Homework #05, due 2/17/10 = , , , , , Additional problems recommended for study: , , 10.2.

Transcription

1 Homework #05, due 2/17/10 = , , , , , Additional problems recommended for study: , , , , , , , , , , , Assume R is a ring with 1 and M is a left R-module. Prove that if A and B are sets with the same cardinality, then the free modules F (A) and F (B) are isomorphic. 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 and f 1 : B A. By Theorem 10.6 the free modules F (A) and F (B) have the universal mapping property. Let us apply this property to f. Note that the range of f is B since f is surjective, and B is a subset of the free module F (B). Thus we have another function (which we call g) 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. 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 A. By the same reasoning, with A and B interchanged and f replaced by f 1, we get another R-module homomorphism ψ : F (B) F (A) extending f 1. The composition of R-module homomorphisms is again an R- module homomorphism, so we obtain the R-module homomorphism ψ ϕ : F (A) F (A). For every a A, (ψ ϕ)(a) = ψ(ϕ(a)) = ψ(f(a)), but f(a) B and ψ extends f 1, so 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, that is, ι(a) = a for every a A. Now, 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, we found that ψ ϕ is a homomorphism of F (A) onto itself that extends the identity map on A, so ψ ϕ must therefore 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 are

2 2 therefore isomorphisms between F (A) and F (B). Thus F (A) = F (B) whenever A = B 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 10.1). 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. Let M be a finite abelian group, so that M is also a Z-module. Let n be the order of M, that is, n = M Z +. Then, for every a M, the order of a divides the order n of the abelian group M, so 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) for every a M, so M Tor(M). The opposite inclusion holds trivially, so M = Tor(M). Let M = i Z Z/2Z. Thus M is the direct product of countably + many copies of the 2-element cyclic group Z/2Z. M is an infinite abelian group whose cardinality is the same as the set of real numbers. Every element of M has order 2, so 2 m = 0 for every m M, so Tor(M) = M Let R be an integral domain. Prove that every finitely generated torsion R-module has a nonzero annihilator, i.e., there is a nonzero r R such that rm = 0 for all m M here r does not depend on m (the annihilator of a module was defined in Exercise 9 of Section 10.1). Give an example of a torsion R-module whose annihilator is the zero ideal. Assume M is a finitely generated torsion R-module. Then there is a finite set A M of nonzero elements such that M = RA. Let A = {a 1,, a n }, n ω. Since M is a torsion module, there exist nonzero elements r 1,, r n R such that r 1 a 1 = 0, r 2 a 2 = 0,..., r n a n = 0. Let q = r 1... r n. Consider an arbitrary element m M. Since M = RA, there are ring elements s 1,, s n R such that m = s 1 a s n a n. Let 1 i n. Since R is an integral domain, R is commutative, so q = pr i where p is the product of all the other factors r j with j i. Then qa i = (pr i )a i = p(r i a i ) = p0 = 0. This

3 shows qa 1 = qa 2 = = qa n = 0, so we have qm = q(s 1 a s n a n ) = q(s 1 a 1 ) + + q(s n a n ) module axiom = (qs 1 )a (qs n )a n module axiom = (s 1 q)a (s n q)a n since R is commutative = s 1 (qa 1 ) + + s n (qa n ) module axiom = s 1 (0) + + s n (0) shown above = = 0 Since m was an arbitrary element of M, we have shown that qm = 0 for every m M. Finally, we observe that q is nonzero because r i 0 and the product on nonzero elements in the integral domain R cannot be zero. For an example of a torsion R-module whose annihilator is the zero ideal, let R = Z, and let M be the direct sum of the finite cyclic groups Z/nZ with 2 n Z +, so M = 2 n Z + Z/nZ. Let m M. Then there are k Z + and a 1,..., a k Z such that m = (a 1 + 2Z,, a k + kz, 0, 0, 0, ), so k! m = 0, so m Tor(M). Then M is a torsion module, but no element of Z annihilates every element of M, for if k Z + then k (..., 1 + (k + 1)Z, 0, 0,... ) Assume R is a ring with 1 and M is a left R-module. Let N be a submodule of M. Prove that if both M/N and N are finitely generated, then M so is M. N is a finitely generated R-module, so there is a finite subset A N such that N = RA. M/N is also finitely generated R-module. The elements of M/N have the form b + N with b M, so there is a finite subset B M such that M/N = R{b + N b B}. Let A = {a 1,, a n } and B = {b 1,, b k }. Then N = {r 1 a r n a n r 1,, r n R} 3

4 4 M/N = {s 1 (b 1 + N) + + s k (b k + N) s 1,, s k R} = {s 1 b s k b k + N s 1,, s k R} We will show M = R(A B). Let m M. Then m + N M/N, so by the second equation above there exist s 1,, s k R such that m+n = s 1 b s k b k + N. Now m m + N, so m s 1 b s k b k + N, so by the first equation above there exist r 1,, r n R such that m = s 1 b s k b k + r 1 a r n a n R(A B), as was to be shown. Now A B is finite because both A and B are finite, so M is therefore finitely generated by the finite set A B An element e R is called a central idempotent if e 2 = e and er = re for all r R. If e a central idempotent in R, prove that M = em (1 e)m. [Recall Exercise 14 in Section 10.1.] We wish to show that em and (1 e)m are submodules and that M is the internal direct sum of the two submodules em and (1 e)m. To see that em is a submodule, first note that em 1 + em 2 = e(m 1 + m 2 ) em for all m 1, m 2 M, so em is closed under addition. To see that em is closed under the action of R, let m M and r R. Then, using the fact that e is in the center and hence er = re, we have r(em) = (re)m = (er)m = e(rm) em. The proof that (1 e)m is also closed under addition is essentially the same, namely, (1 e)m 1 +(1 e)m 2 = (1 e)(m 1 +m 2 ) (1 e)m for all m 1, m 2 M. Next we show that (1 e)m is closed under the action of R. Let m M and r R. Then, using the fact that e is in the center and hence er = re, we have r((1 e)m) = (r(1 e))m = (r1 re)m = (1r er)m = ((1 e)r)m = (1 e)(rm) (1 e)m

5 Next we show em (1 e)m = {0}. Let m em (1 e)m. Then there are m 1, m 2 M such that m = em 1 = (1 e)m 2, so m = em 1 = e 2 m 1 e = e 2 = (ee)m 1 = e(em 1 ) = e((1 e)m 2 ) em 1 = (1 e)m 2 = (e(1 e))m 2 = (e1 e 2 )m 2 = (e e)m 2 e = e 2 = 0m 2 = 0. It therefore follows by Prop that M = em (1 e)m. 5