}. Solution :Let R = Z × Z for all four of these subproblems. (a) Let R = Z × Z and S = {(a, a) | a ∈ Z}. S is not an ideal of R because S does not absorb R. (1, 1) ∈ S but (1, 0)(1, 1) = (1, 0) ∈ / S . (b). We need to verify that S = {(2a, 2b) | a, b ∈ Z} is a subring of R and absorbs R. Let (2a , 2b ), (2a , 2b ) ∈ I and (x, y) ∈ R. (0, 0) ∈ I , so that I is not empty. (2a , 2b ) − (2a , 2b ) = (2(a − a ), 2(b − b )) ∈ S , so that S is an additive subgroup. (2a , 2b )(2a , 2b ) = (4a a , 4b b ) ∈ S shows S is closed under multiplication, so that S is a subring. (2a , 2b )(x, y) = (2a x, 2b y) ∈ S , so that S absorbs R. Since R is a commutative ring, we don’t need to (a) (b) (c) (d)

1

1

1

1

1

1

1

1

2

2

2

2

2

1

2

1 2

1

2

1

2

1 2

1

calculate absorption on the right. Hence S is an ideal of R.

(c). We need to show that S = (2a, 0) a Z is a subring and absorbs R on one side (since R is commutative. . (Absorption on one side is sufficient since R is commutative.) Let (2 a1 , 0), (2a2 , 0) I and (x, y) R. (0, 0) S , so that S is not empty. (2a1 , 0) (2a2 , 0) = (2(a1 a2 ), 0) S , so that S is an additive subgroup. (2a1 , 0)(2a2 , 0) = (4a1 a2 , 0) S shows I is closed under multiplication, so that S is a subring. (2a1 , 0)(x, y) = (2a1 x, 0) S , so that I absorbs R. Hence S is an ideal of R.

{

∈

∈

| ∈ }

∈

−

−

∈

∈

∈ (d) The set S = {(a, −a) | a ∈ Z} is not closed under the multiplication operation: (1, −1) ∈ S but (1, −1)(1, −1) = (1, 1) ∈ / S . So S is not a subring and hence cannot be an ideal.


17

Dummit & Foote Text Exercise 7.3-10: Decide which of the following are ideals of the ring Z [x]: (a) The set of all polynomials whose constant term is a multiple of 3 (b) The set of all polynomials whose coefficient of x 2 is a multiple of 3 (c)The set of all polynomials whose constant term, coefficient of x, and coefficient of x 2 are all zero (d) Z [x2 ] (the set of all polynomials in which only even powers of x appear) (e) The set of all polynomials whose coefficients sum to zero (f ) The set of polynomials p(x) such that p  (0) = 0, where p  (x) is the usual first derivative of p with respect to x. Solution :

(a) Let S be the given set and α = 3a + xp(x), β = 3b + xq (x) S and σ = r + xt(x) Z[x]. We know that 0 S and α β = 3(a b) + x( p(x) q (x)) S . This implies S is a subgroup. S is a subring since αβ = 9ab + x(3bp(x) + 3aq (x) + xp(x)q (x)) S . Lastly, S is closed under multiplication since ασ = 3ar + x(3at(x) + rp(x) + xp(x)t(x)) S . Hence S is an ideal.

∈

−

−

−

∈

∈

(b) This set is not closed under multiplication since x therefore cannnot be an ideal.

∈

∈

∈

2

∈ S but xx = x ∈/ S . Hence S is not a subring and

(c) Let S be the given set and α = x 3 p(x), β = x 3 q (x) S and c(x) Z[x]. We know that 0 S , and α β = x3 ( p(x) q (x)) S . So S is a subgroup. αβ = x6 p(x)q (x) S shows that S is closed under multiplication. Hence t S is a subring. S absorbs Z [x]. since αc(x) = x3 p(x)c(x) S . Hence, S is an ideal.

−

−

∈

∈

∈ ∈

∈

∈

(d) Sinc3 x 2 is in this set, but x 2 x = x3 is not. Since this subset does not absorb Z [x] on the left, it is not an ideal. (e) A polynomial p(x) is in this set S when p(1) = 0. Suppose p, q S and r Z[x]. We know that 0 S . Since ( p q )(1) = p(1) q (1) = 0, implies p q S , S is subgroup. Also, since ( pq )(1) = (1), S is closed under multiplication and hence S is a subring. Since ( pr)(1) = p(1)r(1) = 0 r(1) = 0, S absorbs Z [x]. Hence S is an ideal.

−

−

∈

− ∈

∈

∈

·

(f ) Let S be the given subset and let p(x) = x2 a and q (x) = x. Since p  (x) = 2x, p S . But (qp) = q  p + p q , so that (qp) (0) = q  (0) p(0) + p (0)q (0) = p(0) = a. Since S does not absorb Z[x] on the left, it is not an ideal.

−

∈

−

Dummit & Foote Text Exercise 7.3-11: Let R be the ring of all continuous real valued functions on 1 the closed interval [0, 1]. Prove that the map φ : R R given by φ(f ) = 0 f (x)dx is a homomorphism of additive groups but not a ring homomorphism.



→

Solution :

We do remember from calculus that integration is a linear operation. That is, 1

φ(f + g) =



1

(f + g)(x)dx =

0



1

f (x) + g(x)dx =

0



1

f (x)dx +

0



g(x)dx = φ(f ) + φ(g)

0

. So, φ is an additive group homomorphism. φ does not meet the homomorphic property for multiplication. For example, φ(x3 ) = 1/4 while φ(x)3 = 1/8. // Dummit & Foote Text Exercise 7-3.26: The characteristic char(R) of a ring R is the smallest positive integer n such that 1 + 1 + + 1 = 0 (n times) in R; if no such n exists, then the characteristic of R is said to 0. For example Z /(nZ) is a ring of characteristic n for each positive integer n and Z is a ring of characteristic 0. (a) Prove that the map Z R defined by

· ··

→

 →  − −

k

1+1+ 0 1 1

·· · + 1(k times) − · · · − 1 (k times)

if k > 0 if k = 0 if k < 0

is a ring homomorphism whose kernel is n Z, where n is the characteristic of R (this explains the use of the terminology ”characteristic 0” instead of the archaic phrase ”characteristic for rings in which no sum of 1’s is zero).

∞


18

(b) Determine the characteristics of the rings Q , Z [x], Z /nZ[x]. (c) Prove that if p is a prime and if R is a commutative ring of characteristic p, then (a + b) p = a p + b p for all a, b R.

∈

Solution :

(a). I had help with this part. My approach to the induction argument wasn’t as clean as this one. We will show that φ(a + b) = φ(a) + φ(b) for nonnegative b by induction. For the base case b = 0, we have φ(a + 0) = φ(a) = φ(a) + 0 = φ(a) + φ(0). Now assume that for some b 0, for all a, φ(a + b) = φ(a) + φ(b). If a + b > 0, then φ(a + (b + 1)) = φ((a + b) + 1) = φ(a + b) + 1 = φ(a) + φ(b) + 1 = φ(a) + φ(b + 1). Hence induction hypothesis holds for all positive b. Note the case where a + b < 0 is handled by the next case. Assume b < 0. If a 0, then φ(a + b) = φ(b + a) = φ(b) + φ(a) = φ(a) + φ(b). If a < 0, thena + b < 0, and we have φ(a + b) = φ( a b) = (φ( a) + φ( b)) = φ( a) φ( b) = φ(a) + φ(b). Therefore for all integers a and b, φ(a + b) = φ(a) + φ(b).

≥

≥ − − −

− −

−

− − − −

Again we use induction to show that φ(ab) = φ(a)φ(b) for the non-negative case. For the base case, φ(a0) = φ(0) = 0 = φ(a) 0 = φ(a)φ(0). Now assume that the result holds for all a, for some b 0. Then φ(a(b + 1)) = φ(ab + a) = φ(ab) + φ(a) = φ(a)φ(b) + φ(a) = φ(a)(φ(b) + 1) = φ(a)φ(b + 1). Hence since the induction hypothesis, for all a and for all non-negative b. For the negative case, assume b < 0. Then φ(ab) = φ(( a)( b)) = φ( a)φ( b) = ( φ(a))( φ(b)) = φ(a)φ(b). Thus φ is a ring homomorphism.

·

− −

≥

−

−

−

−

Let n = char R. (ker φ nZ). If a ker φ then φ(a) = 0. Assume that n does not divide a. Then by the division algorithm, a = nq + r for some r with 0 < r < n. We are given φ(n) = 0 by definition and that n is minimal with this property. Since 0 = φ(a) = φ(qn + r) = φ(r), we have a contradiction in that n is not miminal. Thus n divides a, and a nZ.

⊆

∈

∈

(nZ kerφ). If a nZ, then a = nb for some integer b. Since φ(n) = 0 , φ(a) = φ(nb) = φ(n)φ(b) = 0 φ(b) = 0. This implies a kerφ.//

⊆ ∈

·

∈

(b) Q has no non-identity element whose additive order is finite. So its characteristic is zero. If we select the inclusion map, it satisfies the requirements defined in part (a). Let the inclusion map. Then, φ(0) = 0 = ι(0), for a 0, φ(a + 1) = φ(a) + 1 = ι(a) + 1 = a + 1 = ι(a + 1), and fora < 0, φ(a) = φ( a) ι( a) = ( a) = a = ι(a). Since the inclusion map is injective, it’s kernel is trivial. Hence charQ = 0.

≥ − − − −

−−

For Z [x], as with Q , let φ = ι and the result follows: charZ[x] = 0. For Z /nZ[x], φ(n) = n = 0. This implies that n Z kerφ. Next, let a kerφ andassume that n a. By the division algorithm, a = bn + r where 0 < r < n. Since φ is a homomorphism, 0 = φ(bn + r) = φ(bn) + φ(r) = φ(b)φ(n) + φ(r) = φ(r). This is a contradiction of the minimality of n. Therefore r = 0. This implies n a and kerφ nZ. Therefore charZ/nZ = n.

⊆

|

∈

|

⊆

(c) Let R be a commutative ring with characteristic p, and let a, b R. If p is a prime and 0 < k < p, then k! does not divide p nor does ( p k)! divide p!. But by definition of the factorial, p divides p!. So p divides p = kp!! (n k)! for 0 < k < p. Applying the binomial theorem and the fact that R has characterstic p k p gives us (a + b) p = k=0 pk ak b p−k = a p + b p .



