HOMEWORK 1 SOLUTIONS

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1 HOMEWORK 1 SOLUTIONS MATH 121 Problem (10.1.2). Prove that R and M satisfy the two axioms in Section 1.7 for a group action of the multiplicative group R on the set M. Solution. For the rst axiom, we have to check that for r 1, r 2 R and m M, we have r 1 (r 2 m) = (r 1 r 2 )m. For the second axiom, we have to check that for m M, 1m = m. Both of these are included as part of the denition of a module. Problem (10.1.4). Let M be the module R n described in Example 3 and let I 1, I 2,..., I n be left ideals of R. Prove that the following are submodules of M: (a) {(x 1, x 2,..., x n ) x i I i }. (b) {(x 1, x 2,..., x n ) x i R and x 1 + x x n = 0}. Solution. (a) Call the set N. We have to check that N is a subgroup, and that for r R and x N, rx N. Let x = (x 1,..., x n ), y = (y 1,..., y n ) N. Then x + y = (x 1 + y 1,..., x n + y n ) N since the I i 's are ideals and hence closed under sums. Now, let x = (x 1,..., x n ) N and r R. Then rx = (rx 1,..., rx n ) N, again because the I i 's are ideals. Hence N is a submodule. (b) Call this set P. As in part (a), we check that P is a subgroup and is closed under scalar multiplication. Let x = (x 1,..., x n ), y = (y 1,..., y n ) P and r R. Then x + y = (x 1 + y 1,..., x n + y n ). From the above, we know that this is in N, so we need to check that (x 1 + y 1 )+ +(x n +y n ) = 0. This is true because x 1 + +x n = y 1 + +y n = 0. Similarly, rx = (rx 1,..., rx n ) N, and (rx 1 ) + + (rx n ) = r(x x n ) = 0, so rx P. Hence P is a submodule. Date: 10 January,

2 2 MATH 121 Problem (10.1.9). If N is a submodule of M, the annihilator of N in R is dened to be {r R rn = 0 for all n N}. Prove that the annihilator of N in R is a 2-sided ideal of R. Solution. Let ann(n) be the annihilator of N in R. Suppose r ann(n), a R, and n N. We have to show that ar, ra ann(n). We have (ar)(n) = a(rn) = 0 because rn = 0, and (ra)(n) = r(an) = rn = 0 for some n N. Hence ann(n) is a 2-sided ideal of R. Problem ( ). If I is a right ideal of R, the annihilator of I in M is dened to be {m M am = 0 for all a I}. Prove that the annihilator of I in M is a submodule of M. Solution. Let ann(i) be the annihilator of I in M. Suppose that m, m ann(i), r R, and a I. Then we have so m + m ann(i). Furthermore, a(m + m ) = am + am = 0, a(rm) = (ar)m = a m = 0 for some a I. Hence rm ann(i). Thus ann(i) is a submodule of M. Problem ( ). Let M be the abelian group (i.e., Z-module) Z/24Z Z/15Z Z/50Z. (a) Find the annihilator of M in Z (i.e., a generator for this principal ideal). (b) Let I = 2Z. Describe the annihilator of I in M as a direct product of cyclic groups. Solution. (a) An element a Z annihilates M if and only if am = 0 for all m in a set of generators of M. One choice of generators of M is (1, 0, 0), (0, 1, 0), and (0, 0, 1). Hence, ann(m) is the intersection of the annihilators of the submodules generated by these elements. Clearly, the annihilators of these submodules are (24), (15), and (50), respectively, so their intersection is (lcm(24, 15, 50)) = (600). (b) An element m = (m 1, m 2, m 3 ) of M is annihilated by I if and only if (2m 1, 2m 2, 2m 3 ) = 0. This happens if and only if 2m 1 24Z/24Z, 2m 2 15Z/15Z, and 2m 3 50Z/50Z, or, equivalently, if m 1 12Z/24Z, m 2 15Z/15Z, and m 3 25Z/50Z. Hence ann(i) = 12Z/24Z 15Z/15Z 25Z/25Z.

