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MAT7400 Assignment #3

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MAT7400 Assignment #3

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Dummit & Foote Text Exercise 3.4-2: Exhibit all 3 composition series for Q8 and all 7 composition series for D8 . List the composition factors in each case. Solution: In Exercise 3.1-32, we showed every subgroup of Q8 is normal. Of the normal subgroups, hii, hji, and hki are maximal. In addition, these three only have one non-trivial subgroup. Every quotient in each series has order two: Q8 /hii, hii/h−1i/, h−1i/1, Q8 /hji, hji/h−1i/, h−1i/1, Q8 /hki, hki/h−1i/, h−1i/1, So the every quotient is isomorphic to the simple abelian group Z/2Z. This gives us the following three composition series for Q8 : 1 C h−1i C hii C Q8 , 1 C h−1i C hji C Q8 , 1 C h−1i C hki C Q8 . Exercise 3.1-33 informed us that there are three maximal subgroups of D8 which are normal: hs, r2 i, hri, and hrs, r2 i. Each have order four. hs, r2 i has three subgroups of order two: hsi, hr2 i, hr2 si. Since each of these subgroups have index two, they are normal to hs, r2 i. hri has one subgroup of order two: hr2 i and since it’s index is two, it is normal in hri. hrs, r2 i has three subgroups of index two (and hence normal): hr2 i, hrsi, hr3 si. Again, every quotient is isomorphic to the simple abelian group Z/2Z. This gives us the following seven composition series for D8 : 1. 1 C hsi C hs, r2 i C D8 2. 1 C hr2 si C hs, r2 i C D8 3. 1 C hr2 i C hs, r2 i C D8 4. 1 C hr2 i C hri C D8 5. 1 C hr2 i C hrs, r2 i C D8 6. 1 C hrsi C hrs, r2 i C D8 7. 1 C hr3 si C hrs, r2 i C D8 Dummit & Foote Text Exercise 3.4-5: Prove that subgroups and quotient groups of a solvable group are solvable. Solution: A subnormal series of a group G is a chain of subgroups G = Gs > · · · G2 > G1 > G0 such that Gi E Gi+1 for all i. Let G be a solvable group. So G has a subnormal series 1 = G0 E G1 E · · · E Gs−1 E Gs = G with abelian factors: for each i = 1, . . . , s, Gi+1 /Gi is an abelian group. Let H ≤ G. Consider H ∩ Gi and H ∩ Gi+1 . Each is a group. If g ∈ H ∩ Gi and h ∈ H ∩ Gi+1 , then hgh−1 ∈ H since g, h ∈ H. Also, hgh−1 ∈ Gi . Since Gi E Gi+1 , H ∈ Gi C H ∩ Gi+1 . This also implies that H ∩ Gi+1 ≤ NG (H ∩ Gi ). Hence H is solvable and the composition (subnormal) series is 1 E H ∩ G1 E · · · E H ∩ Gs−1 E H ∩ Gs = H ∩ G = H. We can also use the second (diamond) isomorphism theorem to show this as well. Again let H ≤ G and for each i let Hi = Gi ∩ H. Then for every i, Gi1 E Gi , and we have the subnormal series 1 = H0 E H1 E · · · E Hs−1 E Hs = H. By the second (diamond) isomorphism theorem, for every i, Hi /Hi−1 = Hi /(Gi−1 ∩ H) ∼ = (Hi Gi−1 )/Gi−1 ,

MAT7400 Assignment #3

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which is a subgroup of the abelian group Gi /Gi−1 and therefore is abelian. Hence, H has a subnormal series with abelian factors and that implies H is solvable. (I had help with this part of problem.) Let K be a quotient group of G with K = G/N with N E G. Define π : G → K as the projection homomorphism. and last define Ki = π(Hi ). Then for each i, Ki−1 E Ki . We have now obtained the subnormal series: 1 = K0 E K1 E · · · E Ks−1 E Ks = K. Again by the second (diamond) isomorphism theorem we obtain Ki /Ki−1 = (Gi N )/(Gi−1 N ) ∼ = Gi /(Gi ∩ Gi−1 N ) which is a quotient of Gi /Gi−1 by the third isomorphism theorem since Gi−1 E Gi ∩ Gi−1 N . Thus, the quotient groups Ki /Ki−1 are abelian and therefore K is solvable.// Dummit & Foote Text Exercise 3.5-10: Find a composition series for A4 . Deduce that A4 is solvable. Solution: It is quite fortuitous for me as Prof. Okoh had a very similar problem assigned as an exam problem! Let H = {(), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}. We know that A4 = {(), (1, 2, 3), (1, 3, 2), (2, 3, 4), (2, 4, 3), (1, 2, 4), (1, 4, 2), (1, 3, 4), (1, 4, 3), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}. Since H is a finite and nonempty subset of A4 and closed under the binary operation, it is a subgroup of A4 . To show it is a normal subgroup. We need to show that aha−1 ∈ H for all h ∈ H and a ∈ A4 . Since conjugation preserves shapes (see Example 7.2.3 in Beachy and Blair’s text), aHa−1 preserve the cycle structure of H. Since H contains all of elements of A4 that have this cycle structure, aHa−1 = H. Hence H is normal in A4 . The definition of solvable is a chain of normal subgroups G = N0 ⊇ N1 ⊇ N2 ⊇ · · · ⊇ Nn = {e} such that (1) Ni E Ni−1 for i = 1, 2, 3, . . . n (2) Ni−1 /Ni is abelian for i = 1, 2, 3, . . . n (3) Nn = {e}. The chain to verify is A4 ⊇ H ⊇ {e} where G = N0 , A4 = N1 , H = N2 and N3 = {e}. (1) H E A4 (just proven) and {e} E H since {e} is the trivial normal subgroups to all subgroups. (2) By Lagrange, the factor group A4 /H 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), (12)(34), (13)(24), (14)(23)} = H. So H is the identity in the factor group and the identity commutes with itself. So H/{e} is abelian. (3) N3 = {e}. Therefore A4 is solvable.// Dummit & Foote Text Exercise 3.5-11: Prove that Σ4 has no subgroup isomorphic to Q8 . Solution: Assume a subgroup H ≤ Σ4 exists such that H ∼ = Q8 . Q8 contains 6 elements of order 4 namely (i, −i, j, −j, 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 is a subgroup, it is closed under the binary operation. That implies H 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 Q8 .//