−

−

∈

 

Dummit & Foote Text Exercise 7.3-28: Prove that an integral domain has characteristic p, where p is either a prime or 0 (cf. Exercise 26) . Solution :

Proof by contradiction. Assume that R has a characteristic n and n is composite. This implies n = pq for some integers p, q . Let φ be the homomorphism from Exercise 7.3-26 which takes an integer k Z to a sum (k-times) of 1 or -1. This implies φ(a) and φ(b) be non-zero. But φ(a)φ(b) = φ(n) = 0 and that implies φ(a) and φ(b) are zero divisors. This contradicts our assumption that R is an integral domain. Hence, the characteristic of R is not composite. Therefore it is prime or it is zero.//

∈


19

Dummit & Foote Text Exercise 7.3-29: Let R be a commutative ring. Recall (cf. Exercise 13, Section 1) that an element x R is called nilpotent if x n = 0 for some n Z+ . Prove that the set of nilpotent elements form an ideal – called the nilradical of R denoted by N(R) . [Hint: Use the Binomial Theorem to show N(R) is closed under addition.]

∈

∈

If x, y N(R), then for some positive integers n and m, x n = y m = 0. Since we are given the hint to use the Binomial Theorem, consider (x + y)n+m . The Binomial Theorem tells us Solution :

∈

n+m n+m

(x + y)

=

  k=0

n + m k n+m−k x y . k

If k n, x k = 0. If k < n, then y n+m−k = 0. Hence (x + y)n+m = 0 and x + y N(R). So this set is closed. The set has inverses for its elements since we have ( x)n = ( 1)n xn = 0, so that x N(R). The set has an identity since 0 = 01 N(R), N(R) is an additive subgroup of R.

≥

−

∈

∈

−

− ∈

Since R is commutative, if r R then (rx)n = r n xn = 0 and (xr)n = xn rn = 0. So, N(R) absorbs R on the left and the right. Hence N(R) an ideal.//

∈

Dummit & Foote Text Exercise 7.3-30: Prove that if R is a commutative ring and N(R) is its nilradical (cf. the preceding exercise), then zero is the only nilpotent element of R/ N(R) i.e., prove that N(R/N(R)) = 0. Solution :

From Exercise 7.3-29, we know that N(R) is an ideal of R. If x + N(R) N(R/N(R)), then for some positive integer n, (x + N(R))n = x n + N(R) = N(R). Hence x n N(R). Then for some positive integer m, (xn )m = x nm = 0. Hence x N(R). This implies that x + N(R) = N(R). That is x = 0. In other words, the additive subgroup N(R/N(R)) contains only one element, the additive identity. //

∈

∈

∈

Dummit & Foote Text Exercise 7.3-34: Let I , J be ideals of R. (a) Prove that I + J is the ideal of R containing both I and J . (b) Prove that I J is an ideal contained in I J . (c) Give an example where I J = I J . (d) Prove that if R is commutative and if I + J = R, then I J = I J .

∩

 ∩

∩ Solution : I had help with this problem. (a) Let a + b , a + b ∈ I + J for some a ∈ I and b ∈ J . Then (a + b ) − (a + b ) = (a − a ) + (b − b ) ∈ I + J since I and J are closed under subtraction. Let r ∈ R. Then, r(a + b) = ra + rb ∈ I + J since I and J absorb R on the right. (a + b)r ∈ I + J since I and J absorb R on the left. Thus I + J is an ideal of R. Now, let K be an ideal of R that contains I and J . That is, I , J ⊆ K . If a + b ∈ I + J , then a + b ∈ K since K is closed under addition. So I + J ⊆ K . (b) Let x, y ∈ IJ , with x = a b and y = c d , where a , c ∈ I and b , d ∈ J . We have x + y ∈ IJ . Let r ∈ R. Since I is an ideal of R, rx = (ra )b ∈ IJ . Similarly, xr ∈ IJ . Hence I J is an ideal of R. Again let x = a b . Since a ∈ I and I is an ideal, a b ∈ I , and thus x ∈ I . Similarly, x ∈ J . Thus IJ ⊆ I ∩ J .  2 Z ∩ 4Z = 4 Z. (c) If R = Z, I = 2 Z and J = 4Z, then (2Z)(4Z) = 8Z = (d) Assume I + J = R and R is commutative. Part (c) of this exercise tells us that I J ⊆ I ∩ J . We have (I ∩ J )(I + J ) = (I ∩ J )R = I ∩ J . If a ∈ I ∩ J , d = c(a + b) for some a ∈ I and b ∈ J and c ∈ I ∩ J . Then d = ca + cb = ac + cb ∈ IJ . Hence I J = I ∩ J .// 1

1

1

2



2

1

2



i i

i

1

i i

1

2

i

2

i

2



i i

i

i

i

i

i

i

i i

Dummit & Foote Text Exercise 7.4-4: Assume R is commutative. Prove that R is a field if and only if 0 is a maximal ideal. Solution :

( ) Assume R is a field and let I be an ideal of R which properly contains zero. This implies that there exists an element x = 0 in I . Then a is a unit since R is field. Proposition 9 tells us that I = R. Hence, R is the only ideal of R that properly contains zero. This implies that 0 is a maximal ideal in R.

⇒




20

( ) Assume that zero is a maximal ideal in R. Let x R and x = 0. Since zero is maximal and since R is given to by commutative, (a) = R. Note that R contains the multiplicative identity, 1. Therefore, exists elements y and Z inR such that xy = za = 1. This implies that x has a two-sided inverse. Therefore, every nonzero element in R is invertible and that implies R is a field. //

⇐

∈



Dummit & Foote Text Exercise 7.4-7: Let R be a commutative ring with 1. Prove that the principal ideal generated by x in the polynomial ring R[x] is a prime ideal if and only if R is an integral domain. Prove that (x) is a maximal ideal if and only if R is a field. Solution :

If R is a commutative ring then φ : R[x] R defined by φ( p(x)) = p(0) is an surjective ring homomorphism and ker φ = (x). It is clear that φ is surjective. Let p(x) = ai xi and q (x) = bi xi . We then have φ( p + q ) = φ (ai + bi )xi = a 0 + b0 = φ( p) + φ(q )

→

and φ( pq ) = φ

       ai bj )xk

(

k





= a 0 b0 = φ( p)φ(q )

i+j =k

which show that φ is a ring homomorphism. To show that kerφ = (x), assume that p kerφ. We can write p(x) = a 0 + xq (x), and p(0) = a 0 = 0. Thus p(x) = xq (x) (x). So ker φ (x). Conversely, if p(x) = xq (x), then p(0) = 0 q (x) = 0, and φ( p) = 0. (x) ker. Hence ker φ = (x). By the First Isomorphism Theorem for rings, we have R[x]/ker φ = R[x]/(x) = R. Proposition 13 tells us that if R[x] is commutative, then the ideal (x) is a prime ideal in R[x] if and only if the quotient ring R[x]/(x) is an integral domain. Since R[x]/(x) = R, R is an integral domain if and only if R[x]/(x) = R is an integral domain. Proposition 12 tells us that if R[x] is commutative, then the ideal (x) is maximal if and only if the quotient ring R[x]/(x) is a field. We have shown by the first isomorphism theorem that R[x]/(x) = R. So R is a field if and only if R[x]/(x) is a field.//

∈

⊆

∈

⊆

· ∼

∼

∼

∼

Dummit & Foote Text Exercise 7.4-8: Let R be an integral domain. Prove that (a) = (b) for some elements a, b R, if and only if a = ub for some unit uofR.

∈ Solution : (⇒) If (a) = (b), then a ∈ (b). This implies that a = ub for some element u ∈ R. We also have b ∈ (a) and therefore b = va for some element v ∈ R. So a = (uv)a and this implies va = (vu)va. Since since R is an integral domain we have uv = vu = 1. Hence u is a unit.

( ) Let a = ub and u R be a unit. Then by Proposition 9 in the text we obtain (a) = aR = ubR = buR = bR = (b).//

⇐

∈

Dummit & Foote Text Exercise 7.4-9: Let R be the ring of all continuous functions on [0, 1] and let I be the collection of functions f (x) in R with f (l/3) = f (l/2) = 0. Prove that I is an ideal of R but is not a prime ideal. Solution :

I is an ideal. If f , g I , then (f g)(1/2) = f (1/2) g(1/2) = 0 0 = 0 and (f g)(1/3) = f (1/3) g(1/3) = 0 0 = 0. So, f g I . Since 0 I , I is an additive subgroup of R by the subgroup criterion. If h R, then (f h)(1/2) = f (1/2)h(1/2) = 0 h(1/2) = 0 and (f h)(1/2) = (hf )(1/2) = (f h)(1/3) = (hf )(1/3) = 0. So we have shown that I is an ideal of R. I is not a prime ideal. If f (x) = x2 1/4 and g(x) = x2 1/9, (f g)(1/2) = (f g)(1/3) = 0 then f g I . However, f / I and g / I since f (1/3) = 0 and g (1/2) = 0. //

−

−

∈

∈

∈

∈

−

−

−



−

− ∈ 

−

∈

−

·

∈

Dummit & Foote Text Exercise 7.4-10: Assume R is commutative. Prove that if P is a prime ideal of R and P contains no zero divisors then R is an integral domain.

MAT7400 Assignment #3 Solution :

21

Let x, y R such that xy = 0. Since P is a prime ideal and xy P , x P . The following equally valid if y is chosen instead of x. If x = 0 then since P contains no zero divisors in R, y = 0, //

∈

∈



∈

Dummit & Foote Text Exercise 7.4-15: Let x 2 + x + 1 be an element of the polynomial ring E = F2 [x] and use the bar notation to denote passage to the quotient ring F 2 [x]/(x2 + x + 1). (a) Prove that E has 4 elements: 0, 1, x, and x + 1. (b) Write out the 4 4 addition table for E and deduce that the additive group E is isomorphic to the Klein 4-group. × (c) Write out the 4 4 multiplication table for E and prove that E is isomorphic to the cyclic group of order 3. Deduce that E is a field.

× ×

Solution : Note: This

problem is identical to Example 4.3.4 in Beachy & Blair’s text, pages 208-209. (a) The congruence classes modulo x 2 + x + 1 are polynomials that that have degree less than 2 over F 2 . The congruence classes are 0, 1, x, and 1 + x. (b) + 0 1 x 1+x 0

0

1

x

1+x

1

1

0

1+x

x

x

x

1+x

0

1

1+x

1+x

x

1

0

Since the group has order four and each element except the identity has order 2, this group is isomorphic to the Klein four group. (c) 0 1 x 1+x

×

×

0

0

0

0

0

1

0

1

x 1+x

x

0 1+x

0

1

1+x

0 1+x

1

x

×

E = 1, x, 1 + x . Since the order of E is three, it’s isomorphic to a cyclic group of order three. (d) x 2 + x + 1 is irreducible over the field F2 since it has no roots in F 2 . Therefore F2 [x]/(x2 + x + 1) is a field.//

{

}

Dummit & Foote Text Exercise 7.3-16: Let f (x) = x 4 16 be an element of E = Z[x] and use the bar notation to to denote passage to the quotient ring row Z [x]/(x4 16). (a) Find a polynomial of degree 3 which is congruent to g(x) = 7x13 11x9 + 5x5 2x3 + 3 modulo x 4 16. (b) Prove that x + 2 and x 2 are zero divisors in E .

