# Math 210B: Algebra, Homework 4

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1 Math 210B: Algebra, Homework 4 Ian Coley February 5, 2014 Problem 1. Let S be a multiplicative subset in a commutative ring R. Show that the localisation functor R-Mod S 1 R-Mod, M S 1 M, is exact. First, without loss of generality, we assume that 1 S since previous homework assignments have shown that inclusion of units in a localisation changes nothing. Second, we assume that 0 / S, so that S 1 R-Mod contains more than the zero module. Let 0 X f Y g Z 0 be an exact sequence of R-modules. Then we need to show that 0 S 1 X f S 1 Y g S 1 Z 0 is also exact. Suppose that f (x/s) = f(x)/s = 0/1 S 1 Y. This implies that there exists t S so that t f(x) = 0. Since f is an R-module homomorphism, this implies f(t x) = 0. Since f is injective, t x = 0. Therefore we see x/s = 0/1, since t(x s) = t x = 0. Therefore f is injective. Now we need to show that im f = ker g. We have g (f (x/s)) = g(f(x)/s) = (g f)(x)/s = 0, so im f ker g. Conversely, if g (y/s) = 0, then g(y)/s = 0, so there exists t S so that t g(y) = g(t y) = 0. Therefore t y ker g, so t y im f. Let x X so that f(x) = t y. Then f (x/(st)) = f(x)/(st) = (t y)/(st) = y/s, so y/s im f. Therefore the sequence is exact in at the middle term. Finally, we need to show that g is surjective. Let z/s S 1 Z. Then there exists y Y so that g(y) = z. Then g (y/s) = g(y)/s = z/s, so g is surjective. This completes the proof. Problem 2. Prove that if for a module M over a commutative ring R one has M m = 0 for any maximal ideal m R, then M = 0. Here M m is the localisation of M with respect to the multiplicative subset R \ m. 1

2 Suppose that M is not the zero module, so 0 m M. Consider a = ann(m) = {r R : r x = 0}, the annihilator ideal of m. This is a proper ideal of R, so a m for some maximal ideal m. In the localisation M m, we have m/1 = 0/1, so there exists some s / m so that s m = 0. But this is a contradiction since any such element is in a m. Therefore we must have M = 0. Problem 3. Determine the torsion part of the group Z 2 /N, where N is the cyclic subgroup of Z 2 generated by (6, 21). First, we take the exact sequence 0 N Z 2 G 0, where G is the group in question. We can write (6, 21) = 6 (1, 0) + 21 (0, 1) in terms of the generators e 1, e 2 of Z 2. This yields a matrix ( ) 6 A =. 21 We now reduce this matrix until it is diagonal as possible. ( ) ( ) ( ) ( 6 r 2 3r! 6 r 1 2r 2 0 r 2 r This yields a factorisation of the group G as G = Z Z/3Z. Therefore the torsion part of this group is Z/3Z. ). Problem 4. Find the rank of the subgroup in Z 3 generated by (2, 2, 0), (0, 4, 4), and (5, 0, 5). We know that a submodule of a module over a PID is free itself. We know that the above three generators are a generating set, but we do not know if it is a basis. To show this, we need to show that the map f : Z 3 M by f(a, b, c) = a(2, 2, 0) + b(0, 4, 4) + c(5, 0, 5) is an isomorphism. It is surjective by construction. Suppose that f(a, b, c) = 0. Then (2a + 5c, 4b 2a, 4b 5c) = (0, 0, 0). 2

3 Solving these equations, we see a = 2b, so replacing we have 4b + 5c = 0, 4b 5c = 0. Therefore we see b = 5 and c = 4 is a solution, so in total f(10, 5, 4) = (0, 0, 0). Additionally, any integer multiple f(α(10, 5, 4)) = αf(10, 5, 4) = 0 since f is a Z-module homomorphism.. Therefore ker f = Z is a free module of rank 1. Therefore M is a free module of rank 2. Problem 5. Let f : A A be an endomorphism of an abelian group A such that f(f(a)) = a for all a A. Show that there is a structure of a Z[i]-module on A such that ia = f(a) for all a A. We let the action of a + bi on g A be given by (a + bi) g = a g + b f(g). We need to check that this action is appropriately distributive, since it is clear that 1 g = g and A is an abelian group. First, all nice properties hold for the action of Z Z[i] on A. Now, we have (a + bi) g + (a + bi) h = a g + b f(g) + a h + b f(h) = a (g + h) + b [f(g) + f(h)] = a (g + h) + b f(g + h) = (a + bi) (g + h), where we use that f is a group homomorphism. Similarly, (a + bi) [(c + di) g] = (a + bi) (c g + d f(g)) = a (c g) + a (d f(g)) + b f(c g) + b f(d f(g)) = (ac) g + (ad + bc) f(g) + bd f(f(g)) = (ac bd) g + (ad + bc) f(g) = (ac + adi + bci + bdi 2 ) g = [(a + bi)(c + di)] g. Therefore A is an Z[i]-module with this structure. Problem 6. Classify all finitely generated modules over Z/nZ. Recall from the last homework that if M is a Z-module such that nz M = 0, then M is naturally a Z/nZ-module. Conversely, if M is a Z/nZ-module, then we may view M as a Z-module via a m = [a]ṁ, where [a] is the equivalence class of a modulo n. 3