3 HOMEWORK 1 SOLUTIONS 3 Problem (10.2.5). Exhibit all Z-module homomorphisms from Z/30Z to Z/21Z. Solution. A Z-module homomorphism from a cyclic module to any module is determined by where a generator is sent. (See Problem below.) Let φ : Z/30Z Z/21Z be a Z-module homomorphism. Then we must have 30φ(1) = 0. The elements y Z/21Z so that 30y = 0 are y = 7k (mod 21) for k = 0, 1, 2, so there are three such homomorphisms, given by 1 7, 1 14, and 1 0. Problem (10.2.9). Let R be a commutative ring. Prove that Hom R (R, M) and M are isomorphic as left R-modules. [Show that each element of Hom R (R, M) is determined by its value on the identity of R.] Solution. We dene a map ψ : Hom R (R, M) M given by φ φ(1). Let's show that this is a homomorphism of left R-modules. For φ 1, φ 2 Hom R (R, M), we have ψ(φ 1 + φ 2 ) = (φ 1 + φ 2 )(1) = φ 1 (1) + φ 2 (1) = ψ(φ 1 ) + ψ(φ 2 ), and for φ Hom R (R, M) and r R, we have ψ(rφ) = (rφ)(1) = rφ(1) = rψ(φ), as desired. Now let's show that ψ is injective. Suppose that ψ(φ) = 0. This means that φ(1) = 0. Now, let r R. We have φ(r) = rφ(1) = r0 = 0 since φ is an R-module homomorphism. Hence φ = 0, as desired. Finally, we show that ψ is surjective. Pick m M. We need to nd φ Hom R (R, M) so that φ(1) = m. We dene φ(r) = rm, but we need to check that this is actually an R-module homomorphism. Let r 1, r 2 R. We have φ(r 1 + r 2 ) = (r 1 + r 2 )m = r 1 m + r 2 m = φ(r 1 ) + φ(r 2 ) and φ(r 1 r 2 ) = (r 1 r 2 )m = r 1 (r 2 m) = r 1 φ(r 2 ). Hence φ is indeed an R-module homomorphism, and so ψ is an isomorphism. Problem ( ). Let R be a commutative ring. Prove that Hom R (R, R) and R are isomorphic as rings. Solution. We showed in Problem that Hom R (R, R) and R are isomorphic as R-modules, so it suces to check that our map ψ from that problem is in fact a ring map. Hence, we have to check that ψ(φ 1 φ 2 ) = ψ(φ 1 )ψ(φ 2 ) and ψ(1) = 1. For the rst one, we have ψ(φ 1 φ 2 ) = (φ 1 φ 2 )(1) = φ 1 (φ 2 (1)) = φ 1 (1)φ 2 (1) = ψ(φ 1 φ 2 ). For the second, we have ψ(id) = id(1) = 1, as desired. Thus ψ is a ring isomorphism.

4 4 MATH 121 Problem ( ). Let A 1, A 2,..., A n be R-modules and let B i be a submodule of A i for each i = 1, 2,..., n. Prove that (A 1 A n )/(B 1 B n ) = (A 1 /B 1 ) (A n /B n ). [Recall Exercise 14 in Section 5.1.] Solution. In Problem 5.1.4, we showed that there is an isomorphism (A 1 A n )/(B 1 B n ) = (A 1 /B 1 ) (A n /B n ) of groups. We must now show that it is compatible with the R-module structure. Recall that the map is given by φ((a 1,..., a n ) mod B 1 B n ) = (a 1 mod B 1,..., a n mod B n ). Now, let r R and (a 1,..., a n ) A 1 A n. We have φ(r(a 1,..., a n ) mod B 1 B n ) = φ((ra 1,..., ra n ) mod B 1 B n ) = (ra 1 mod B 1,..., ra n mod B n ) = r(a 1 mod B 1,..., a n mod B n ) = rφ((a 1,..., a n ) mod B 1 B n ) Problem ( ). 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. Let n N. By hypothesis, we can nd some m M so that ϕ(m) = n + a i n i, where a i I and n i N. For each i, we can nd some m i so that ϕ(m i) = n i + a ijn ij. Hence ( ϕ m ) a i m i = n i,j a i a ijn ij N + I 2 N. Hence we've shown that the induced map M N/I 2 N is surjective. Continuing in the same manner, we can see that all the induced maps M N/I k N are surjective. Since some I r = 0, this shows that the original map ϕ : M N is surjective. Note that we didn't use the fact that I is nilpotent until the last step. Consider the more down-to-earth example that sets aside that hypothesis. Let R = Z, M = N = Z, I = (3), and suppose that ϕ : M N is multiplication by 5. Now, the induced map M/IM N/IN is surjective, which means that every integer is a multiple of 5 up to a multiple of 3. For example, if n = 4, then n is a multiple of 5 up to a multiple of 3 because = 10 is a multiple of 5. The argument above shows that we