−

−

Solution :

−

−

−

≤

−

(a). Using the division algorithm (and Matlab), we obtain the following: g(x) = (7 x9 + 101 x5 + 1621 x) (x4

3

∗ ∗ ∗ · − 16) + (−2 ∗ x + 25936 ∗ x + 3) and thus, −2 ∗ x + 25936 ∗ x + 3 is the desired polynomial. (b)x + 2 and x − 2 are zero divisors since they are factors of x − 16. That is, x − 16 = (x − 2)(x + 2)(x + 4) implies 0 ≡ (x − 2)(x + 2)(x + 4) mod x − 16. 3

4

4

2

2

4


22

Dummit & Foote Text Exercise 7.4-18: Prove that if R is an integral domain and R[[x]] is the ring of formal power series in the indeterminate x then the principal ideal generated by x is a prime ideal (cf. Exercise 3, Section 2). Prove that the principal ideal generated by x is a maximal ideal if and only if R is a field. Solution We

approach this problem in the same manner as Exercise 7.4-7. If R is a commutative ring then φ : R[[x]] R defined by ri xi = r 0 is an surjective ring homomorphism and ker φ = (x). It is clear that φ is surjective. Let p(x) = ai xi and q (x) = then have φ( p + q ) = φ (ai + bi )xi = a 0 + b0 = φ( p) + φ(q )



→

and φ( pq ) = φ

       ai bj )xk

(

k

→





bi xi . We

= a 0 b0 = φ( p)φ(q )

i+j =k

which show that φ is a ring homomorphism. Not concering ourselves regarding convergence makes me a bit worried that the above is not entirely true. Now to show kerφ = (x). (kerφ (x)) Assume ri xi ker φ. Then r 0 = 0 (using the definition that 00 = 1), and ri xi = x ri+1 xi (x). ((x) kerφ) If y = x ai xi , then y (x) and that implies φ(y) = 0, so that y ker φ. By the First Isomorphism Theorem for rings, we have R[[x]]/ker φ = R[[x]]/(x) = R. Proposition 13 tells us that if R[[x]] is commutative, then the ideal (x) is a prime ideal in R[[x][ if and only if the quotient ring R[[x]]/(x) is an integral domain. Since R[[x]]/(x) = R, R is an integral domain if and only if R[[x]]/(x) = R is an integral domain. Proposition 12 tells us that if R[[x]] is commutative, then the ideal (x) is maximal if and only if the quotient ring R[[x]]/(x) is a field. We have shown by the first isomorphism theorem that R[[x]]/(x) = R. So R is a field if and only if R][x]]/(x) is a field.//



∈



⊆



∈

∈

⊆



∈

∼

∼

∼

∼

Dummit & Foote Text Exercise 7.4-19: Let R be a finite commutative ring with identity. Prove that every prime ideal of R is a maximal ideal. Solution :

If P is a prime ideal of a ring R that is commutative and contains 1, then R/P is an integral domain. Suppose that P is prime, and let a + P and b + P be elements of R/P . Suppose that their product ab + P = 0 + P , the zero element of R/P . This implies ab0 P or equivalently ab P . Since P is prime, either a P or b P . If a P , then a + P = 0 + P and if b P , then b + P = 0 + P . Either way, the only way a product equals zero is when one of the factors equals zero, which along with R/P is commutative implies R/P is an integral domain. If R is a finite integral domain, then R is a field. Prof. Robert Ash of UIUC had a very simple proof this. If a R and a = 0,the map x ax, x R,is injective because R is an integral domain (no zero divisors). If R is finite,the map is surjective as well, so that ax = 1 for some x. Hence R is a field. Since R/P is a finite integral domain, it is a field and therefore P R is maximal.//

∈

∈



∈

∈ ∈

∈

→

∈

∈

⊆

Dummit & Foote Text Exercise 7.4-26: Prove that a prime ideal in a commutative ring R contains every nilpotent element (cf. Exercise 13, Section 1). Deduce that the nilradical of R (cf. Exercise 29, Section 3) is contained in the intersection of all the prime ideals of R. (It is shown in Section 15.2 that the nilradical of R is equal to the intersection of all prime ideals of R.) Let P R be a prime ideal and let r R be nilpotent with r n = 0. Let 1 m n be minimal such that r m P . R/P is an integral domain and x m + P = 0 R/P . If m 2, we have (r + P )(rm−1 + P ) = 0, so that r + P is a zero divisor in R/P , a contradiction since R/P is an integral domain. Therefore m = 1, and r P . Hence N(R) P . If P(R) is the collection of all prime ideals of R, we have N (R) P(R). Solution :

∈

⊆

⊆

∈

∈

∈

⊆

≥

≤ ≤


23

Dummit & Foote Text Exercise 7.4-28: Prove that if R is a commutative ring and N = (a1 , a2 , . . . , an ) where each a i is a nilpotent element, then N is a nilpotent ideal (cf. Exercise 37, Section 3). Deduce that if the nilradical of R is finitely generated then it is a nilpotent ideal. Solution :

First we complete Exercise 7.4-12 as we will need the result from it. Assume R is commutative and suppose I = (a1 , a2 , . . . , an ) and J = (b1 , b2 . . . bm ) are two finitely generated ideals in R. Prove that the product ideal I J is finitely generated by the elements a i bj for i = 1, 2, . . . , n and j = 1.2, . . . , m. Proof: Let H be the set containing the elements a i bj for i = 1, 2, . . . , n and j = 1.2, . . . , m. (IJ H ) Let x = i ri si IJ , where r i = j t i,j aj and s i = k ui,k bk . Then since R is commutative, x = i ( j t i,j aj )( k ui,k bk ) = i j k ti,j ui,k aj bk (H ). (IJ H ) Let x = ri,j ai bj (H ). Since I is an ideal, x IJ . Back to the given problem. Assume a m = 0 for all a i A for some positive integer m i . M = mi . From i Exercise 7.4-12 (which we just completed), we have N M = X where X = aki ki = M . So for each k M element ai X , some k i is at least m i and therefore X = 0 and that implies N = 0. Thus N is a nilpotent ideal. If N(R) is finitely generated, then each generator is (by definition) nilpotent. Thus N(R) is a nilpotent ideal.//

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⊆



⊇



i

 

∈

∈

∈

∈

i

∈



{

∈



 |

}

i

Dummit & Foote Text Exercise 7.4-30: Let I be an ideal of the commutative ring R and define n

radI = r

+

{ ∈ R | r ∈ I for some n ∈ Z }

called the radical of I . Prove that radI is an ideal containing I and that (radI )/I = N(R/I ) (cf. Exercise 29, Section 3). For all a I , a 1 I , so I binomial theorem tells us that Solution :

∈

⊆ radI . Hence radI is nonempty. Let a, b ∈ radI with a

∈

m+n m+n

(a + b)

=

n

, bm

∈ I . The

 

m + n k m+n−k a b . k

k=0

Since I is an ideal, if k n, then a k I , and if k < n, then b m+n−k I . Since R is commutative, this implies every term of (a + b)n+m I , and consequently (a + b)m+n I since I is a group under addition. Next, for all r R, (ra)n = r n an I , and ( a)n R. Hence radI is an ideal of R. //

≥

∈ ∈ ∈ ∈ ∈ ∈ − ∈ x + I ∈ radI/I ⇔ x ∈ radI ⇔ x ∈ I for some n ≥ 1 ⇔ x + I = 0 ∈ R/I for some n ≥ 1 ⇔ (x + I ) R/I for some n ≥ 1 ⇔ x + I ∈ N(R/I ). Hence radI/I = N(R/I ) . // n

n

n

=0

∈

Dummit & Foote Text Exercise 7.4-31: An ideal I of the commutative ring R is called a radical ideal if rad I = I . (a) Prove that every prime ideal of R is a radical ideal. (b) Let n > 1 be an integer. Prove that 0 is a radical ideal in Z/nZ is radical if and only if n is a product of distince primes to the first power (i.e., n is squarefree). Deduce that (n) is a radical ideal in Z if and only if n the product of distinct primes in Z. (a) Let P R be a prime ideal. Assume x radP . Then for some n 1, x n P . Assume that n is minimal with this property. If n 2, then xx n−1 P . Since P is prime, then either x P or xn−1 P which contradicts our assumption that n is minimal. Therefore n = 1 and x P . (b) ( ) Let the prime factorization of n be n = pki . Assume that 0 is a radical ideal in Z/nZ, and assume some k i 2. Then pi is nonzero in Z /nZ, and ( pi )max k = 0, a contradiction. This implies n is squarefree. ( ) Assume n is squarefree and let a Z/nZ with a k = 0 for some positive integer k. If some prime p divides n but not a, then no power of a can be divisible by n, a contradiction since a k 0 mod n and p divides n. Hence n divides a, and rad 0 = 0. Solution :

∈ ⇒

⇐

⊆

≥

∈ ∈

≥





∈

≥

∈

∈

i



i

≡

∈


24

I had help with this part of the problem. Let R be a commutative ring , let I R be an ideal, and let J R be an ideal containing I . Then J/I is radical ideal in R/I if and only if J is a radical ideal in R. Proof: ( Assume J/I is a radical ideal in R/I , and let x rad J . Then x m J for some m 1. Next (x + I )m = x m + I J/I , so x + I rad J/I = J/I . Thus x J , and J is a radical ideal in R. ( ) Assume J is radical in R and let x + I rad J/I . Then (x + I )m = xm + I J/I for some m 1, so that x m J . Thus x rad J = J , and x + I J/I Hence J/I is a radical ideal. If (n) is an ideal of Z , then (n) is a radical ideal in Z if and only if 0 is a radical ideal in Z/(n) if and only if n is square free. //

⊆

⇒

⇐

∈

∈

∈

∈

∈ ∈

∈

⊆ ∈

∈

≥

∈

≥

Dummit & Foote Text Exercise 7.4-33: Let R be the ring of all continuous functions from the closed interval [0, 1] to R and for each c [0, 1] let M c = f R f (c) = 0 (recall that M c was shown to be a maximal ideal of R). (a) Prove that if M is any maximal ideal of R then there is a real number c [0, 1] such that M = M c . (b) Prove that if b and c are distinct points in [0, 1] then M b = M c . (c) Prove that M c is not equal to the principal ideal generated by x c. (d) Prove that M c is not a finitely generated ideal.