4 Therefore all finitely generated modules over Z/nZ are Z-modules which are annihilated by n Z. Let n = p k 1 1 p km m be the (distinguished) prime factorisation of n. Using the classification theorem of finitely generated abelian groups, such Z-modules are exactly: M = Z/q α 1 1 Z Z/q α l l Z, where q i = p j for some prime and α i k j, so that n annihilates each term. These are all such modules. Problem 7. Classify all finitely generated modules over S 1 Z where S = {p n, n 0} and S = Z \ pz (p is a prime integer). We claim that the localisation of a PID is a PID. Let I S 1 Z be an ideal. Then by previous results, there exists J Z so that I = JS 1 Z. Let J = (n). Then we claim J = (n/1). let x I be any element. Then x (n)s 1 Z = x = (n/1) (a/b) (n/1). So S 1 Z is a PID. Therefore we may apply the classification theorem of modules over a PID. To do so, we need only determine the prime elements of S 1 Z in each case. In the first case, S = {p n }. In S 1 Z, this p is no longer a prime element, since it is invertible. However, all other prime elements are still prime, since S 1 (Z/(q)) = S 1 Z/(q/1) is still a domain, as the localisation of a domain is a domain by another previous result. Therefore modules over this S 1 Z are of the form S 1 Z r S 1 Z/(q k 1 1 /1) S 1 Z/(q kn n /1), where q p, the localised prime. On the other hand, if S = Z \ pz, then p is the only prime element of S 1 Z, since every other prime element is invertible. Hence modules of this PID are S 1 Z r S 1 Z/(p k 1 /1) S 1 Z/(p kn /1). This completes the proof. Problem 8. Let N be a submodule of a free finitely generated module F over a PID R. Show that N is a direct summand of F if and only if N af = an for all a R. If N is a submodule of F = R n, then N = n N i, 4

5 where each N i R is a submodule, i.e. an ideal. Further, N = R r, since it is a submodule of a free module. Clearly r n, and indeed we see r + k n by Problem 9. Hence N is a direct summand of F if and only if N i = R or 0 at every place. Suppose now that this is so. Then up to reordering, N is generated by e 1,..., e r, the standard basis of F = R n. Then N af = span{ae 1,..., ae r } = an. Problem 9. Show that a submodule N of a module M generated by n elements over a PID also can be generated by n elements. Let {x 1,..., x n } be a basis for M, and consider the canonical projection π : M N. Then let π(x i ) = x i for all i. We claim this is a basis of N. To see this, we have a surjective homomorphism R n M given by n (r 1,..., r n ) r i x i. Then we may compose these (R-module) homomorphisms to get a total homomorphism R n N by n (r 1,..., r n ) r i x i. This homomorphism is still surjective, hence { x 1,... x n } generate N. Problem 10. Let M (resp. N) be a product (resp. a direct sum) of countably many copies of the group Z. Show that the sequence of abelian groups is not split. 0 N M M/N 0 If the above sequence is split, then in particular there exists an injective splitting map M/N M. We claim no such map exists. We examine so-called infinitely-divisible elements. If A is an arbitrary abelian group, we consider the subgroup n N 2 n A, whose elements we call 2 -divisible. Note that in Z, the only 2 -divisible element is 0, hence in M the only 2 -divisible element is the infinite product of 0. However, M/N has nontrivial 2 -divisible elements. Let N N be an arbitrary large number, and let α = (2 0, 2 1,..., 2 n,...) + N M/N. Then 2 N α = (2 N, 2 N+1,...) = (2 0, 2 1,..., 2 N, 2 N+1,...) = α mod N since modulo N means we may modify any finite number of elements. Since an Z-linear injection M/N M must preserve the 2 -divisible elements, we conclude that no injection exists since α 0. Therefore the sequence is not split. 5

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