MAT7400 Assignment #3

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

and

(−1)2 + (−1)

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

(−1) + (−1)2

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

Dummit & Foote Text Exercise 7.1-2: Prove that if u is a unit in R then so is −u. Solution: 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 T R be a ring and let X be a nonempty T set of subrings of R. T We know that X ⊆ R is a subgroup. To show that X is closed T under multiplication, letTa, 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. X is closed under addition. If ab , dc ∈ X, then b and d are odd. Therefore bd is odd. Next ab + dc = ad+bc bd 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 0 a a a 0 a 1 is the additive identity element for X since 1 + b = b = b + 1 for all b ∈ X. a −a a −a 0 0 −a a 0 0 For every b ∈ X, b ∈ X. Because b + b = b = 1 and b + b = b = 1 , every element of X has an additive inverse. So X is a group under addition. if f racab 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

6∈ 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. ( 31 )2 ∈ X but ( 31 )2 + ( 13 )2 = 29 6∈ X since 2 6∈ Q and therefore cannot be in X.

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(e) This set X is not closed under addition. 15 + 51 = 25 6∈ X but 51 ∈ X. ad−bc (f ) Let Y be the given set. If ab and dc are in lowest terms and a and c are even, then ab + −c d = bd . 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 a db = b , so that cab = dab. Thus c = d. Since c must be even and d odd, we have a contradiction. So Y is a 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 = a1˙ = 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 = ak 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 = ak b, where k ≥ 1. Then (ab)k = ak bk = (ak b)bk−1 = nbk−1 ≡ 0 mod n. (b) (⇒) Assume a ∈ Z/(n) is nilpotent. Then am = nk for some m and k. If p is a prime that divides n, then p divides am , so that it divides a. Hence every prime that divides n divides a. (⇐) Let n = pb11 · · · pbkk and a = pc11 · · · pckk m, where 1 ≤ bi , ci for all i and m is an integer. Let t = max{bi }. Then at = (pc11 · · · pckk m)t = pc1i t · · · pcki t mt , where ci t ≥ bi for each i. Hence at = nd for some integer d, and we have at ≡ 0 mod n. Since 72 has prime factorization 23 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}. (c) Proof by contradiction. Assume f ∈ R is nilpotent. If f 6= 0, then there exists x ∈ X such that f (x) 6= 0. Let m be the least positive integer such that f (x)m = 0. Then f (x)f (x)m−1 = 0 and f (x) 6= 0 and f (x)m−1 6= 0. We have our desired contradiction as F contains zero divisors. Therefore no nonzero element of R is nilpotent.//

MAT7400 Assignment #3

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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. Solution: (a) Let n be the least positive integer such that xn = 0. If n = 1, then x = 0. If n > 1, then x 6= 0, xn−1 6= 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 = rn xn = rn · 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 6= 0. We know that 1 = (1 − xn ) = (1 + x)(1 − x + x2 + · · · (−1)n−1 xn−1 ). We also know from the closure properties of a ring, (1 − x + x2 + · · · xn−1 ) ∈ 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.// Dummit & Foote Text Exercise Q 7.1-19: Let I be any nonempty index set and let Ri 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. Q Q Q Q IfQ( ai ), ), ( ci ) ∈Q I RQ i , then Q( biQ Q Q Q Q Q Q Q ( ai )(( bi )( ci )) = ( ai )( (bi ci )) = ai (bi ci ) = (ai bi )ci = ( ai bi )( ci ) = (( ai )( bi ))( ci ), so multiplication is associative. Q Q Q Q Q Q Q Q Q Next, bi ) + (Q ci )) = ( ai )( (bi + ci )) = ai (bi + ci ) = (ai bi + ai ci ) = ( ai bi ) + ( ai ci ) = Q ( Q ai )(( Q ( ai )( bi ) + ( ai )( ci ). So, multiplication distributes over addition on the left. Q Q Q Q Q Q Q Q Q ((Q ai )Q + bi )(Q ci ) = Q( (ai + bi ))( ci )) = (ai + bi )(ci ) = (ai ci + bi ci ) = ( ai ci ) + ( bi ci ) = ( ai )( ci ) + ( bi )( ci ). So, multiplication distributes over addition on the right as well. Q Thus I Ri is 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 6= 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 thatv 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 ∈ R and consider x − y. If x 6= 0 and y = 0, then v(x − y) = v(x) ≥ 0. If x = 0 and y 6= 0, then

MAT7400 Assignment #3

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v(x − y) = v(−y) = v(y) ≥ 0. If x = y = 0, then x − y = 0 ∈ R. If x, y 6= 0, then either x − y = 0 ∈ R or x − y 6= 0, and then v(x − y) ≥ min(v(x), v(−y)) = min(v(x), v(y)) ≥ 0. Now to check for multiplicative closure. If x, y ∈ R and xy = 0, then xy ∈ R. If xy 6= 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) = −v(u−1 ), and both v(u) and v(u−1 ) are nonnegative. Hence v(u) = v(u−1 ) = 0. (⇐) If v(u) = 0, then −0 = 0 = −v(u) = v(u−1 ), so that u−1 ∈ R and u ∈ R is a unit.// 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 a a c = α where = pα , p 6 |c and p 6 |d. vp : Q× → Z by vp 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 vp ( ab ) ≥ 0, and let a = pm c and b = pn d, where p does not divide c or d. By definition, vp ( ab ) = m − n ≥ 0. Take the case that ab is in lowest terms then either m or n is 0. If n 6= 0, then m = 0, and vp ( ab ) < 0, a contradiction. Hence n = 0 and gcd(b, p) = 1. ({a/b ∈ Q | a = 0 or gcd(b, p) = 1} ⊆ R). If ab ∈ Q and p does not divide b, then Hence vp (f racab) ≥ 0, and we have ab ∈ R.//

a b

= pα ) dc for some α ≥ 0.