∈

{ ∈ |

}

∈



−

Solution : I

had a lot of help on this problem. It was good to apply material from MAT5600. (a) Proof by contradiction. Let M R be a maximal ideal and assume M = M c c [0, 1]. Then for all such c, there exists a function f c M such that f c (c) = 0. Assume that f c (c) > 0. Since f c is continuous, there exists a positive real number ε c > 0 such that f c [(c εc , c + εc )] = 0. The set (c εc , c + εc ) [0, 1] c [0, 1] covers[0, 1]. Since [0, 1] is compact (closed and bounded subset of R , this cover has a finite subcover by the definition of compact. Let K [0, 1] be the finite subcover K is finite and (c εc , c + εc ) [0, 1] c K covers [0, 1]. For each c K , define u c (x) = 1 + (x c)/εc if x (c εc , c) [0, 1], u c (x) = 1 + (c x)/εc if x [c, c + εc ) [0, 1], and u c (x) = 0 otherwise. Then u c R and u c is zero outside of (c εc , c + εc ). Next consider g = uc f c . Since f c M , we have g M . However, for all x [0, 1], g(x) is positive since each u c (x)f c (x) is nonnegative and some u c (x)f c (x) is positive. Thus g(x) > 0 for all x, and therefore 1/g R. This implies that M contains a unit, a contradiction. Thus M = M c for some c [0, 1]. (b) Assume b = c. Then x b M b but (x b)(c) = c b = 0. This implies x b / M c . Therefore, M b = M c . x c (c) Assume M c = (x c). Then x c = f (x)(x c) for some f (x) R. If x = c, then f (x) = . As x c x approaches c from the right, f (x) approaches , while f(x) approaches as x approaches c from the left. So, no extension of f is in R. Therefore, M c is not generated by x c. (d) Proof by contradiction. Assume M c = (A) is finitely generated, with A = ai (x) 1 i n . Let f = ai . Then f is continuous on [0, 1] and f M c . Thus f = ri ai for some continuous functions r i R. If we let r = ri , we have f (x) = ri (x)ai (x) ri (x) ai (x) r(x)f (x). For each b = c, there must exist a function a i such that a i (b) = 0, otherwise h(b) = 0 for all h M c and x c M c . Thus c is the only zero of f . From f (x) r(x)f (x), for x = c we have r(x) 1/ f (x). As x approaches c, f (x) approaches 0, so that 1/ f (x) is unbounded and that implies r(c) does not exist which is a contradiction since r(x) R. Thus M c is not finitely generated.

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{ − { − ∈ −





− ∈

−

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∈

∈ ∩

⊆

∈

− 

∈

−

√ ∈ √ √ ≤ 

| |



√

∞

 ≤ | 



∈

∈

∈

− ∈

− −∞

√

−

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∈

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∀ ∈

−

∈

| − |

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| − | −

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{

| ≤ ≤ } || | ≤ ∈ ≥

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Dummit & Foote Text Exercise 7.4-37: A commutative ring R is called a local ring if it has a unique maximal ideal. Prove that if R is a local ring with maximal ideal M then every element of R M is a unit. Prove conversely that if R is a commutative ring with 1 in which the set of nonunits forms an ideal M , then R is a local ring with unique maximal ideal M .

−

Solution :

Let x R M . If the ideal (x) is proper then it must be contained in some maximal ideal, and M is the only maximal ideal. Thus (x) M , a contradiction since x / M . Thus (x) = R, and for some y R we have xy = yx = 1. This implies x is a unit. Assume the set M of nonunits in R form an ideal. To show that M is maximal, assume M  I for some ideal I . Then I contains a unit, so that I = R. To show that M is the unique maximal ideal, assume there

∈

∈ −

⊆

∈


25

is an ideal N R such that N M . Then N contains some element x not in M , which is a unit. Thus N = R. In particular, every proper ideal of R is contained in M . Thus M is the unique proper ideal of R, and R is a local ring.

⊆

⊆

Dummit & Foote Text Exercise 7.4-38: Prove that the ring of all rational numbers whose denominators is odd is a local ring whose unique maximal ideal is the principal ideal generated by 2. R is a commutative ring with 1 = 0. Note that 1 = 11 . Let M R be the set of nonunits. Since R Q, every element a/b R has an inverse in Q , b/a. If a/b is invertible in R then its inverse is b/a. However, a/b is not invertible in R when a is even (including zero). Thus M = 2a/b a, b Z, b = 0 . So, M (2). Since 2 itself is a nonunit in R, M = (2), that is, M is an ideal. By the previous exercise (7.4-37), R is local with maximal ideal M = (2). Solution :

⊆ ⊆



∈

⊆

{

|

∈

 }

Dummit & Foote Text Exercise 7.4-39: Following the notation of Exercise 26 in Section 1, let K be a field, let v be a discrete valuation on K and let R be the valuation ring of v . For each integer k 0 define Ak = r R v(r) k 0 . (a) Prove that A k is a principal ideal and that A 0 A1 A2 . (b) Prove that if I is any nonzero ideal of R, then I = Ak for some k 0. Deduce that R is a local ring with unique maximal ideal A 1 .

{ ∈ |

≥

≥ }∪{ }

⊇ ⊇ ⊇ · · · ≥ Solution : (a) It is clear that A ⊆ A for all k. Claim: A is an ideal. Since 0 ∈ A , A is nonempty. Let a, b ∈ A . If one or both of a and b is 0, then a + b ∈ A . If a + b = 0, then a + b ∈ A . If a + b  = 0, we have v (a + b) ≥ min(v(a), v(b)) ≥ k, so that a + b ∈ A . Next, v (−a) = v(a) and that implies −a ∈ A . If r ∈ R then v (r) ≥ 0 and v (ra) = v(r) + v(a) ≥ k, so that ra ∈ A . Since K is commutative, A is an ideal k +1

k

k

k

k

k

k

k

k

k

k

k

of R. To show that A k is principal, choose x Ak such that v (x) = k. We can choose one because that element exists since v is surjective. Note that x −1 exists in K , and that v(x−1 ) = v(x). Let r Ak . Then v(r) k and v (x−1 r) = v(x−1 ) + v(r) = v(r) v(x) 0. That implies x −1 r R and r = xx−1 r. Thus r (x), and we have A k (x), and thus A k = (x). In particular, A k is generated by any element of valuation k. (b) Let I Ak be a nonzero ideal of R. Let k be minimal among v (r) for r I . Let a I such that v(a) = k. In particular, we have I Ak . Moreover, since A k = (a), we have A k I . Hence I = A k . From part (a), every proper ideal of R is contained in A 1 and therefore A 1 is the unique maximal ideal of R.

∈

−

⊆ ⊆

≥

∈

−

∈

∈

⊆

⊆

∈

≥

∈

Dummit & Foote Text Exercise 7.4-40: Assume R is commutative. Prove that the following are equivalent: (see also Exercises 13 and 14 in Section 1) (i) R has exactly one prime ideal (ii) every element of R is either nilpotent or a unit (iii) R/(R) is a field (cf. Exercise 29, Section 3). Solution :

(i) (ii) Assume R has exactly one prime ideal. Since every maximal ideal of R is prime, R is local. By Exercise 7.4-26, N(R) is the unique maximal ideal. By Exercise 7.4-37, every element of R N(R) is a unit, and (by definition) the remaining elements are nilpotent. (ii) (iii) Assume x + N(R) is nonzero. Then x / N(R). Since x is not nilpotent in R, x is a unit. Then x−1 exists in R, and we have (x + N(R))(x−1 + N(R)) = 1. Thus every nonzero element of R/ N(R) is a unit. Since R is commutative, R/ N(R) is a field. (iii) (i) Assume R/ N(R) is a field. By Exercise 7.4-26, N(R) is contained in every prime ideal of R. By the Lattice Isomorphism Theorem for rings, the only possible proper prime ideal is N(R). Also, N(R) is prime since it is maximal in R. Thus N(R) is the unique prime ideal of R.

− ⇒

⇒

∈

⇒

Dummit & Foote Text Exercise 7.5-3: Let F be a field. Prove that F contains a unique smallest subfield F 0 and that F 0 is isomorphic to either Q or Z /pZ for some prime p (F 0 is called the prime subfield of F ). [See Exercise 26, Section 3.]


26

Solution :

We need to make use of the following fact and we will prove it: Let F be a field and let be a set of subfields of F . Then is a subfield of F . Proof: is a subring −1 of F , contains 1 since 1 E for all E . Let r F . Then r E for each E , and so r E . Then r−1 F , and hence F is a field. // Let F be a field, and let denote the set of all subfields of F . Then F 0 = is a subfield of F and is contained in every other subfield. Consider the ring homomorphism φ : Z F 0 with φ(1) = 1. The induced map ψ : Z/(n) F 0 is an injective ring homomorphism, Exercises 7.4-26 and 7.4-28 informs us that n is a prime or 0. For all nonzero a Z/(n), ψ(a) is a unit in F 0 . Thus we have an monomorphism (injective homomorphism)Φ : F (Z/(n)) F 0 , where F (Z/(n)) denotes the field of fractions. Note that im Φ is a subring of F which is a field, and thus im Φ = F 0 . Thus Φ is an isomorphism. If n = 0, then F 0 = Q, and if n = p is prime, then F 0 = Z/( p).//

X ∈

∈

∈ X

 X

∈

∈

∈ X

 F

F

→

∈

 X

∈

→

→

∼

∼

Dummit & Foote Text Exercise 7.5-4: Prove that any sub field of R must contain Q . Solution :

Exercise 7.5-3 showed us that , R contains a unique inclusion-smallest subfield which is isomorphic either to Z /( p) for a prime p or to Q . Assume that the unique smallest subfield of R is isomorphic to Z/( p), and let a in this subfield be nonzero. Then pa = 0 R, and since p R is a unit implies a = 0, which is a contradiction. Hence the unique smallest subfield of R is isomorphic to Q . Any subfield of R contains a subfield isomorphic to Q.

∈

∈

In Section 10.1’s exercises R is a ring with 1 and M is a left R-module. Dummit & Foote Text Exercise 10.1-2: Prove that R x and M satisfy the two axioms in Section 1.7 for a group action of the multiplicative group R x on the set M . R× is a group under the multiplication. The restriction of the group action to R × M is a mapping R M M . Since R is a multiplicative group, it contains the multiplicative identity, 1. For all m M , the second axiom is satisfied, 1 m = m. For all r 1 , r2 R× , (r1 r2 ) m = r 1 (r2 m) and thereby satisfies the first axiom.// Solution :

×

∈

·

× →

·

∈

·

·

×

·

Dummit & Foote Text Exercise 10.1-5: For any left ideal I of R define IM =

{



finite

ai mi a i

· | ∈ I, m ∈ M i

to be the collection of all finite elements of the form am where a submodule of M . Solution :

∈ I and m ∈ M . Prove that I M is a

Apply the definition of the submodule criterion! Since 0 0 = 0 IM , I M is not empty. Let IM and letr R. Then i ai mi , k b k nk ( i ai mi ) + r ( k bk nk ) = i ai mi + k (rbi ) nk IM . Hence I M M is a submodule. //



·

·

 · ∈  · ·

∈ ·

·



· ∈

∈ ⊆

Dummit & Foote Text Exercise 10.1-15: If M is a finite abelian group then M is naturally a Z-module. Can this action be extended to make M into a Q-module?. Solution :

Proof by contradiction. Let M be a (multiplicative) finite abelian group of order k. Assume that M is a Q -module and that the action of Q on M extends the natural action of Z given by k m = m k . Let x M be any nonidentity (and non-zero) element. Then k1 x = y for some y M . That leads to k ( k1 x) = k y. So x = y k = 1, a contradiction. Therefore no such module structure exists.

