Feb 17, 2010 ... 10.3.1 Assume R is a ring with 1 and M is a left R-module. Prove that if A and B
are sets ... 10.3.5 Let R be an integral domain. Prove that every ...
Homework #05, due 2/17/10 = 10.3.1, 10.3.3, 10.3.4, 10.3.5, 10.3.7, 10.3.15 Additional problems recommended for study: 10.2.1, 10.2.2, 10.2.3, 10.2.5, 10.2.6, 10.2.10, 10.2.11, 10.3.2, 10.3.9, 10.3.12, 10.3.13, 10.3.14. 10.3.1 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 Rmodule 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
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therefore isomorphisms between F (A) and F (B). Thus F (A) ∼ = F (B) whenever |A| = |B|. 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 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 ). Q 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 . 10.3.5 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 = {a1 , · · · , an }, n ∈ ω. Since M is a torsion module, there exist nonzero elements r1 , · · · , rn ∈ R such that r1 a1 = 0, r2 a2 = 0, . . . , rn an = 0. Let q = r1 · . . . · rn . Consider an arbitrary element m ∈ M . Since M = RA, there are ring elements s1 , · · · , sn ∈ R such that m = s1 a1 + · · · + sn an . Let 1 ≤ i ≤ n. Since R is an integral domain, R is commutative, so q = pri where p is the product of all the other factors rj with j 6= i. Then qai = (pri )ai = p(ri ai ) = p0 = 0. This
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shows qa1 = qa2 = · · · = qan = 0, so we have qm = q(s1 a1 + · · · + sn an ) = q(s1 a1 ) + · · · + q(sn an )
module axiom
= (qs1 )a1 + · · · + (qsn )an
module axiom
= (s1 q)a1 + · · · + (sn q)an
since R is commutative
= s1 (qa1 ) + · · · + sn (qan )
module axiom
= s1 (0) + · · · + sn (0) = 0 + ··· + 0 =0
shown above
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 ri 6= 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 M= Z/nZ. 2≤n∈Z+
Let m ∈ M . Then there are k ∈ Z+ and a1 , . . . , ak ∈ Z such that m = (a1 + 2Z, · · · , ak + 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, . . . ) 6= 0. 10.3.7 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 = {a1 , · · · , an } and B = {b1 , · · · , bk }. Then N = {r1 a1 + · · · + rn an |r1 , · · · , rn ∈ R}
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M/N = {s1 (b1 + N ) + · · · + sk (bk + N )|s1 , · · · , sk ∈ R} = {s1 b1 + · · · + sk bk + N |s1 , · · · , sk ∈ R} We will show M = R(A ∪ B). Let m ∈ M . Then m + N ∈ M/N , so by the second equation above there exist s1 , · · · , sk ∈ R such that m+N = s1 b1 + · · · + sk bk + N . Now m ∈ m + N , so m ∈ s1 b1 + · · · + sk bk + N , so by the first equation above there exist r1 , · · · , rn ∈ R such that m = s1 b1 + · · · + sk bk + r1 a1 + · · · + rn an ∈ 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. 10.3.15 An element e ∈ R is called a central idempotent if e2 = 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 em1 + em2 = e(m1 + m2 ) ∈ eM for all m1 , m2 ∈ 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)m1 +(1−e)m2 = (1−e)(m1 +m2 ) ∈ (1 − e)M for all m1 , m2 ∈ 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
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Next we show eM ∩ (1 − e)M = {0}. Let m ∈ eM ∩ (1 − e)M . Then there are m1 , m2 ∈ M such that m = em1 = (1 − e)m2 , so m = em1 = e2 m1
e = e2
= (ee)m1 = e(em1 ) = e((1 − e)m2 )
em1 = (1 − e)m2
= (e(1 − e))m2 = (e1 − e2 )m2 = (e − e)m2 = 0m2 = 0.
e = e2
It therefore follows by Prop. 10.5 that M ∼ = eM ⊕ (1 − e)M .