Dummit & Foote Text Exercise 7.2-2 Let p(x) = an 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) = bm xm + bm−1 xm−l + · · · + b0 be a nonzero polynomial of minimal degree such that g(x)p(x) = 0. Show that bm an = 0 and so an g(x) is a polynomial of degree less than m that also gives 0 when multiplied by p(x). Conclude that an g(x) = 0. Apply a similar argument to show by induction on i that an−i g(x) = 0 for i = 0, 1, . . . , n and show that this implies bm 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. Pm (⇒) 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 ai q(x) = 0 for all 0 ≤ i ≤ n. I obtained help for this part of the problem.  Pn+m P The base case: p(x)q(x) = k=0 a b xk = 0. The coefficient of xn+m in this product is an bm i j i+j=k on the left side and 0 on the right side. Thus an bm = 0. Now an q(x)p(x) = 0, and the coefficient of xm in q is an bm = 0. Thus the degree of an 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, an q(x) = 0. So, an bi = 0 for all 0 ≤ i ≤ m. For the inductive step, suppose that for some 0 ≤ s < 4, we have ar q(x) = 0 for all s < r ≤ n. Now  Pn+m P P 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 P 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 as bm = 0. As in the base case, as q(x)p(x) = 0 and as q(x) has degree strictly less than that of q(x), so that by minimality, as q(x) = 0. By induction, ai q(x) = 0 for all 0 ≤ i ≤ n. In particular, ai bm = 0. Hence bm p(x) = 0.

MAT7400 Assignment #3

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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 ∞ X

an xn = a0 + 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 polynomial addition and multiplication to power series as though they were ”polynomials of infinite degree”: ∞ X

n

an x +

n=0 ∞ X

∞ X

∞ X

n

bn x =

n=0 ∞ X

an xn ×

n=0

(an + bn )xn

n=0

bn xn =

n=0

∞ n X X n=0

! xn .

ak bn−k

k=0

(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 1P− x is a unit in R[[x]] with inverse 1 + x + x2 + · · · . ∞ (c) Prove that n=0 an xn is a unit in R[[x]] if and only if a0 is a unit in R. P∞ P∞ P∞ Solution: (a) Let r = n=0 an xn , s = n=0 bn xn , and v = n=0 cn xn . Addition is Associative: P∞ P∞ P∞ (r + s) + v = (( n=0 an xn ) + ( n=0 bn xn )) + ( n=0 cn xn ) P∞

+ bn )xn ) + (

P∞

=

(

=

P∞

+ bn ) + cn )xn

=

P∞

+ (bn + cn ))xn

=

(

P∞

an xn ) + (

=

(

P∞

an xn ) + ((

n=0 (an

n=0 ((an n=0 (an

n=0

n=0

P∞

n=0 cn x

n=0 (bn

P∞

n

)

+ cn )xn )

n=0 bn x

n

)+(

P∞

n=0 cn x

= r + (s + v). Addition is Commutative: r+s =

∞ X

! an x

n

+

n=0

=

∞ X

∞ X

! bn x

n

an x

n

n=0

(an + bn )xn

n=0

=

∞ X

(bn + an )xn

n=0

=

∞ X n=0

=s+r P∞

n=0

0 · xn = 0 is the additive identity.

! bn x

n

+

∞ X n=0

!

n

))

MAT7400 Assignment #3

9

r+0

P∞

=

(

=

P∞

=

(

=

P∞

=

(

=

0+r

n=0

an xn ) + (

n=0 (an

P∞

n=0

n=0

n=0

0 · xn )

+ 0)xn

an xn ) = α

n=0 (0

P∞

P∞

+ an )xn

0 · xn ) + (

P∞

n=0

an xn )

P∞ R[[x]] contains left and right inverses. Define r ∈ R[[x]] as r = n=0 (−an )xn . Then we have P∞ P∞ r + r = ( n=0 an xn ) + ( n=0 (−an )xn ) =

P∞

=

P∞

=

0

=

P∞

=

P∞

=

(

n=0 (an

n=0

n=0

− an )xn

0 · xn

0 · xn

n=0 (−an

+ an )xn

P∞

n=0 (−an )x

n

)+(

P∞

n=0

an xn )

= r + r. 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: ! ∞ !! ∞ ! ∞ X X X n n n (rs)v = an x bn x cn x n=0

 = 

∞ X

ai bj  x



∞ X

∞ X

n



 X

 X



t+k=n

∞ X

cn x



ai bj  ck  xn

i+j=t



 X

!

n=0

X

 n=0

n=0



i+j=n

 n=0

=

 X

n=0

=

n=0



t+k=n i+j=t

ai bj ck  xn

n

MAT7400 Assignment #3

10

=

∞ X



∞ X

i+j+k=n



 X

X

ai bj ck  xn

 n=0

=

ai bj ck  xn

 n=0

=

 X

∞ X

i+s=n j+k=s



 X

ai 

 n=0

i+s=n

∞ X

=

=

! an xn

∞ X





 X

j+k=n

∞ X

! n

n=0



bj c k  x n 

 n=0

an x

bj ck  xn

j+k=s

n=0 ∞ X

 X

∞ X

! n

bn x

n=0

!! cn x

n

n=0

= r(sv). Distributivity: r(s + v)

=

∞ X

! an x

n

n=0

=

∞ X

=

! an x

n

=



ai bj + ai cj  xn

i+j=n



 X

∞ X

∞ X



X





ai bj  xn  + 

i+j=n

! an x

n=0

= rs + rv.

n

ai cj  xn

i+j=n





 X

ai bj  + 

i+j=n

 n=0

=

 X

n=0

= 

(bn + cn )x

ai (bj + cj ) xn

 