∈ · ·

·

·

∈

·

Dummit & Foote Text Exercise 10.1-18: Let F = R, let V = R2 and let T be the linear transformation from V to V which is rotation clockwise about the origin by π/2 radians. Show that V and 0 are the only F [x]-submodules for this T .


27

Solution :

I had help on this problem as I didn’t quite understand fully F [x] modules. Note that (y, x). Let N be an F [x]-submodule of V . Assume that there exists a T : V V is given by (x, y) nonzero element (a, b) N . This implies either a = 0 or b = 0 and therefore a 2 + b2 > 0. Let (x, y) V , ax + by bx ay r1 = 2 , and r 2 = 2 . Then 2 a +b a + b2

→

∈

→ −



−

(r1 + r2 x) (a, b) = = = =



∈

r1 (a, b) + r2 (x (a, b)) (r1 a, r1 b) + (r2 b, r2 a) (r1 a + r2 b, r1 b r2 a) (r1 a + r2 b, r1 b r2 a)

·

·

  

=

=

=

· · − − −

axa + bya bxb ayb axb + byb + 2 , 2 a2 + b2 a + b2 a + b2

−

a2 x + b2 x b 2 y + a2 y , 2 a2 + b2 a + b2 a2 + b2 b2 + a2 x 2 ,y a + b2 a2 + b2

−

bxa aya a2 + b2

−







= (x, y). Hence (x, y) N and N = V since N V . So, every nonzero F [x]-submodule of V is V . Therefore, 0 andV arethe only F [x]-submodules of V .

∈

⊆

Dummit & Foote Text Exercise 10.1-19: Let F = R, let V = R2 and let T be the linear transformation from V to V which is projection onto the y-axis. Show that V , 0, the x-axis, and the y-axis are the only F [x]-submodules for this T . Solution :

Let N V be an F [x] submodule. Let the x-axis be represented by X = R 0 and the y axis be represented as Y = 0 R. Assume there exists an element (a, b) N where a, b = 0 and let (x, y) V . Then ( xa + ( yb xa )x) (a, b) = (x, y). Thus V N , and since N V , that implies N = V . Next assume either a = 0 or b = 0 for every element (a, b) N and assume that (a, 0) N with a = 0 and there exists (0, b) N with b = 0. Then (a, b) N with a, b = 0, a contradiction. Thus N X . Now let (x, 0) X . Since xa (a, 0) = (x, 0), X N . Hence N = X . Next assume (0, b) N with b = 0. As before, N Y . If (0, y) Y , then yb (0, b) = (0, y), and so Y N . Thus N = Y . Last, if for all (a, b) N a = b = 0, then N = 0. So, the only possible submodules of V are 0, X , Y , and V . 0 and V are trivial submodules. So, we need to verify that X and Y are submodules by the submodule criterion. Sincee (0, 0) X , X is nonempty. Let (a, 0), (b, 0) X , p(x) F [x], and p(x) = p 0 + xp (x). Then x (b, 0) = (0, 0) and

⊆

−

−

− ×

·

∈ · ∈

∈



×

∈

⊆

∈

⊆





∈

⊂ ∈ 

⊆

∈

∈

⊆

·



⊆

∈

∈

·

∈

∈

(a, 0) + p(x) (b, 0) = = (a, 0) + p0 (b, 0) + p (x) (x (b, 0)) = (a, 0) + ( p0 b, 0) + (0, 0) = (a + p0 b, 0) X

·

·

· ·

∈

By the submodule criterion, X V is a submodule. We repeat the process for Y . Since (0, 0) Y , Y is nonempty. Let (0, a), (0, b) X , p(x) F [x] and p(x) = pi xi . Then T 2 (v) = T (v). In particular, x k v = x v for all k 1. If v Y , then T (v) = v. We obtain the following:

⊆



∈

·

·

≥

∈

∈

∈


28

(0, a) + p(x) (0, b) = = = =

·

So, Y

(0, a) + ( pi xi ) (0, b) (0, a) + pi (xi (0, b)) (0, a) + pi (0, b) (0, a + pi b) Y.

 ·  ·· · ∈

⊆ V is a submodule.

Dummit & Foote Text Exercise 10.2-1 : Use the submodule criterion to show that kernels and images of R-module homomorphisms are submodules. Solution :

Let R be a ring with 1 and let M and N be left R-modules, and let φ : M N be an R-module homomorphism. φ(0) = 0 implies that ker φ and im φ are both nonempty. Let a, b ker φ and let r R. Then φ(a + r b) = φ(a) + r φ(b) = 0, so that a + r b ker φ. Hence ker φ is an R-submodule of M by the submodule criterion. Let x, y M and let r R. Then φ(x) + r φ(y) = φ(x + r y). Since φ(x) and φ(y) are arbitrary in imφ, im φ is an R-submodule of N by the Submodule criterion.

→

∈

∈

∈

·

∈

·

·

· ∈

·

Dummit & Foote Text Exercise 10.2-2 : Show that the relation ”is R-module isomorphic to” is an equivalence relation on any set of R-modules. Solution :

∼

M = N if there exists an R-module isomorphism φ : M N . To prove an equivalence relation, we need to show that the relation is reflexive, symmetric, and transitive. Let S be a set of left R-modules. (Reflexive). Let M S . The identity mapping id : M M is an R-module isomorphism since id(x + r y) = x + r y = id (x) + r id(y) for all x, y M and r R. Henee M = M , and so the relation = is reflexive. (Symmetric). Assume that M , N S and that M = N . This implies that there exists an R module homomorphism from φ : M N . Since φ is a bijection, φ has in inverse that is well defined. Let r R and x, y N . Let x = φ(a) and y = φ(b) for some a, b M . Then

→

∈ ·

·

· ∈

→

∈

∈ ∼

→

∼

∈

∼

−

∈

φ−1 (x + r y) = = = =

·

∈

φ−1 (φ(a) + r φ(b)) φ−1 (φ(a + r b)) a+r b φ−1 (x) + r φ−1 (y).

·

·

·

·

∼

So φ −1 : N M is an R-module isomorphism. Hence N = M and the relation is symmetric. (Transitive). Let M , N , P S and M = N and N = P . Then there exist R-module isomorphisms φ : M N and σ : N P . We know that σ φ : M T is a bijection (since function composition of bijections is a bijection). To show that it is also an R-module homomorphism, let x, y M and r R. Then

→

→

→

∼

∈

◦

(σ φ)(x + r y) = = = =

◦

·

∼ →

∈

∈

σ(φ(x + r y)) σ(φ(x) + r φ(y)) σ(φ(x)) + r σ(φ(y)) (σ φ)(x) + r (σ φ)(y)

·

◦

·

·

∼

· ◦

So σ φ : M P is an R-module isomorphism, and we have M = P . Hence the relation is transitive. Since all three requirement of an equivalence relation have been met, ”is R-module isomorphic to” is an equivalence relation.

◦

→

Dummit & Foote Text Exercise 10.2-3: Give an explicit example of a map from one R-module to another which is a group homomorphism but not an R-module homomorphism. Solution :

Let R be a non-commutative ring with 1. Choose an element a not in the center of R and then consider R as a left module over itself via multiplication. That is, M = R. If φ(x) = a x defines the map

·


29

φ : M M , we have a group homomorphism (an endomorphism) since φ(x + y) = a (x + y) = a x + a y = φ(x) + φ(y). However, if b R does not commute with a, then φ(b) = a b = ab but b φ(1) = b (a 1) = b a = ba = ab. Hence φ, although a group homomorphism, is not an R-module homorphism.

→

·

·

·

·

·

· ·

·

∈



Dummit & Foote Text Exercise 10.2-13: Let I be a nilpotent ideal in a commutative ring R (cf. Exercise 37, Section 7.3), let M and N be R-modules and let φ : M N be an R-module homomorphism. Show that if the induced map φ : M/IM N/IN is surjective, then φ is surjective.

→

→

Solution :

The induced map φ is given by φ(m + IM ) = φ(m) + IN . Since φ is surjective, N/IN = φ[M/IM ] = (φ[M ] + IN )/IN . By the Lattice Isomorphism theorem for modules, we have N = φ[M ] + IN . We prove the statement N = φ[M ] + I t N for all t 1 by induction. The base case t = 1 has been shown to be true. Assume for the induction step the equation holds for some t > 1. Then

≥

N = = = =

φ[M ] + I t N φ[M ] + I t (φ[M ] + IN ) φ[M ] + I t φ[M ] + I t+1 N φ[M ] + I t+1 N since I t φ[M ]

⊆ φ[M ].

The induction hypothesis is proved. Since I k = 0, N = φ[M ] + I k N = φ[M ] and implies φ is surjective. Dummit & Foote Text Exercise 10.2-14: Let R = Z[x] be the ring of polynomials in x and let A = Z[t1 , t2 , . . . , ] be the ring of polynomials in the independent indeterminates t 1 , t2 , . . . . Define an action of R on A as follows: 1) let 1 R act on A as the identity, 2) for n 1, let x n 1 = t n , let x n ti = t n+i for i = 1, 2, . . . and let x n act as 0 on monomials in A of (total) degree at least two, and 3) extend Z-linearly, i.e., so that the module axioms 2(a) and 2(c) are satisfied. (a) Show that x p+q ti = x p (xq ti ) = t p+q+i and use this to show that under this action the ring A is a (unital) R-module. (b)Show that the map φ : R A defined by φ(r) = r 1A is an R-module homomorphism of the ring R into the ring A mapping 1R to 1A , but is not a ring homomorphism from R to A.

∈

◦

◦

≥

◦

◦

◦

→

◦

(a) x p+q ti = t p+q+i = x p (tq+i ) = x p (xq ti ). Hence makes A into a unital left R-module. (b) Let a, b, r R. Then φ(a + r b) = (a + r b) 1 = a 1 + r (b 1) = φ(a) + r φ(b) Solution :

◦

∈

◦

◦

·

◦

◦

· · · · ·

·

so that φ is an R-module homomorphism. To show it’s not a ring homomorphism, we use contradiction. Assume that φ is a ring homomorphism. Then φ(x2 ) = x2 1 = t 2 but φ(x2 ) = φ(x)φ(x) = (x 1)(x 1) = t 1 2 contradiction. Hence φ is not a ring homomorphism.