5 HOMEWORK 1 SOLUTIONS 5 can improve the situation to saying that 4 is a multiple of 5 up to a multiple of 9, which is true because = 40 is a multiple of 5. Similarly, 4 is a multiple of 5 up to a multiple of 27, because = 85 is a multiple of 5, and so on. The only dierence here is that we cannot conclude that ϕ is actually surjective because no power of I is 0. Problem (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 nite abelian group is a torsion Z-module. Give an example of an innite abelian group that is a torsion Z-module. Solution. Suppose that A is a nite abelian group (or Z-module) of order n. Then by Lagrange's Theorem, for each a A, na = 0. Hence A is a torsion Z-module. An example of an innite abelian group that is a torsion Z-module is (Z/2Z). i=1 Problem (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 and M is a cyclic module with any nonzero element as its generator. Determine all irreducible Z-modules. Solution. Suppose that M is an irreducible R-module. Let m be a nonzero element of M. Then Rm is a submodule of M, so by hypothesis, we must have M = Rm. Hence, M is cyclic, and any nonzero element of M si a generator. For the other direction, suppose M is a cyclic module, and that any nonzero element is a generator. Now suppose that N is a nonzero submodule of M. Then for any nonzero n N, Rn must be all of M, because n is also a nonzero element of M. As any submodule of M contains some Rn, M must be irreducible. The irreducible Z-modules, then, are exactly the cyclic groups of prime order. Problem ( ). 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 End R (M) is a division ring (this result is called Schur's Lemma.) [Consider the kernel and image.] Solution. Let φ : M 1 M 2 be a nonzero R-module homomorphism between two irreducible R-modules. Then ker φ and im φ are submodules of M 1 and M 2, respectively, so we must have ker φ = 0 and im φ = M 2. Hence φ is an isomorphism. Now, let φ End R (M) be nonzero. By the previous argument, φ is an automorphism, so it has an inverse. Since φ was arbitrary, every nonzero element of End R (M) is invertible, so End R (M) is a division ring.

6 6 MATH 121 Problem ( ). Let I be a nonempty index set and for each i I let M i be an R-module. The direct product of the modules M i is dened 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 dened 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, prod i I M i, which consists of the elements i I m i such that only nitely many of the components m i are nonzero; the action of R on the direct product or direct sum is given by r i I m i = i i rm i (cf. Appendix I for the denition of Cartesian products of innitely 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.] Solution. and (a) We have (m i ) i I + (m i) i I = (m i + m i) i I M i r(m i ) i I = (rm i ) i I M i, so M i is an R-module. To show that M i is a submodule of M i, we must show that if all but nitely many of the m i and m i are zero, then this is also true of the m i + m + i and rm i. For the sum, there are only nitely many indices i so that at least one of m i and m i is nonzero, so there can only be nitely many indices in which the sum is nonzero. Similarly, for the scalar product, there are only nitely many i in which m i is nonzero, and these are the only positions in which rm i can be nonzero. Hence the direct sum is indeed a submodule of the direct product. (b) One way to do this is to observe that M i is a torsion group, whereas M i is not. To see that M i is torsion, let (m i ) be any element of M i, and suppose that n is large enough so that m i = 0 for i > n. Then n!(m i ) = 0, so (m i ) is a torsion element. Since (m i ) was arbitrary, M i is a torsion group. Now, let's illustrate a non-torsion element of M i. Take x = (1, 1, 1,...) = M i. Suppose that nx = 0. Then in order to annihilate the 1 in the i th place, n must be a multiple of i. Hence, n must be a multiple of i for each i Z +. The only integer with this property is 0, so x is torsion-free. Another way to do this is to note that M i is countable (as a set), whereas Mi is uncountable.

7 HOMEWORK 1 SOLUTIONS 7 Problem ( ). Let R be a Principal Ideal Domain, let M be a torsion R-module (cf. Exercise 4) and let p be a prime in R (do not assume M is nitely generated, hence it need not have a nonzero annihilator cf. Exercise 5). The p-primary component of M is the set of all elements of M that are annihilated by some positive power of p. (a) Prove that the p-primary component is a submodule. [See Exercise 13 in Section 1.] (b) Prove that this denition of p-primary component agrees with the one given in Exercise 18 when M has a nonzero annihilator. (c) Prove that M is the (possibly innite) direct sum of its p-primary components, as p runs over all primes of R. Solution. (a) Let M[p] denote the p-primary component of M. Suppose that m, n M[p] and r R. Suppose that m and n are annihilated by p a and p b, respectively. Then so m + n M[p]. Also, p a+b (m + n) = p a+b m + p a+b n = 0, p a (rm) = r(p a m) = r0 = 0, so rm M[p]. Hence M[p] is a submodule of M. (b) In this case, any element of M[p] is annihilated by p αp, so the denitions coincide. (c) We dene a map φ : M M[p], as follows. Let m M. Then m is contained in a nitely generated R-submodule of M, say N = Rm. Then N has an annihilator ann(n), so by Exercise , N is the direct sum of its p-primary components, θ : N N[p]. Now, we embed each N[p] inside the corresponding M[p] in the natural way: ι p : N[p] M[p]. Now, we dene φ(m) = ( ι p θ)(m). One can check that φ is in fact an R-module isomorphism.