! n

i+j=n

n=0

=

∞ X

cn x

n=0

 X



∞ X

+

!! n

n=0



∞ X

bn x



n=0

∞ X

! n

n=0

n=0 ∞ X

∞ X

∞ X n=0

! n

bn x

+

∞ X



 X 

n=0

i+j=n

∞ X

n

n=0

! an x



ai cj  xn  ∞ X n=0

! cn x

n

MAT7400 Assignment #3

(r + s)v

11

∞ X

=

∞ X

! an x

n

+

n=0

=

∞ X

=

! (an + bn )x

n

=



 X

ai cj + bi cj  xn

i+j=n



 X

n=0

= 

∞ X

∞ X



 X

ai cj  + 

i+j=n

 X





ai cj  xn  + 

i+j=n ∞ X

! an x

bi cj  xn

i+j=n

 

n=0

=

(aj + bj )ci  xn

 

cn x

i+j=n

n=0

=

! n

 X



∞ X

cn x

n=0



∞ X

! n

n=0

∞ X



n=0

bn x

n=0

n=0 ∞ X

∞ X

!! n

n

! cn x

n

+



n=0

i+j=n

∞ X

n

! bn x

Hence multiplication distributes over addition on the left and right. If R is commutative for multiplication, then so is R[[x]]: ! ∞ ! ∞ X X n n rs = an x bn x

=

∞ X

n=0



∞ X

i+j=n



bj ai  xn

 ∞ X n=0

=

 X

n=0

=

ai bj  xn

 n=0

=

 X

sr.

j+i=n

! bn x

n

∞ X n=0

! an x

n



bi cj  xn  ∞ X n=0

= rv + sv.

n=0

 X

n=0

n=0

n=0



∞ X

! cn x

n

MAT7400 Assignment #3

12

P∞ If R has a multiplicative identity, then so does R[[x]]. Define 1 = n=0 en xn by e0 = 1 and ei+1 = 0. That gives us ! ∞ ! ∞ X X n n r·1 = an x en x n=0

=

∞ X

 X

ai ej  xn

 n=0

=

n=0



∞ X

i+j=n

an e0 xn

n=0

=

∞ X

an xn

n=0

= r 1·r

∞ X

=

! en x

n

n=0

=



=

∞ X

an x

 X

n=0

! n

n=0



∞ X

∞ X

ej ai  xn

i+j=n

e0 an xn

n=0

=

∞ X

an xn

n=0

= r Hence R[[x]] has the multiplicative identity 1. P∞ P∞ (b). Let 1 − x = n=0 bn xn where b0 = 1, b1 = −1, and bi = 0 for i ≥ 2. Let r = n=0 xn . Taking the product gives   ! ∞ ! ∞ ∞ X X X X  (1 − x)r = bn x n xn = bi  xn . n=0

So if n ≥ 1, then

n=0

n=0

i+j=n

P

di = 0, and d0 = 1. Hence (1 − x)r = 1. P∞ (c) (⇒) Suppose u ∈ R[[x]] is a unit, with inverse u−1 = n=0 an xn . So,   ! ∞ ! ∞ ∞ X X X X  1 = uu−1 = an xn an xn = ai aj  xn . i+j=n

n=0

n=0

n=0

i+j=n

0

The coefficient of x in this power series is 1 on the left hand side and a0 a0 on the right hand side. Thus a0 a0 = 1 ∈ R. We also have   ! ∞ ! ∞ ∞ X X X X  1 = u−1 u = an xn an xn = ai aj  xn . n=0

n=0

n=0

i+j=n

MAT7400 Assignment #3

13

Again, the coefficient of x0 in this power series is 1 on the left hand side and a0 a0 on the right hand side. Hence a0 a0 = 1 ∈ R and a0 is a unit in R. I had help on the following part (could not keep the indices on the consistent). Psums ∞ (⇐) Suppose a ∈ R is a unit, with a a = a a = 1. Define u = b xn with b0 = a0 and 0 0 0 0 0 n n=0 P bk+1 = −a0 i+j=k+1,j≤k ai bj . Then   ! ∞ ! ∞ ∞ X X X X  uu = an xn bn x n = a i bj  x n n=0

If n = 0,

P

i+j=n

n=0

n=0

ai bj = a0 a0 = 1. If n ≥ 1, then 

X i+j=n

i+j=n

ai bj = a0 bn +

X

 X

ai bj = a0 −a0

i+j=n,j 0. Since fc is continuous, there exists a positive real number εc > 0 such that fc [(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 uc (x) = 1 + (x − c)/εc if x ∈ (c − εc , c) ∩ [0, 1], uc (x) = 1 + (c − x)/εc if x ∈ [c, c + εc )P ∩ [0, 1], and uc (x) = 0 otherwise. Then uc ∈ R and uc is zero outside of (c − εc , c + εc ). Next consider g = uc fc . Since fc ∈ M , we have g ∈ M . However, for all x ∈ [0, 1], g(x) is positive since each uc (x)fc (x) is nonnegative and some uc (x)fc (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 = Mc for some c ∈ [0, 1]. (b) Assume b 6= c. Then x − b ∈ Mb but (x − b)(c) = c − b 6= 0. This implies x − b ∈ / Mc . Therefore, Mb 6= Mc . |x − c| . As (c) Assume Mc = (x − c). Then |x − c| = f (x)(x − c) for some f (x) ∈ R. If x 6= c, then f (x) = 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, Mc is not generated by x − c. (d) Proof by contradiction. Assume Mc = (A) is√finitely generated, √ √ withPA = {ai (x) | 1 ≤ i ≤ n}. Let P f= |ai |. Then f is continuous on [0, 1] and f ∈ M . Thus f = P ri ai for some continuous c √ P P functions ri ∈ 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 6= c, there must exist a function ai such that a (b) = 6 0, otherwise h(b) = 0 for all h ∈ Mp i c and √ x − c ∈ Mc . Thus c is the only zero of f . From f (x) ≤ r(x)f (x), for x = 6 c we have r(x) ≥ 1/ f (x). As x p 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 Mc is not finitely generated. 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