·

·

·

Dummit & Foote Text Exercise 10.3-1: Prove that if A and B are sets of the same cardinality, then the free modules F (A) and F (B) are isomorphic. Solution :

I had help on this problem. Prof. R.D. Maddux of Iowa State University had the most lucid explanation for this problem: http://orion.math.iastate.edu/maddux/505-Spring-2010/hw05.3.pdf Since A and B have the same cardinality, there exists a bijection f : A B. Since f is a bijection, its inverse is also a bijection, that is, f −1 : B A. A. By Theorem 10.6 the free modules F (A) and F (B) have the universal mapping property. We now apply this property to f . Note that the image of f is B since f is surjective, and B is a subset of the free module F (B). Thus we have another function that maps A into F (B) and is equal to f on all elements of A, that is, g : A F (B) and g(a) = f (a) for all a A.

→

→

→

∈


30

By the universal mapping property there is a unique R-module homomorphism φ : F (A) F (B) which extends g (and f ). That is, f (a) = g(a) = φ(a) for all a = inA. Using the same reasoning, with A and B interchanged and f replaced by f −1 , we obtain another R-module homomorphism ψ : F (B) F (A) extending f −1 . The composition of R-module homomorphisms is again an R-module homomorphism: ψ φ : F (A) F (A). For every a A, (ψ φ)(a) = ψ(φ(a)) = ψ(f (a). Since f (a) B and ψ extends f −1 , ψ(f (a)) = f −1 (f (a)) = a. This shows that ψ φ is an extension of the identity map ι : A F (A) which sends every element of A to itself (i.e, ι(a) = a for every a A. By the universal mapping property, ι has a unique extension to an R-module homomorphism from F (A) to itself. The identity map from F (A) to F (A) is an R-module homomorphism of F (A) onto itself, and by the universal mapping property of F (A) it is the only R-module homomorphism of F (A) onto itself. However, ψ φ is a homomorphism of F (A) onto itself that extends the identiy map on A so ψ φ must be the identity map from F (A) to F (A). Similarly, φ ψ is the identity map from F (B) to F (B). Thus φ and ψ are R-module homomorphisms between F (A) and F (B), and they are inverses of each other, so they are both injective and surjective, and therefore are isomorphisms between F (A) and F (B). Thus, F (A) = F (B) whenever A = B . //

→

→

◦

→

∈

◦

∈

◦

◦

◦

◦

→

∈

∼

| | | | Dummit & Foote Text Exercise 10.3-2: Assume R is commutative. Prove that R ∼ = R if and only n

m

if n = m, i.e., two free R-modules of finite rank are isomorphic if and only if they have the same rank. [Apply Exercise 12 of Section 2 with I a maximal ideal of R. You may assume that if F is a field, then F n = F m if and only if n = m, i.e., two finite dimensional vector spaces over F are isomorphic if and only if they have the same dimension - this will be proved later in Section 11.1.]

∼

Solution :

There are two solutions. One easy and one a bit more involved. Simpler way. Assume that R n = Rm . Let I be a maximal ideal of R. Then R n /IRn = R m /IRm . Exercise 10.2-12 (not assigned) tells that R n /IR n =R (R/IR) (R/IR) (R/IR) (n times). This implies n m (R/I ) = (R/I ) . Since these are vector spaces over the field R/I , we have n = m. // The more complicated way. It is clear that if n = m, then the result holds. Assume hat R m = Rn , where Rφ : R m Rn is an m n n R-module isomorphism. Let I R be a maximal ideal. Next π φ : R R /IR , where π denotes the natural projection. Claim: ker π φ = I Rm . (IRm ker π φ): If x = ai (ri,j ) IRm , then φ( ai (ri,j )) = ai φ((ri,j )) IRn . Thus (π φ)(x) = 0, and we have I Rm ker π φ. (ker π φ IRm ): Assume x = (ri ) ker π φ. Then 0 = (π φ)(x) = φ(x) + IRn . Thus φ(x) IRn . Let φ(x) = ai (ri,j ), where (ri,j ) Rn . Since φ is surjective, we have (ri,j ) = φ((si,j )) for some (si,j ) Rm . Then φ(x) = eo ai φ((si,j )) = φ( ai (si,j )). Since φ is injective, we have x = ai (si,j ), and thus x IRm as desired. Certainly π φ is also surjective. By the first isomorphism theorem, the induced map ψ : Rm /IRm Rn /IR n is an R-module isomorphism. The result of Exercise 10.2-12 implies (as with the easier method) that ( R/I )m =R (R/I )n . Since I is maximal, R/I is a field, and therefore m = n.//

∼

∼

∼

×

×···

∼ ◦

⊆

◦

∈

◦ ◦

⊆



◦ ⊆

 ·

·

∼

⊆

∈ ◦

∈

∈

◦

◦





→

→



·

◦

→

·

∈

∈



·

∈

∼

Dummit & Foote Text Exercise 10.3-4: An R-module M is called a torsion module if for each m M there is a nonzero element r R such that rm = 0, where r may depend on m (i.e., M = Tor(M ) in the notation of Exercise 8 of Section 1). Prove that every finite abelian group is a torsion Z module. Give an example of an infinite abelian group that is a torsion Z module.

∈

∈

−

−

Solution :

Let M be a finite abelian group. This implies it is also a Z -module. Let n be the order of M . Then for every a M , by Lagrange, the order of a divides the order of n of the the abelian group M . Hence, na = 0. Since n Z + and na is the result of the action of n on a in the module M , we have a Tor(M ), so M Tor(M ). Tor(M ) M is obviously true. Therefore M = Tor(M ) . // Let N = i∈N Z/2Z and this implies that N is a direct product of countably many copies of the 2-element cyclic group Z /2Z. But N is an infinite abelian group whose cardinality is the same as the set of real numbers (by Cantor’s Diagonal Argument). Every element of N has order 2 so 2 n = 0 for every n N . Hence Tor(N ) = N .

∈



∈ ∈ ⊆

⊆

·

∈


31

Dummit & Foote Text Exercise 10.3-6 Prove that if M is a finitely generated R-module that is generated by n elements then every quotient of M may be generated by n (or fewer) elements. Deduce that quotients of cyclic modules are cyclic. Solution :

We need to show M /N = (ai + N i 1, 2, 3 , n 1, n ). ((ai + N i 1, 2, 3 , n 1, n ) M/N ) holds by the definition of a quotient. (M/N ) (ai + N i 1, 2, 3 , n 1, n ). Assume x + N M/N where x = i ri ai . Then x + n = ri (ai + N ) as required. Hence ai + N ni=1 is an R-module generating set for M /N .// If M is cyclic, then it has a generating set consisting of a single element. Moreover, the quotient of M is then generated by at most one element. Since every module is generated by at least one element, every quotient of M is also cyclic.

| ∈ { · ·· − } ⊇ ⊇ | ∈ { · ·· − ·



| ∈ {

} {

· ·· −

}



∈

}

·

Dummit & Foote Text Exercise 10.3-7: Let N be a submodule of M . Prove that if both M /N and N are finitely generated then so is M . Solution :

If M /N = (ai + N 1 i m) and N = (bj 1 j n), then M = (ai , bj 1 i m, 1 j n). M (ai , bj 1 i m, 1 j n)) Assume m M . Then m + N = i ri (ai + N ) = ( i ri ai ) + N . Then m N and that implies m i ri ai i ri ai = j s j bj . This leads to m = i ri ai + j sj bj (ai , bj 1 i m, 1 j n). Hence M is finitely generated.// (ai , bj 1 i m, 1 j n) M ) holds.

| ≤ ⊆ | ≤ − · · | ≤ ≤

| ≤ ≤ | ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ ∈ ∈ − · · ∈ | ≤ ≤ ≤ ≤ ≤ ≤ ⊆

  









·

·

Dummit & Foote Text Exercise 10.3-9: An R-module M is called irreducible if M = 0 and if 0 and M are the only submodules of M . Show that M is irreducible if and only if M = 0 M is a cyclic module with any nonzero element as generator. Determine all the irreducible Z -modules.





Solution :

( ) If M is irreducible, then M = 0 by definition. Let x M be nonzero. Then Rx M is a nonzero submodule since 1 x = x Rx. Since M is irreducible, Rx = M . Hence any nonzero element of M generates M . ( ) Assume M = 0 and that if x = 0, then Rx = M . Let N M be a nonzero submodule. Then there exists some nonzero x N , and M = Rx N . Thus N = M . Since the only submodules of M are 0 and M , M is irreducible.// We know that Z -modules are abelian groups, and that Z-submodules are the subgroups of abelian groups. If M is an irreducible Z -module, it is a abelian group whose subgroups that has no nontrivial proper subgroups). Hence M is cyclic and therefore it is of prime order.

⇒

⇐

·



∈



∈



∈

⊆

⊆

⊆

Dummit & Foote Text Exercise 10.3-10: Assume R is commutative. Show that an R-module M is irreducible if and only if M is isomorphic (as an R-module) to R/I where I is a maximal ideal of R. [By the previous exercise, if M is irreducible there is a natural map R M defined by r rm, where m is any fixed nonzero element of M .]

→

Solution :

→

( ) Assume M is irreducible, let m M a fixed element and be nonzero, and define φm : R Mbyφm (r) = r m. Then for all x, y R and r R, φm (x + r y) = (x + r y) m = x m + r (y m) = φ m (x) + r φm (y). So φ m is an R-module homomorphism. Since m = 0 and M is irreducible, by Exercise 10.3.9, M = Rm. If b M , then there exists a R such that b = a m = φ m (a). Hence φ m is surjective. We need to show that ker φm is a maximal ideal. Let x + ker φ m be nonzero. Then x m = 0. Since M is irreducible, M = R(x m). This implies m = y (x m) = (yx) m for some y R. Then 1 m (yx) m = 0, so that 1 yx ker φ m . That is, (y + ker φ m )(x + ker φ m ) = 1 + ker φ m . So every nonzero element of R/ ker φm has a left inverse. Because R/ ker φ m is commutative, R/ ker φm is a field, so that ker φ m is a maximal ideal of R. By the First Isomorphism Theorem, =R R/ ker φm , where ker φ m is a maximal ideal.

→

⇒

· · · 

· ∈

· −

·

·

·

· ·

∈ ∈

∈

·

∈

·

− ∈

∼

· ·

·

∈

· 


32

( ) Assume I R is a maximal ideal. Then R/I is a field. Let x + I be nonzero. Then there exists y R such that (y + I )(x + I ) = (1 + I ) . If r + I R/I , then r + I = (ry + I )(x + I ) which implies R/I = R(x + I ). That is, R/I is generated (as an R-module) by any nonzero element. By Exercise 10.3-9, R/I is an irreducible R-module.

⇐

⊆

∈

∈

Dummit & Foote Text Exercise 10.3-11: Show that if M 1 and M 2 are irreducible R-modules, then any nonzero R-module homomorphism from M 1 to M 2 is an isomorphism. Deduce that if M is irreducible then EndR (M ) is a division ring (this result is called Schur’s Lemma). [Consider the kernel and the image.] Solution :

Assume φ : M 1 M 2 is an R-module homomorphism with M 1 and M 2 irreducible. In Exercise 10.2-1, we proved that ker φ and im φ are submodules of M 1 and M 2 respectively. Since φ is not the zero homomorphism, its kernel is not all of M 1 , and since M 1 is irreducible ker φ = 0. Thus φ is injective. Similarly, imφ is not the zero submodule, and consequently M 2 is not the zero submodule. Hence φ is surjective. Therefore φ is an R-module isomorphism. Let M be an irreducible unital left R-module. We know that EndR (M ) is a ring under pointwise addition and composition. If φ EndR (M ) is nonzero, then by the first part of this exercise, it is an isomorphism and so has an inverse φ −1 which is also in EndR (M ). So EndR (M ) is a division ring. We are not given any additional information that would lead us to conclude that EndR (M ) is commutative and therefore a field.

→

∈

Dummit & Foote Text Exercise 10.3-12: Let R be a commutative ring and let A, B , and M be R-modules. Prove following isomorphisms of R-modules: (a) HomR (A B, M ) = HomR (A, M ) HomR (B, M ) (b) HomR (M, A B) = HomR (M, A) HomR (M, B).

×

∼ × ∼

× ×

Solution :

I had A LOT of help with this problem. (a) Let f : A M and g : B M , be R-module homomorphisms defined by φ(f, g) : A φ(f, g)(a, b) = f (a) + g(b). Then for all a 1 , a2 A, b 1 , b2 B, and r R, we have

→

→

∈

φ(f, g)((a1 , b1 ) + r (a2 , b2 )) = = = =

·

∈

∈ φ(f, g)((a + r · a , b + r · b )) f (a + r · a ) + g(b + r · b ) f (a ) + r · f (a ) + g(b ) + r · g(b ) φ(f, g)(a , b ) + r · φ(f, g)(a , b ). 1

1

2

2

1

1

2

1

2

2

1

× B → M by

1

2

1

2

2

So φ(f, g) : A B M is an R-module homomorphism. HomR (A B, M ). We will show that Φ is an Now consider the map Φ : HomR (A, M ) HomR (B, M ) R-module homomorphism. Let f 1 , f 2 : A M and g 1 , g2 : B M be R-module homomorphisms and let r R. Then Φ((a1 , b1 ) + r (a2 , b2 ))(a, b) = Φ(f 1 + r f 2 , g1 + r g2 )(a, b) = (f 1 + r f 2 )(a) + (g1 + r g2 )(b) = f 1 (a) + r f 2 (a) + g1 (b) + r g2 (b) = Φ(f 1 , g1 )(a, b) + r Φ(f 2 , g2 )(a, b) = (Φ(f 1 , g1 ) + r Φ(f 2 , g2 ))(a, b).

× →

× →

∈

→

·

×

→ · · ·

·

·

·

·

·

So Φ is an R-module homomorphism. Next we show Φ is an injection. Assume (f, g) ker Φ. Then 0 = Φ(f, g)(a, 0) = f (a) for all a A, so that f = 0. Similarly, g = 0. Thus (f, g) = 0 and ker Φ = 0. Hence Φ is an injection. Next we show Φ is an surjection. Let τ : A B M be an R-module homomorphism, and define f τ : A M and g τ : B M by f τ (a) = τ (a, 0) and g τ (b) = τ (0, b). Then

∈

∈

→

× →

→

f τ (a1 + r a2 ) = τ (a1 + r a2 , 0 = τ (a1 , 0) + r τ (a2 , 0) = f τ (a1 ) + r f τ (a2 ).

·

·

·

·


33

So f τ is an R-module homomorphism. Similarly, g τ is an R-module homomorphism. Since Φ(f τ , gτ )(a, b) = f τ (a) + gτ (b) = τ (a, 0) + τ (0, b) = τ (a, b), we have Φ(f τ , gτ ) = τ . Hence Φ is a surjection. Thus, we have HomR (A B, M ) =R HomR (A, M ) HomR (B, M ) . // (b) Now let π A : A B A and π B : A B B be the left and right coordinate projections respectively and these are R-module homomorphisms. Define Ψ : HomR (M, A B) HomR (M, A) HomR (M, B) by Ψ(φ) = (πA φ, πB φ). If φ, ψ HomR (M, A B) and r R, then

∼

× × → ◦

◦

×

∈ Ψ(φ + r · ψ)

= = = =

× → × → × × ∈ (π ◦ (φ + r · ψ), π ◦ (φ + r · ψ)) (π ◦ φ + r · (π ◦ ψ), π ◦ φ + r · (π ◦ ψ)) (π ◦ φ, π ◦ ψ) + r · (π ◦ φ, π ◦ ψ) Ψ(φ) + r · Ψ(ψ). A

B

A

A

A

B

A

B

B

B

Thus Ψ is an R-module homomorphism. We now show that Ψ is an injection. Assume φ m M , then 0 = Ψ(φ)(m, m) = ((πA φ)(m), (πb ψ)(m)) = (πA (φ(m)), πB (φ(m)) = φ(m).

∈

◦

∈ ker Ψ. If

◦

So φ = 0, hence ker Ψ = 0, and thus Ψ is an injection. Next we show that P si is a surjection. Assume (f, g) HomR (M, A) HomR (M, B). Define φf,g : M A B by φ f,g (m) = (f (m), g(m). φf,g is an R-module homomorphism. Moreover, Ψ(φf,g )(m) = ((πA φf,g )(m), (πB φf,g )(m)) = (f (m), g(m)). Thus Ψ(φf,g ) = (f, g), and so Ψ is a surjection . Hence Hom R (M, A B) =R HomR (M, A) HomR (M, B).//

∈

→ ×

◦

◦

× ∼

×

×

Dummit & Foote Text Exercise 10.3-13: Let R be a commutative ring and let F be a free R-module of finite rank. Prove the following isomorphism of R-modules: HomR (F, R) = F .

∼

Let F be a free R-module of finite rank. That is, let F be free on the set ai ni=1 . Then every element of F can be written uniquely as ri ai for some r i R. Next, define Φ : HomR (F, R) F by Φ(φ) = φ(ai ) ai . Φ is an R-module homomorphism. If φ, ψ HomR (F, R) and r R, then Solution :





·

·

∈



∈

Φ(φ + r ψ) = (φ + r ψ)(ai ) ai = ( φ(ai ) ai ) + r ( = Φ(φ) + r Φ(ψ)

·

· · ·

{ } ∈

·

·



→

ψ(ai ) ai )

·

shows that Φ is an R-module homomorphism. Φ is injective. Assume φ ker Φ. Then 0 = Φ(φ) = φ(ai ) ai . Since F is free on the a i , we have φ(ai ) = 0 for all a i , and thus φ = 0. So ker Φ = 0, and thus Φ is injective. Φ is surjective. Let ti ai F . Define g : F R by g ( ri ai ) = ri ti . So , g is an R-module homomorphism. Moreover, g(ai ) = t i , so that Φ(g) = ti ai . Hence Φ is surjective. Thus Φ is an R-module isomorphism, so that we have HomR (F, R) = F .//

∈



· ∈

→

 ·   · · ∼

Dummit & Foote Text Exercise 10.3-20: Let I be a no empty index set and for each i I let M i be an R-module. The direct product of the modules M i is defined to be their direct product as abelian groups (cf. Exercise 15 in Section 5.1) with the action of R componentwise multiplication. The direct sum of the modules M i is defined to be the restricted direct product of the abelian groups M i (cf. Exercise 17 in Section 5.1) with the action of R componentwise multiplication. In other words, the direct sum of the M i ’s is the subset of the direct product, i∈I M i , which consists of all elements i∈I m i such that only finitely many of the components m i are nonzero; the action of R on the direct product or direct sum is given by

∈










34

r i∈I m i = i∈I rm i (cf. Appendix I for the definition of Cartesian products of infinitely many sets). The direct sum will be denoted by i∈I M i . (a) Prove that the direct product of the M i ’s is an R-module and the direct sum of the M i ’s is a submodule of their direct product. (b) Show that if R = Z, I = Z+ and M i is the cyclic group of order i for each i, then the direct sum of the M i ’s is not isomorphic to their direct product. [Look at torsion.]





(a) We know from Exercise 5.1-15, that i∈I M i is an abelian group under point-wise addition and multiplication. So we just need to verify the three left-module axioms. Let r, s R and let (mi ), (ni ) i∈I M i . Then Solution :

∈

∈



(r + s) (mi ) = = = =

·

(rs) (mi ) = = = =

·

r ((mi ) + (ni )) = = = = =

·

    ∈

((r + s) mi ) (r mi + s mi ) (r mi ) + (s mi ) r (mi ) + s (mi ),

·

·

· ·

·

·

((rs) (r (s r (s r (s

·

·m ) · · m )) · ·m ) · · (m )), r · (m + n ) (r · (m + n )) (r · m + r · n ) (r · m ) + (r · n ) r · (m ) + r · (n ). i

i

i

i

i

i

i

i

i

i

i

i

i

i

Thus i∈I M i is a left R-module. If R has a 1 and each M i is unital, then 1 (mi ) = (1 mi ) = (mi ), for all (mi ), so that I M i is unital. To show that i∈I M i is a submodule of i∈I M i , we apply the submodule criterion. i∈I M i is nonempty since 0 R. Let J, K I such that x j = 0 i∈I M i . Now let (xi ), (yi ) i∈I M i and let r for j / J and y k = 0 for k / K . Consider (xi ) + r (yi ) = (xi + r yi ). If i / J K , then x i + r yi = 0. Since J K I is finite, (xi ) + r (yi ) i∈I M i . By the submodule criterion, i∈I M i i∈I M i is an R-submodule. (b) Consider the two Z -modules: M = N Z/nZ and N = N Z/nZ (they’re both abelian groups). The goal is to show that these two modules are not isomorphic by verifying that one is torsion while the other is not. Let (ai ) M . By definition of the direct sum, there is a natural number L such that, for all k L, L ak = 0. Now let t = k=1 n . Then t (ai ) = 0 and that implies every element of M is torsion, so that M is torsion. However, t (1) = (t) is nonzero for all integers t so N is not torsion. To put it another way, if there existed r Z such that r(1, 1, 1, . . . ) = (0, 0, 0, . . . ), then we have a contradiction since r 1 = r = 0 Z(n + 1)Z. Hence the direct product is not a torsion module. Thus M and N are not isomorphic as Z -modules.