MAT7400 Assignment #3

25

is an ideal N ⊆ R such that N 6⊆ 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. Solution: R is a commutative ring with 1 6= 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 6= 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). 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 Ak is a principal ideal and that A0 ⊇ 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 A1 . Solution: (a) It is clear that Ak+1 ⊆ Ak for all k. Claim: Ak is an ideal. Since 0 ∈ Ak , Ak is nonempty. Let a, b ∈ Ak . If one or both of a and b is 0, then a + b ∈ Ak . If a + b = 0, then a + b ∈ Ak . If a + b 6= 0, we have v(a + b) ≥ min(v(a), v(b)) ≥ k, so that a + b ∈ Ak . Next, v(−a) = v(a) and that implies −a ∈ Ak . If r ∈ R then v(r) ≥ 0 and v(ra) = v(r) + v(a) ≥ k, so that ra ∈ Ak . Since K is commutative, Ak is an ideal of R. To show that Ak 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 Ak ⊆ (x), and thus Ak = (x). In particular, Ak 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 Ak = (a), we have Ak ⊆ I. Hence I = Ak . From part (a), every proper ideal of R is contained in A1 and therefore A1 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 F0 and that F0 is isomorphic to either Q or Z/pZ for some prime p (F0 is called the prime subfield of F ). [See Exercise 26, Section 3.]

MAT7400 Assignment #3

26

Solution: We need to make use of the following fact and we will prove it: T T Let F be a field and let X be a set of subfields of F . Then X is a subfield of F . Proof: X is a subring of F , contains 1 since 1 ∈ E for all E ∈ X . Let r ∈ F . Then r ∈ E for each E ∈ X , and so r−1 ∈ E. Then r−1 ∈ F , and hence F is a field. // T Let F be a field, and let F denote the set of all subfields of F . Then F0 = F is a subfield of F and is contained in every other subfield. Consider the ring homomorphism φ : Z → F0 with φ(1) = 1. The induced map ψ : Z/(n) → F0 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 F0 . Thus we have an monomorphism (injective homomorphism)Φ : F (Z/(n)) → F0 , 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 = F0 . Thus Φ is an isomorphism. If n = 0, then F0 ∼ = Q, and if n = p is prime, then F0 ∼ = Z/(p).// 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 Rx and M satisfy the two axioms in Section 1.7 for a group action of the multiplicative group Rx on the set M . Solution: 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 r1 , r2 ∈ R× , (r1 r2 ) · m = r1 · (r2 · m) and thereby satisfies the first axiom.// Dummit & Foote Text Exercise 10.1-5: For any left ideal I of R define X IM = { ai · mi | ai ∈ I, mi ∈ M finite

to be the collection of all finite elements of the form am where a ∈ I and m ∈ M . Prove that IM is a submodule of M . Solution: Apply the definition of the submodule criterion! Since 0 · 0 = 0 ∈ IM , IM is not empty. Let P P a · m , b · nk ∈ IM andP letr ∈ R. Then i i k i k P P P ( i ai · mi ) + r · ( k bk · nk ) = i ai · mi + k (rbi ) · nk ∈ IM . Hence IM ⊆ 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 = mk . 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 .

MAT7400 Assignment #3

27

Solution: I had help on this problem as I didn’t quite understand fully F [x] modules. Note that T : V → V is given by (x, y) 7→ (y, −x). Let N be an F [x]-submodule of V . Assume that there exists a nonzero element (a, b) ∈ N . This implies either a 6= 0 or b 6= 0 and therefore a2 + b2 > 0. Let (x, y) ∈ V , bx − ay ax + by , and r2 = 2 . Then r1 = 2 a + b2 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 bxa − aya + 2 , 2 − 2 a 2 + b2 a + b2 a + b2 a + b2

 =

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

=

 2  a + b2 b2 + a2 x 2 ,y a + b2 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 6= 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 6= 0 and there exists (0, b) ∈ N with b 6= 0. Then (a, b) ∈ N with a, b 6= 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 6= 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) = p0 + xp0 (x). Then x · (b, 0) = (0, 0) and (a, 0) + p(x) · (b, 0)

= = (a, 0) + p0 · (b, 0) + p0 (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 P 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, xk · v = x · v for all k ≥ 1. If v ∈ Y , then T (v) = v. We obtain the following:

MAT7400 Assignment #3

28

(0, a) + p(x) · (0, b)

P = (0, a) + (P pi xi ) · (0, b) = (0, a) + P pi · (xi · (0, b)) = (0, a) +P pi · (0, b) = (0, a + pi b) ∈ Y.

So, 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) ∼ P . Hence the relation is transitive. So σ ◦ φ : M → P is an R-module isomorphism, and we have M = 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

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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 6= 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 t1 , 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 xn ◦ 1 = tn , let xn ◦ ti = tn+i for i = 1, 2, . . . and let xn 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 xp+q ◦ ti = xp ◦ (xq ◦ ti ) = tp+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. Solution: (a) xp+q ◦ ti = tp+q+i = xp ◦ (tq+i ) = xp ◦ (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) 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 = t2 but φ(x2 ) = φ(x)φ(x) = (x · 1)(x · 1) = t1 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.

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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|. // ∼ Rm if and only Dummit & Foote Text Exercise 10.3-2: Assume R is commutative. Prove that Rn = 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 Fn ∼ = 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 Rn ∼ = Rm . Let I be a maximal ideal of R. Then Rn /IRn ∼ = Rm /IRm . Exercise (R/IR) × (R/IR) × · · · (R/IR) (n times). This implies 10.2-12 (not assigned) tells that Rn /IRn ∼ =R m (R/I)n ∼ (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 Rm ∼ = Rn , where Rφ : Rm → Rn is an R-module isomorphism. Let I ⊆ R be a maximal ideal. Next π ◦ φ : Rm → Rn /IRn , where π denotes the natural projection. Claim: ker π ◦ φ = IRm . P P P (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 IRm ⊆ ker π ◦ φ. (ker π ◦P φ ⊆ IRm ): Assume x = (ri ) ∈ ker π ◦ φ. Then 0 = (π ◦ φ)(x) = φ(x) + IRn . Thus φ(x) ∈ IRn . Let m φ(x) = ai · (ri,j Rn . Since φ is surjective, we have (ri,j ) = φ((s P), where (ri,j ) ∈ P Pi,j )) for some (si,j ) ∈ R . 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 /IRn 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 ). // Q 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 .

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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. P (M/N ) ⊇P(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). P P M ⊆ (ai , bjP| 1 ≤ i ≤ m, 1 ≤ j ≤ n)) Assume m P ∈ M . ThenP m + N = i ri · (ai + N ) = ( i ri · ai ) + N . ThenP m − i ri ·P ai ∈ N and that implies m − i ri · ai = j sj 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 6= 0 and if 0 and M are the only submodules of M . Show that M is irreducible if and only if M 6= 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 6= 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 6= 0 and that if x 6= 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 7→ 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 → M byφm (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) = φm (x) + r · φm (y). So φm is an R-module homomorphism. Since m 6= 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 = φm (a). Hence φm is surjective. We need to show that ker φm is a maximal ideal. Let x + ker φm be nonzero. Then x · m 6= 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 φm . That is, (y + ker φm )(x + ker φm ) = 1 + ker φm . So every nonzero element of R/ker φm has a left inverse. Because R/ker φm is commutative, R/ker φm is a field, so that ker φm is a maximal ideal of R. ∼R R/ker φm , where ker φm is a maximal ideal. By the First Isomorphism Theorem, =

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(⇐) 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 M1 and M2 are irreducible R-modules, then any nonzero R-module homomorphism from M1 to M2 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 : M1 → M2 is an R-module homomorphism with M1 and M2 irreducible. In Exercise 10.2-1, we proved that ker φ and im φ are submodules of M1 and M2 respectively. Since φ is not the zero homomorphism, its kernel is not all of M1 , and since M1 is irreducible ker φ = 0. Thus φ is injective. Similarly, imφ is not the zero submodule, and consequently M2 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 × B → M by φ(f, g)(a, b) = f (a) + g(b). Then for all a1 , a2 ∈ A, b1 , b2 ∈ B, and r ∈ R, we have φ(f, g)((a1 , b1 ) + r · (a2 , b2 ))

= = = =

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

So φ(f, g) : A × B → M is an R-module homomorphism. Now consider the map Φ : HomR (A, M ) × HomR (B, M ) → HomR (A × B, M ). We will show that Φ is an R-module homomorphism. Let f1 , f2 : A → M and g1 , g2 : B → M be R-module homomorphisms and let r ∈ R. Then Φ((a1 , b1 ) + r · (a2 , b2 ))(a, b) = Φ(f1 + r · f2 , g1 + r · g2 )(a, b) = (f1 + r · f2 )(a) + (g1 + r · g2 )(b) = f1 (a) + r · f2 (a) + g1 (b) + r · g2 (b) = Φ(f1 , g1 )(a, b) + r · Φ(f2 , g2 )(a, b) = (Φ(f1 , g1 ) + r · Φ(f2 , 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 fτ : A → M and gτ : B → M by fτ (a) = τ (a, 0) and gτ (b) = τ (0, b). Then fτ (a1 + r · a2 )

= = =

τ (a1 + r · a2 , 0 τ (a1 , 0) + r · τ (a2 , 0) fτ (a1 ) + r · fτ (a2 ).

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So fτ is an R-module homomorphism. Similarly, gτ is an R-module homomorphism. Since Φ(fτ , gτ )(a, b)

= fτ (a) + gτ (b) = τ (a, 0) + τ (0, b) = τ (a, b),

we have Φ(fτ , gτ ) = τ . Hence Φ is a surjection. Thus, we have HomR (A × B, M ) ∼ =R HomR (A, M ) × HomR (B, M ). // (b) Now let πA : A × B → A and πB : 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)

= (πA ◦ (φ + r · ψ), πB ◦ (φ + r · ψ)) = (πA ◦ φ + r · (πA ◦ ψ), πB ◦ φ + r · (πB ◦ ψ)) = (πA ◦ φ, πA ◦ ψ) + r · (πB ◦ φ, πB ◦ ψ) = Ψ(φ) + r · Ψ(ψ).

Thus Ψ is an R-module homomorphism. We now show that Ψ is an injection. Assume φ ∈ ker Ψ. If m ∈ M , then 0 = Ψ(φ)(m, m) = ((πA ◦ φ)(m), (πb ◦ ψ)(m)) = (πA (φ(m)), πB (φ(m)) = φ(m). 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 φf,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 HomR (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. n Solution: Let F be a free R-module of finite P rank. That is, let F be free on the set {ai }i=1 . Then every element P of F can be written uniquely as ri · ai for some ri ∈ 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

Φ(φ + r · ψ)

P = P(φ + r · ψ)(ai ) · aiP = ( φ(ai ) · ai ) + r · ( ψ(ai ) · ai ) = Φ(φ) + r · Φ(ψ)

shows that Φ is an R-module homomorphism. P Φ is injective. Assume φ ∈ ker Φ. Then 0 = Φ(φ) = φ(ai ) · ai . Since F is free on the ai , we have φ(ai ) = 0 for all ai , and thus φ = 0. So ker Φ = 0, and thus Φ is injective. P P P Φ is surjective. Let ti · ai ∈ F . Define g : F → R by P g( ri · ai ) = ri ti . So , g is an R-module homomorphism. Moreover, g(ai ) = ti , 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 Mi be an R-module. The direct product of the modules Mi 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 Mi is defined to be the restricted direct product of the abelian groups Mi (cf. Exercise 17 in Section 5.1) with the action of R componentwise multiplication. In other words, the direct sum of the Mi ’s Q Q is the subset of the direct product, i∈I Mi , which consists of all elements i∈I mi such that only finitely many of the components mi are nonzero; the action of R on the direct product or direct sum is given by