∈

∪ ⊆

∈

·

 ∈

∈

·

·

∈

  ∈ · ∈  

·



·

∈

∈ ∪

·

⊆





⊆

·

≥

·

Dummit & Foote Text Exercise 10.3-23 : Show that any direct sum of free R-modules is free. Solution :

I had help on this problem. Let R be a ring with 1 and let F i I be a family of free unital left R-modules. Let A i = ai,j be a free basis for each F i and assume that the A i are disjoint. Let A = I Ai . Let F (A) denote the free R-module on A. There is the natural inclusion A F (A) and a natural inclusion ι : A (bt ), where I F i , where a i,j bt = a i,j if t = i and 0 otherwise. By the universal property of free modules, there is a unique R-module homomorphism Ψ : F (A) I F i such that Ψ(ai,j ) = ι(ai,j ). Since we’ve established that Ψ is a R-module homomorphism, we show that Ψ is an R-module isomorphism. Ψ is injective. Assume x ker Ψ and let x = ri,j ai,j . Then 0 = Ψ(x)i = j ri,j ai,j F i . Since F i is free on A i , r i,j = 0 for all i and j . This implies x = 0 and ker Ψ = 0. Hence Ψ is injective.

{ }

{ }

→

→

∈



→





→



∈




35

 ∈  

Ψ is surjective. Assume ( j ri,j ai,j ) I F i . Since only finitely many of the r i,j are nonzero, F (A). Then Ψ( i,j ri,j ai,j ) = ( j ri,j ai,j ), so that Ψ is surjective. i,j r i,j ai,j Therefore I F i =R F (A) and I F i is free as an R-module.



 ∈

∼

Dummit & Foote Text Exercise 10.3-24: (An aribitrary direct product of free modules need not be free) For each positive integer I , let m i be the free Z -module Z , and let M be the direct product Z+ M i (cf. Exercise 20). Each element of M can be written uniquely in the form (a1 , a2 , a3 , . . . ) with a i Z for all i. Let N be the submodule of M consisting of all such tuples with only finitely many nonzero a i . Assume M is a free Z -module with basis . (a) Show that N is countable. (b) Show that there is a countable subset 1 of such that N is contained in the submodule N 1 generated by 1 . Show also that N 1 is countable. (c) Let M = M/N 1 . Show that M is a free Z-module. Deduce that if x is any nonzero element of M then there are only finitely many distinct positive integers k such that x = km some m M (depending on k). (d) Let = (b1 , b2 , b3 , . . . ) b i = i! for all i . Prove that is uncountable. Deduce that there exists some s such that s / N 1 . (e) Show that the assumption M is free leads to a contradiction: By (d) we may choose s with s / N 1 . Show that for each positive integer k there is some m M with s = km, contrary to (c). [Use the fact that N N 1 .]



∈

B

B B

B

S { ∈ S

∈

|

±

}

∈

S

∈ S

∈

⊆

∈

Solution :

I had a lot of help on this problem. Especially part (3). Part (e) was the most difficult to understand. (a) For each n Z+ , there is a natural inclusion ι n : Zn N given by ι n (a)i = a i+1 if 0 i < n and 0 otherwise. Each element of N is in the image of some ι n . So, N = n∈N+ im ι n . Hence, N is a countable union of countable sets, and therefore is countable. (b) For each x N , there is a finite subset B x such that x = b∈B rb b. Let 1 = x∈N Bx . As the countable union of finite sets, 1 is countable. Since x N and x = b∈B rb b b∈B1 Rb, N N 1 = b∈B1 Rb and N 1 is also countable. (c) The natural inclusion ι : M/N 1 is given by b b + N 1 . By the universal property of free 1 modules, there is a unique R-module homomorphism Φ : F ( M/N 1 such that Φ(b) = b + N 1 for 1) all b . 1 Now to show that Φ is an isomorphism. Φ is injective. Let x ker P hi and let x = ri bi . Then 0 = ri bi + N 1 , so that ri bi N 1 . Thus ri = 0, and we have x = 0. So Φ is injective. Φ is surjective. Assume x + N 1 M/N 1 . Since M is free on , we have x + N 1 = ri bi + N 1 for some r i and b i ri bi ) = x and that implies Φ is surjective. 1 . Then Φ( Therefore Φ is an R-module isomorphism, and M /N 1 is free as a Z-module. (d) There exists a bijection φ : +1 if b i > 0 and b i 1 if b i < 0. A N 1, 1 = T . That is b i diagonalization argument (a modification of Cantor’s) shows that T is uncountable. Assume f : N T is a bijection. Construct g = (ei ) T as follows: ei = 1 if f (i)i = 1 and 1 otherwise. Since g differs from every element in im f = T in some component, g / im T , a contradiction. Thus is uncountable. Since N 1 is countable, there must exist s such that s / N 1 . (e) Let k Z be positive. Define m M by m i = i! if i! < k and m i = σ i i!/k if i k, where σ i = 1 if si > 0 and 1 if s i < 0. Now (s km)i = 0 for all i > k , so that s km N N 1 . Thus s + N 1 = k(m + N 1 ). This is a contradiction of part (d) above. Thus M is not free as a Z -module.

∈

≤

∈

⊆

→

⊆ B

B



B − B →



∈



∈ B − B

S→

∈

∈ S

∈ −

−

{

∈

∈

·

x

x

·

→ B − B →

∈B−B

∈



  B ∈



 

B

−}

→

∈ ∈

∈

→ −

−

−

→

S

∈ ⊆

≥

Dummit & Foote Text Exercise 10.3-25: In the construction of direct limits, Exercise 8 of Section 7.6, show that if all A i are R-modules and the maps ρ ij are R-module homomorphisms, then the direct limit A = lim Ai may be given the structure of an R-module in a natural way such that the maps ρi : A i A are all R module homomorphisms. Verify the corresponding universal property (part (e)) for R- module homomorphisms φ : A i C commuting with the ρ ij .

→ −→

−

→

MAT7400 Assignment #3 Solution :

36

?

Dummit & Foote Text Exercise 10.3-26: Carry out the analysis of the preceding exercise corresponding to inverse limits to show that an inverse limit of R-modules is an R-module satisfying the appropriate universal property (cf. Exercise 10 of Section 7.6). Solution :

?

Dummit & Foote Text Exercise 10.4-2: Show that the element ”2 non-zero in 2Z Z Z/2Z.

⊗

⊗ 1” is 0 in Z ⊗

Z

Z/2Z but

Solution :

Dummit & Foote Text Exercise 10.4-3: Show that C are not isomorphic as R -modules.

C and C

⊗

Dummit & Foote Text Exercise 10.4-4: Show that Q Z Q and Q Q-modules. [Show they are both 1-dimensional vector spaces over Q.]

⊗

⊗

R

C

C are both left R -modules but

Solution :

⊗

Q

Q are isomorphic left

Solution :

Dummit & Foote Text Exercise 10.4-6: If R is any integral domain with quotient field Q, prove that (Q/R) R (Q/R) = 0.

⊗

Solution :

Dummit & Foote Text Exercise 10.4-13: Prove that the usual dot product of vectors defined by letting (a1 , . . . , an ) (b1 , . . . , bn ) be a 1 b1 + an bn is a bilinear map from R n Rn to R .

·

· ··

×

Solution :

Dummit & Foote Text Exercise 10.4-14: Let I be an arbitrary nonempty index set and for each i I let N i be a left R-module. Let M be a right R-module. Prove the group isomorphism : M ( i∈I N i ) = (M N i ), where the direct sum of an arbitrary collection of modules is defined in Exercise 20, Section 3. [Use the same argument as for the direct sum of two modules, taking care to note where the direct sum hypothesis is needed- cf. the next exercise.]

∈

⊗ ⊕

∼ ⊗

Solution :

Dummit & Foote Text Exercise 10.4-15: Show that tensor products do not commute with direct products in general. [Consider the extension of scalars from Z to Q of the direct product of the modules M i = Z/2i Z, i = 1, 2, . . . ]. Solution :

Dummit & Foote Text Exercise 10.4-18: Suppose I is a principal ideal in the integral domain R. Prove that the R-module I R I EO has no nonzero torsion elements (i.e., rm = 0 with 0 = r R and m I R I implies that m = 0).

∈ ⊗

Solution :

⊗

 ∈

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38

Dummit & Foote Text Exercise 10.5-7: Let A be a nonzero finite abelian group. (a) Prove that A is not a projective Z -module. (b) Prove that A is not an injective Z -module. Solution :

Dummit & Foote Text Exercise 10.5-9: Assume R is commutative with 1. (a) Prove that the tensor product of two free R-modules is free. [Use the fact that tensor products commute with direct sums.] (b) Use (a) to prove that the tensor product of two projective R-modules is projective. Solution :

Dummit & Foote Text Exercise 10.5-12: Let A be an R-module, let I be any nonempty index set and for each i I let B i be an R-module. Prove the following isomorphisms of abelian groups; when R is commutative prove also that these are R-module isomorphisms. (Arbitrary direct sums and direct products of modules are introduced in Exercise 20 of Section 3.) (a) HomR ( I Bi , A) = I HomR (Bi , A) (b) HomR (A, I Bi ) = I HomR (A, Bi ).

∈

Solution :



∼



∼





Dummit & Foote Text Exercise 10.5-14: Let 0

φ

ψ

→ L → M → N → 0

be a sequence of R-modules. (a) Prove that the associated sequence 0

R (D,

→ Hom

L)

φ

R (D,

→ Hom

M )

ψ

R (D,

→ Hom

N )

→0

is a short exact sequence of abelian groups for all R-modules D if and only if the original sequence is a split short exact sequence. [To show the sequence splits, take D = N and show the lift of the identity map in HomR (N, N ) to HomR (N, M ) is a splitting homomorphism for φ .] (b) Prove that the associated sequence 0

R (N,

→ Hom

D)

ψ

R (M,

→ Hom

D)

φ

R (L, D)

→ Hom

→0

is a short exact sequence of abelian groups for all R-modules D if and only if the original sequence is a split short exact sequence. Solution :

Dummit & Foote Text Exercise 10.5-16: This exercise proves Theorem 38 that every left R-module M is contained in an injective left R-module. (a) Show that M is contained in an injective Z -module Q. [M is a Z -module use Corollary 37.] (b) Show that HomR (R, M ) HomZ (R, M ) HomZ (R, Q). (c) Use the R-module isomorphism M = HomR (R, M ) (Exercise 10) and the previous exercise to conclude that M is contained in an injective module.

⊆

Solution :

∼

⊆

Dummit and Foote Soln

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