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Q Q r i∈I mi = i∈I rmi (cf. Appendix I for the definition of Cartesian products of infinitely many sets). The L direct sum will be denoted by i∈I Mi . (a) Prove that the direct product of the Mi ’s is an R-module and the direct sum of the Mi ’s is a submodule of their direct product. (b) Show that if R = Z, I = Z+ and Mi is the cyclic group of order i for each i, then the direct sum of the Mi ’s is not isomorphic to their direct product. [Look at torsion.] Q Solution: (a) We know from Exercise 5.1-15, that i∈I Mi is an abelian group under point-wise addition and multiplication. So we just need Q to verify the three left-module axioms. Let r, s ∈ R and let (mi ), (ni ) ∈ i∈I Mi . Then (r + s) · (mi )

(rs) · (mi )

r · ((mi ) + (ni ))

= ((r + s) · mi ) = (r · mi + s · mi ) = (r · mi ) + (s · mi ) = r · (mi ) + s · (mi ), = ((rs) · mi ) = (r · (s · mi )) = r · (s · mi ) = r · (s · (mi )), = = = = =

r · (mi + ni ) (r · (mi + ni )) (r · mi + r · ni ) (r · mi ) + (r · ni ) r · (mi ) + r · (ni ).

Q Thus i∈I MiQis a left R-module. If R has a 1 and each Mi is unital, then 1 · (mi ) = (1 · mi ) = (mi ), for all (mi ), so that I Mi is unital. L Q L To show that i∈I M we apply the submodule criterion. i∈I Mi is Li is a submodule of i∈I Mi ,L nonempty since 0 ∈ i∈I Mi . Now let (xi ), (yi ) ∈ i∈I Mi and let r ∈ R. Let J, K ⊆ I such that xj = 0 for j ∈ / J and yk = 0 for k ∈ / K. Consider L (xi ) + r · (yi ) = (xi + r · yi ). If i ∈ / J ∪L K, then xi + Qr · yi = 0. Since J ∪ K ⊆ I is finite, (xi ) + r · (yi ) ∈ i∈I Mi . By the submodule criterion, i∈I Mi ⊆ i∈I Mi is an R-submodule. L Q (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, QL 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 6= 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 {Fi }I be a family of free unital Sleft R-modules. Let Ai = {ai,j } be a free basis for each Fi and assume that the Ai are disjoint. Let A = I Ai . Let F (A) denote the free R-module on A. L There is the natural inclusion A → F (A) and a natural inclusion ι : A → I Fi , where ai,j 7→ (bt ), where bt = ai,j if t = i and 0 otherwise. L By the universal property of free modules, there is a unique R-module homomorphism Ψ : F (A) → I Fi 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. P P Ψ is injective. Assume x ∈ ker Ψ and let x = ri,j ai,j . Then 0 = Ψ(x)i = j ri,j ai,j ∈ Fi . Since Fi is free on Ai , ri,j = 0 for all i and j. This implies x = 0 and ker Ψ = 0. Hence Ψ is injective.

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P L Ψ ri,j ai,j ) ∈ I FP i . Since only finitely many of the ri,j are nonzero, P is surjective. Assume ( jP r a ∈ F (A). Then Ψ( r a ) = ( i,j i,j i,j i,j i,j i,j j ri,j ai,j ), so that Ψ is surjective. L L ∼ Therefore I Fi =R F (A) and I Fi is free as an R-module. Dummit & Foote Text Exercise 10.3-24: (An aribitrary direct product of free modules need not Q be free) For each positive integer I, let mi be the free Z-module Z, and let M be the direct product Z+ Mi (cf. Exercise 20). Each element of M can be written uniquely in the form (a1 , a2 , a3 , . . . ) with ai ∈ Z for all i. Let N be the submodule of M consisting of all such tuples with only finitely many nonzero ai . Assume M is a free Z-module with basis B. (a) Show that N is countable. (b) Show that there is a countable subset B1 of B such that N is contained in the submodule N1 generated by B1 . Show also that N1 is countable. (c) Let M = M/N1 . 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 S = {(b1 , b2 , b3 , . . . ) | bi = ±i! for all i}. Prove that S is uncountable. Deduce that there exists some s ∈ S such that s ∈ / N1 . (e) Show that the assumption M is free leads to a contradiction: By (d) we may choose s ∈ S with s ∈ / N1 . Show that for each positive integer k there is some m ∈ M with s = km, contrary to (c). [Use the fact that N ⊆ N1 .] 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 ιn : Zn → N given bySιn (a)i = ai+1 if 0 ≤ i < n and 0 otherwise. Each element of N is in the image of some ιn . So, N = n∈N+ im ιn . Hence, N is a countable union of countable sets, and therefore is countable. P S (b) For each x ∈ N , there is a finite subset Bx ⊆ B such that x = P B1 = x∈N Bx . As the b∈Bx rb · b. LetP countable union P of finite sets, B1 is countable. Since x ∈ N and x = b∈Bx rb · b ∈ b∈B1 Rb, N ⊆ N1 = b∈B1 Rb and N1 is also countable. (c) The natural inclusion ι : B − B1 → M/N1 is given by b 7→ b + N1 . By the universal property of free modules, there is a unique R-module homomorphism Φ : F (B − B1 ) → M/N1 such that Φ(b) = b + N1 for all b ∈ B − B1 . Now to show that Φ is an isomorphism. P P P Φ is injective. Let x ∈ ker P hi and let x = ri bi . Then 0 = ri bi + N1 , so that ri bi ∈ N1 . Thus ri = 0, and we have x = 0. So Φ is injective. P Φ is surjective. Assume xP+ N1 ∈ M/N1 . Since M is free on B, we have x + N1 = ri bi + N1 for some ri and bi ∈ B − B1 . Then Φ( ri bi ) = x and that implies Φ is surjective. Therefore Φ is an R-module isomorphism, and M/N1 is free as a Z-module. Q (d) There exists a bijection φ : S → N {1, −1} = T . That is bi 7→ +1 if bi > 0 and bi 7→ −1 if bi < 0. A 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 S is uncountable. Since N1 is countable, there must exist s ∈ S such that s ∈ / N1 . (e) Let k ∈ Z be positive. Define m ∈ M by mi = i! if i! < k and mi = σi i!/k if i ≥ k, where σi = 1 if si > 0 and −1 if si < 0. Now (s − km)i = 0 for all i > k, so that s − km ∈ N ⊆ N1 . Thus s + N1 = k(m + N1 ). This is a contradiction of part (d) above. Thus M is not free as a Z-module. Dummit & Foote Text Exercise 10.3-25: In the construction of direct limits, Exercise 8 of Section 7.6, show that if all Ai are R-modules and the maps ρij 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 : Ai → A are all R−module homomorphisms. Verify the corresponding universal property (part (e)) for R- module homomorphisms φ : Ai → C commuting with the ρij .

MAT7400 Assignment #3

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Solution: ? 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 ⊗ 1” is 0 in Z ⊗Z Z/2Z but non-zero in 2Z ⊗Z Z/2Z. Solution: Dummit & Foote Text Exercise 10.4-3: Show that C ⊗R C and C ⊗C C are both left R-modules but are not isomorphic as R-modules. Solution: Dummit & Foote Text Exercise 10.4-4: Show that Q ⊗Z Q and Q ⊗Q Q are isomorphic left Q-modules. [Show they are both 1-dimensional vector spaces over Q.] 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 a1 b1 + · · · an bn is a bilinear map from Rn × 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 Ni be a left R-module. Let M be a right R-module. Prove the group isomorphism : M ⊗ (⊕i∈I Ni ) ∼ = (M ⊗ Ni ), 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 Mi = 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 IEO has no nonzero torsion elements (i.e., rm = 0 with 0 6= r ∈ R and m ∈ I ⊗R I implies that m = 0). Solution:

MAT7400 Assignment #3

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Dummit & Foote Text Exercise 10.4-19: Let I = (2, x) be the ideal generated by 2 and x in the ring R = Z[x] as in Exercise 17. Show that the nonzero element 2 ⊗ x − x ⊗ 2 in I ⊗R I is a torsion element. Show in fact that 2 ⊗ x − x ⊗ 2 is annihilated by both 2 and x and that the submodule of I ⊗R I generated by 2 ⊗ x − x ⊗ 2 is isomorphic to R/I. Solution: Dummit & Foote Text Exercise 10.4-21: Suppose R is commutative and let I and J be ideals of R. (a) Show there is a surjective R-module homomorphism from I ⊗R J to the product ideal IJ mapping i ⊗ j to the element ij. (b) Give an example to show that the map in (a) need not be injective (cf. Exercise 17). Solution: Dummit & Foote Text Exercise 10.4-24: Prove that the extension of scalars from Z to the Gaussian integers Z[i] of the ring R is isomorphic to C as a ring: Z[i]]] ⊗Z R ∼ = C as rings. Solution: Dummit & Foote Text Exercise 10.5-1: . Solution: Dummit & Foote Text Exercise 10.5-2: . Solution: Dummit & Foote Text Exercise 10.5-3: Let P1 and P2 be R-modules. Prove that P1 ⊕ P2 is a projective R-module if and only if both P1 and P2 are projective. Solution: Dummit & Foote Text Exercise 10.5-4: Let Q1 and Q2 be R-modules. Prove that P1 ⊕ P2 is an injective R-module if and only if both Q1 and Q2 are injective. Solution: Dummit & Foote Text Exercise 10.5-5: Let A1 and A2 be R-modules. Prove that A1 ⊕ A2 is a P flat R-module if and only if both A1 and A2 are flat. More generally, prove that an arbitrary direct sum Ai of R-modules is flat if and only if each A is flat. [Use the fact that tensor product commutes with arbitrary direct sums.] Solution: Dummit & Foote Text Exercise 10.5-6: Prove that the following are equivalent for a ring R: (i) Every R-module is projective. (ii) Every R-module is injective. Solution:

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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 Bi 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 L are introduced Q in Exercise 20 of Section 3.) Q Q (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 φ

ψ

0→L→M →N →0 be a sequence of R-modules. (a) Prove that the associated sequence φ0

ψ0

0 → HomR (D, L) → HomR (D, M ) → HomR (D, 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 ψ0

φ0

0 → HomR (N, D) → HomR (M, D) → HomR (L, D) → 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:

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Dummit & Foote Text Exercise 10.5-19: If F is a field, prove that the injective hull of F is F . Solution: Dummit & Foote Text Exercise 10.5-22: Suppose that R is a commutative ring and that M and N are flat R-modules. Prove that M ⊗R N is a flat R-module. [Use the previous exercise.] Solution: Dummit & Foote Text Exercise 10.5-26: Suppose R is a P.I.D. This exercise proves that A is a flat R-module if and only if A is torsion free R-module (i.e., if a ∈ A is nonzero and r ∈ R, then ra = 0 implies r = 0). (a) Suppose that A is flat and for fixed r ∈ R consider the map ψr : R → R defined by multiplication by r : ψr (x) = rx. If r is nonzero show that ψr is an injection. Conclude from the flatness of A that the map from A to A defined by mapping a to ra is injective and that A is torsion free. (b) Suppose that A is torsion free. If I is a nonzero ideal of R, then I = rR for some nonzero r ∈ R. Show φ ι that the map ψr in (a) induces an isomorphism R ∼ = I of R modules and that the composite R → I → R of ψr with the inclusion ι : I ⊆ R is multiplication by r. Prove that the composite 1⊗ψr 1⊗ι A ⊗R R → A ⊗R I → A ⊗R R corresponds to the map a 7→ ra under the identification A ⊗R R = A and that this composite is injective since A is torsion free. Show that 1 ⊗ ψr is an isomorphism and deduce that 1 ⊗ ι is injective. Use the previous exercise to conclude that A is flat. Solution: Dummit & Foote Text Exercise : . Solution: Dummit & Foote Text Exercise : . Solution: Dummit & Foote Text Exercise : . Solution: