# 11. Finitely-generated modules

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1 11. Finitely-generated modules 11.1 Free modules 11.2 Finitely-generated modules over domains 11.3 PIDs are UFDs 11.4 Structure theorem, again 11.5 Recovering the earlier structure theorem 11.6 Submodules of free modules 1. Free modules The following definition is an example of defining things by mapping properties, that is, by the way the object relates to other objects, rather than by internal structure. The first proposition, which says that there is at most one such thing, is typical, as is its proof. Let R be a commutative ring with 1. Let S be a set. A free R-module M on generators S is an R-module M and a set map i : S M such that, for any R-module N and any set map f : S N, there is a unique R-module homomorphism f : M N such that The elements of i(s) in M are an R-basis for M. f i = f : S N [1.0.1] Proposition: If a free R-module M on generators S exists, it is unique up to unique isomorphism. Proof: First, we claim that the only R-module homomorphism F : M M such that F i = i is the identity map. Indeed, by definition, [1] given i : S M there is a unique ĩ : M M such that ĩ i = i. The identity map on M certainly meets this requirement, so, by uniqueness, ĩ can only be the identity. Now let M be another free module on generators S, with i : S M as in the definition. By the defining property of (M, i), there is a unique ĩ : M M such that ĩ i = i. Similarly, there is a unique ĩ such that ĩ i = i. Thus, i = ĩ i = ĩ ĩ i [1] Letting letting i : S M take the role of f : S N in the definition. 167

2 168 Finitely-generated modules Similarly, i = ĩ i = ĩ ĩ i From the first remark of this proof, this shows that ĩ ĩ = identity map on M ĩ ĩ = identity map on M So ĩ and ĩ are mutual inverses. That is, M and M are isomorphic, and in a fashion that respects the maps i and i. Further, by uniqueness, there is no other map between them that respects i and i, so we have a unique isomorphism. Existence of a free module remains to be demonstrated. We should be relieved that the uniqueness result above assures that any successful construction will invariably yield the same object. Before proving existence, and, thus, before being burdened with irrelevant internal details that arise as artifacts of the construction, we prove the basic facts about free modules. [1.0.2] Proposition: A free R-module M on generators i : S M is generated by i(s), in the sense that the only R-submodule of M containing the image i(s) is M itself. Proof: Let N be the submodule generated by i(s), that is, the intersection of all submodules of M containing i(s). Consider the quotient M/N, and the map f : S M/N by f(s) = 0 for all s S. Let ζ : M M/N be the 0 map. Certainly ζ i = f. If M/N 0, then the quotient map q : M M/N is not the zero map ζ, and also q i = f. But this would contradict the uniqueness in the definition of M. For a set X of elements of an R-module M, if a relation x X r x x = 0 with r x R and x M (with all but finitely-many coefficients r x being 0) implies that all coefficients r x are 0, say that the elements of X are linearly independent (over R). [1.0.3] Proposition: Let M be a free R-module on generators i : S M. Then any relation (with finitely-many non-zero coefficients r s R) r s i(s) = 0 s S must be trivial, that is, all coefficients r s are 0. That is, the elements of i(s) are linearly independent. Proof: Suppose s r s i(s) = 0 in the free module M. To show that every coefficient r s is 0, fix s o S and map f : S R itself by { 0 (s so ) f(s) = 1 (s = s o ) Let f be the associated R-module homomorphism f : M R. Then 0 = f(0) = f( s r s i(s)) = r so This holds for each fixed index s o, so any such relation is trivial.

3 Garrett: Abstract Algebra 169 [1.0.4] Proposition: Let f : B C be a surjection of R-modules, where C is free on generators S with i : S C. Then there is an injection j : C B such that [2] f j = 1 C and B = (ker f) j(c) [1.0.5] Remark: The map j : C B of this proposition is a section of the surjection f : B C. Proof: Let {b s : s S} be any set of elements of B such that f(b s ) = i(s). Invoking the universal property of the free module, given the choice of {b x } there is a unique R-module homomorphism j : C B such that (j i)(s) = b s. It remains to show that jc ker f = B. The intersection jc ker f is trivial, since for s r s j(s) in the kernel (with all but finitely-many r s just 0) ( ) C 0 = f r s j(s) = r s i(s) s s We have seen that any such relation must be trivial, so the intersection f(c) kerf is trivial. Given b B, let f(b) = s r s i(s) (a finite sum), using the fact that the images i(s) generate the free module C. Then f(b j(f(b))) = f(b s r s b s ) == f(b) s r s f(b s ) = s r s i(s) s r s i(s) = 0 Thus, j(c) + ker f = B. We have one more basic result before giving a construction, and before adding any hypotheses on the ring R. The following result uses an interesting trick, reducing the problem of counting generators for a free module F over a commutative ring R with 1 to counting generators for vector spaces over a field R/M, where M is a maximal proper ideal in R. We see that the number of generators for a free module over a commutative ring R with unit 1 has a well-defined cardinality, the R-rank of the free module. [1.0.6] Theorem: Let F be a free R-module on generators i : S F, where R is a commutative ring with 1. Suppose that F is also a free R-module on generators j : T F. Then S = T. Proof: Let M be a maximal proper ideal in R, so k = R/M is a field. Let and consider the quotient E = M F = collection of finite sums of elements mx, m M, x F V = F/E with quotient map q : F V. This quotient has a canonical k-module structure (r + M) (x + M F ) = rx + M F We claim that V is a free k-module on generators q i : S V, that is, is a vector space on those generators. Lagrange s replacement argument shows that the cardinality of the number of generators for a vector space over a field is well-defined, so a successful comparison of generators for the original module and this vector space quotient would yield the result. [2] The property which we are about to prove is enjoyed by free modules is the defining property of projective modules. Thus, in these terms, we are proving that free modules are projective.

4 170 Finitely-generated modules To show that V is free over k, consider a set map f : S W where W is a k-vectorspace. The k-vectorspace W has a natural R-module structure compatible with the k-vectorspace structure, given by r (x + M F ) = rx + M F Let f : F W be the unique R-module homomorphism such that f i = f. Since m w = 0 for any m M and w W, we have 0 = m f(s) = m f(i(s)) = f(m i(s)) so Thus, f : V W defined by ker f M F f(x + M F ) = f(x) is well-defined, and f (q i) = f. This proves the existence part of the defining property of a free module. For uniqueness, the previous argument can be reversed, as follows. Given f : V W such that f (q i) = f, let f = f q. Since there is a unique f : F W with f i = f, there is at most one f. Finally, we construct free modules, as a proof of existence. [3] Given a non-empty set S, let M be the set of R-valued functions on S which take value 0 outside a finite subset of S (which may depend upon the function). Map i : S M by letting i(s) be the function which takes value 1 at s S and is 0 otherwise. Add functions value-wise, and let R act on M by value-wise multiplication. [1.0.7] Proposition: The M and i just constructed is a free module on generators S. In particular, given a set map f : S N for another R-module N, for m M define f(m) N by [4] f(m) = s S m(s) f(s) Proof: We might check that the explicit expression (with only finitely-many summands non-zero) is an R-module homomorphism: that it respects addition in M is easy. For r R, we have f(r m) = s S (r m(s)) f(s) = r m(s) f(s) = r f(m) And there should be no other R-module homomorphism from M to N such that f i = f. Let F : M N be another one. Since the elements {i(s) : s S} generate M as an R-module, for an arbitrary collection {r s R : s S} with all but finitely-many 0, ( ) F r s i(s) = r s F (i(s)) = ( ) r s f(s) = f r s i(s) s S s S s S s S s S so necessarily F = f, as desired. [3] Quite pointedly, the previous results did not use any explicit internal details of what a free module might be, but, rather, only invoked the external mapping properties. [4] In this formula, the function m on S is non-zero only at finitely-many s S, so the sum is finite. And m(s) R and f(s) N, so this expression is a finite sum of R-multiples of elements of N, as required.

5 [1.0.8] Remark: For finite generator sets often one takes Garrett: Abstract Algebra 171 S = {1, 2,..., n} and then the construction above of the free module on generators S can be identified with the collection R n of ordered n-tuples of elements of R, as usual. 2. Finitely-generated modules over a domain In the sequel, the results will mostly require that R be a domain, or, more stringently, a principal ideal domain. These hypotheses will be carefully noted. [2.0.1] Theorem: Let R be a principal ideal domain. Let M be a free R-module on generators i : S M. Let N be an R-submodule. Then N is a free R-module on at most S generators. [5] Proof: Induction on the cardinality of S. We give the proof for finite sets S. First, for M = R 1 = R a free module on a single generator, an R-submodule is an ideal in R. The hypothesis that R is a PID assures that every ideal in R needs at most one generator. This starts the induction. Let M = R m, and let p : R m R m 1 be the map p(r 1, r 2, r 3,..., r m ) = (r 2, r 3,..., r m ) The image p(n) is free on m 1 generators, by induction. From the previous section, there is always a section j : p(n) N such that p j = 1 p(n) and N = ker p N j(p(n)) Since p j = 1 p(n), necessarily j is an injection, so is an isomorphism to its image, and j(p(n)) is free on m 1 generators. And ker p N is a submodule of R, so is free on at most 1 generator. We would be done if we knew that a direct sum M 1 M 2 of free modules M 1, M 2 on generators i 1 : S i M 1 and i 2 : S 2 M 2 is a free module on the disjoint union S = S 1 S 2 of the two sets of generators. We excise that argument to the following proposition. [2.0.2] Proposition: A direct sum [6] M = M 1 M 2 of free modules M 1, M 2 on generators i 1 : S i M 1 and i 2 : S 2 M 2 is a free module on the disjoint union S = S 1 S 2 of the two sets of generators. [7] Proof: Given another module N and a set map f : S N, the restriction f j of f to S j gives a unique module homomorphism f j : M j M such that f j i j = f j. Then f(m 1, m 2 ) = (f 1 m 1, f 2 m 2 ) [5] The assertion of the theorem is false without some hypotheses on R. For example, even in the case that M has a single generator, to know that every submodule needs at most a single generator is exactly to assert that every ideal in R is principal. [6] Though we will not use it at this moment, one can give a definition of direct sum in the same mapping-theoretic style as we have given for free module. That is, the direct sum of a family {M α : α A} of modules is a module M and homomorphisms i α : M α M such that, for every family of homomorphisms f α : M α N to another module N, there is a unique f : M N such that every f α factors through f in the sense that f α = f i α. [7] This does not need the assumption that R is a principal ideal domain.

6 172 Finitely-generated modules is a module homomorphism from the direct sum to N with f i = f. On the other hand, given any map g : M N such that g i = f, by the uniqueness on the summands M 1 and M 2 inside M, if must be that g i j = f j for j = 1, 2. Thus, this g is f. For an R-module M, for m M the annihilator Ann R (m) of m in R is Ann R (m) = {r R : rm = 0} It is easy to check that the annihilator is an ideal in R. An element m M is a torsion element of M if its annihilator is not the 0 ideal. The torsion submodule M tors of M is M tors = {m M : Ann R (m) {0}} A module is torsion free if its torsion submodule is trivial. [2.0.3] Proposition: For a domain R, the torsion submodule M tors of a given R-module M is an R-submodule of M, and M/M tors is torsion-free. Proof: For torsion elements m, n in M, let x be a non-zero element of Ann R (m) and y a non-zero element of Ann ( n). Then xy 0, since R is a domain, and And for r R, Thus, the torsion submodule is a submodule. (xy)(m + n) = y(xm) + x(yn) = y 0 + x 0 = 0 x(rm) = r(xm) = r 0 = 0 To show that the quotient M/M tors is torsion free, suppose r (m + M tors ) M tors for r 0. Then rm M tors. Thus, there is s 0 such that s(rm) = 0. Since R is a domain, rs 0, so m itself is torsion, so m + M tors = M tors, which is 0 in the quotient. An R-module M is finitely generated if there are finitely-many m 1,..., m n such that i Rm i = M. [8] [2.0.4] Proposition: Let R be a domain. [9] Given a finitely-generated [10] R-module M, there is a (not necessarily unique) maximal free submodule F, and M/F is a torsion module. Proof: Let X be a set of generators for M, and let S be a maximal subset of X such that (with inclusion i : S M) the submodule generated by S is free. To be careful, consider why there is such a maximal subset. First, for φ not to be maximal means that there is x 1 X such that Rx 1 M is free on generator {x 1 }. If {x 1 } is not maximal with this property, then there is x 2 X such that Rx 1 + Rx 2 is free on generators {x 1, x 2 }. Since X is finite, there is no issue of infinite ascending unions of free modules. Given x X but not in S, by the maximality of S there are coefficients 0 r R and r s R such that rx + s S r s i(s) = 0 [8] This is equivalent to saying that the m i generate M in the sense that the intersection of submodules containing all the m i is just M itself. [9] The hypothesis that the ring R is a domain assures that if r i x i = 0 for i = 1, 2 with 0 r i R and x i in an R-module, then not only (r 1 r 2 )(x 1 + x 2 ) = 0 but also r 1 r 2 0. That is, the notion of torsion module has a simple sense over domains R. [10] The conclusion is false in general without an assumption of finite generation. For example, the Z-module Q is the ascending union of the free Z-modules N 1 Z, but is itself not free.

7 Garrett: Abstract Algebra 173 so M/F is torsion. [2.0.5] Theorem: Over a principal ideal domain R a finitely-generated torsion-free module M is free. Proof: Let X be a finite set of generators of M. From the previous proposition, let S be a maximal subset of X such that the submodule F generated by the inclusion i : S M is free. Let x 1,..., x n be the elements of X not in S, and since M/F is torsion, for each x i there is 0 r i R be such that r i x i F. Let r = i r i. This is a finite product, and is non-zero since R is a domain. Thus, r M F. Since F is free, rm is free on at most S generators. Since M is torsion-free, the multiplication by r map m rm has trivial kernel in M, so M rm. That is, M is free. [2.0.6] Corollary: Over a principal ideal domain R a finitely-generated module M is expressible as M M tors F where F is a free module and M tors is the torsion submodule of M. Proof: We saw above that M/M tors is torsion-free, so (being still finitely-generated) is free. The quotient map M M/M tors admits a section σ : M/M tors M, and thus M = M tors σ(m/m tors ) = M tors free as desired. [2.0.7] Corollary: Over a principal ideal domain R, a submodule N of a finitely-generated R-module M is finitely-generated. Proof: Let F be a finitely-generated free module which surjects to M, for example by choosing generators S for M and then forming the free module on S. The inverse image of N in F is a submodule of a free module on finitely-many generators, so (from above) needs at most that many generators. Mapping these generators forward to N proves the finite-generation of N. [2.0.8] Proposition: Let R be a principal ideal domain. Let e 1,..., e k be elements of a finitely-generated free R-module M which are linearly independent over R, and such that M/(Re Re k ) is torsion-free, hence free Then this collection can be extended to an R-basis for M. Proof: Let N be the submodule N = Re Re k generated by the e i. The quotient M/N, being finitely-generated and torsion-less, is free. Let e k+1,..., e n be elements of M whose images in M/N are a basis for M/N. Let q : M M/N be the quotient map. Then, as above, q has a section σ : M/N M which takes q(e i ) to e i. And, as above, M = ker q σ(m/n) = N σ(m/n) Since e k+1,..., e n is a basis for M/N, the collection of all e 1,..., e n is a basis for M. 3. PIDs are UFDs We have already observed that Euclidean rings are unique factorization domains and are principal ideal domains. The two cases of greatest interest are the ordinary integers Z and polynomials k[x] in one variable over a field k. But, also, we do have

8 174 Finitely-generated modules [3.0.1] Theorem: A principal ideal domain is a unique factorization domain. Before proving this, there are relatively elementary remarks that are of independent interest, and useful in the proof. Before anything else, keep in mind that in a domain R (with identity 1), for x, y R, Rx = Ry if and only if x = uy for some unit u R Indeed, x Ry implies that x = uy, while y Rx implies y = vx for some v, and then y = uv y or (1 uv)y = 0. Since R is a domain, either y = 0 (in which case this discussion was trivial all along) or uv = 1, so u and v are units, as claimed. Next recall that divisibility x y is inclusion-reversion for the corresponding ideals, that is Rx Ry if and only if x y Indeed, y = mx implies y Rx, so Ry Rx. Conversely, Ry Rx implies y Rx, so y = mx for some m R. Next, given x, y in a PID R, we claim that g R such that Rg = Rx + Ry is a greatest common divisor for x and y, in the sense that for any d R dividing both x and y, also d divides g (and g itself divides x and y). Indeed, d x gives Rx Rd. Thus, since Rd is closed under addition, any common divisor d of x and y has Rx + Ry Rd Thus, g Rg Rd, so g = rd for some r R. And x Rg and y Rg show that this g does divide both x and y. Further, note that since a gcd g = gcd(x, y) of two elements x, y in the PID R is a generator for Rx + Ry, this gcd is expressible as g = rx + sy for some r, s R. In particular, a point that starts to address unique factorization is that an irreducible element p in a PID R is prime, in the sense that p ab implies p a or p b. Indeed, the proof is the same as for integers, as follows. If p does not divide a, then the irreducibility of p implies that 1 = gcd(p, a), since (by definition of irreducible) p has no proper divisors. Let r, s R be such that 1 = rp + sa. Let ab = tp. Then and, thus, b is a multiple of p. b = b 1 = b (rp + sa) = br p + s ab = p (br + st) [3.0.2] Corollary: (of proof) Any ascending chain I 1 I 2... of ideals in a principal ideal domain is finite, in the sense that there is an index i such that That is, a PID is Noetherian. I i = I i+1 = I i+2 =... Proof: First, prove the Noetherian property, that any ascending chain of proper ideals I 1 I 2...

9 Garrett: Abstract Algebra 175 in R must be finite. Indeed, the union I is still a proper ideal, since if it contained 1 some I i would already contain 1, which is not so. Further, I = Rx for some x R, but x must lie in some I i, so already I = I i. That is, I i = I i+1 = I i+2 =... Let r be a non-unit in R. If r has no proper factorization r = xy (with neither x nor y a unit), then r is irreducible, and we have factored r. Suppose r has no factorization into irreducibles. Then r itself is not irreducible, so factors as r = x 1 y 1 with neither x 1 nor y 1 a unit. Since r has no factorization into irreducibles, one of x 1 or y 1, say y 1, has no factorization into irreducibles. Thus, y 1 = x 2 y 2 with neither x 2 nor y 2 a unit. Continuing, we obtain a chain of inclusions Rr Ry 1 Ry 2... with all inclusions strict. This is impossible, by the Noetherian-ness property just proven. [11] ring elements have factorizations into irreducibles. That is, all The more serious part of the argument is the uniqueness of the factorization, up to changing irreducibles by units, and changing the ordering of the factors. Consider p e pem m = (unit) q f qfn n where the p i and q j are irreducibles, and the exponents are positive integers. The fact that p 1 ab implies p 1 a or p 1 b (from above) shows that p 1 must differ only by a unit from one of the q j. Remove this factor from both sides and continue, by induction. 4. Structure theorem, again The form of the following theorem is superficially stronger than our earlier version, and is more useful. [4.0.1] Theorem: Let R be a principal ideal domain, M a finitely-generated free module over R, and N an R-submodule of M. Then there are elements [12] d 1... d t of R, uniquely determined up to R, and an R-module basis m 1,..., m t of M, such that d 1 e 1,..., d t e t is an R-basis of N (or d i e i = 0). Proof: From above, the quotient M/N has a well-defined torsion submodule T, and F = (M/N)/T is free. Let q : M (M/N)/T be the quotient map. Let σ : F M be a section of q, such that M = ker q σ(f ) Note that N ker q, and (ker q)/n is a torsion module. The submodule ker q of M is canonically defined, though the free complementary submodule [13] σ(f ) is not. Since σ(f ) can be described as a sum of a uniquely-determined (from above) number of copies R/ 0, we see that this free submodule in M complementary to ker q gives the 0 elementary divisors. It remains to treat the finitely-generated torsion module (ker q)/n. Thus, without loss of generality, suppose that M/N is torsion (finitely-generated). [11] Yes, this proof actually shows that in any Noetherian commutative ring with 1 every element has a factorization into irreducibles. This does not accomplish much, however, as the uniqueness is far more serious than existence of factorization. [12] Elementary divisors. [13] Given a submodule A of a module B, a complementary submodule A to A in B is another submodule A of B such that B = A A. In general, submodules do not admit complementary submodules. Vector spaces over fields are a marked exception to this failure.

10 176 Finitely-generated modules For λ in the set of R-linear functionals Hom R (M, R) on M, the image λ(m) is an ideal in R, as is the image λ(n). Let λ be such that λ(n) is maximal among all ideals occurring as λ(n). [14] Let λ(n) = Rx for some x R. We claim that x 0. Indeed, express an element n N as n = i r ie i for a basis e i of M with r i R, and with respect to this basis define dual functionals µ i Hom R (M, R) by µ i ( j s j e j ) = e i (where s j R) If n 0 then some coefficient r i is non-zero, and µ i (n) = r i. Take n N such that λ(n) = x. Claim µ(n) Rx for any µ Hom R (M, R). Indeed, if not, let r, s R such that rλ(n) + sµ(n) is the gcd of the two, and (rλ + sµ)(n) is a strictly larger ideal than Rx, contradiction. Thus, in particular, µ i (n) Rx for all dual functionals µ i for a given basis e i of M. That is, n = xm for some m M. Then λ(m) = 1. And M = Rm ker λ since for any m M λ(m λ(m )m) = λ(m ) λ(m ) 1 = 0 Further, for n N we have λ(n ) Rx. Let λ(n ) = rx. Then λ(n r n) = λ(n ) r λ(n) = λ(n ) rx = 0 That is, Thus, with n = xm. And N = Rn ker λ N M/N Rm/Rn (ker λ)/(ker λ N ) Rm/Rn = Rm/Rxn R/Rx = R/ x The submodule ker λ is free, being a submodule of a free module over a PID, as is ker λ N. And the number of generators is reduced by 1 from the number of generators of M. Thus, by induction, we have a basis m 1,..., m t of M and x 1,..., x t in R such that n i = x i m i is a basis for N, using functional λ i whose kernel is Rm i Rm t, and λ i (n i ) = x i. We claim that the above procedure makes x i x i+1. By construction, n i+1 ker λ i and n i ker λ i+1 Thus, with r, s R such that rx i + sx i+1 is the greatest common divisor g = gcd(x i, x i+1 ), we have (rλ i + sλ i+1 )(n i + n i+1 ) = r λ i (n i ) + r λ i (n i+1 ) + +s λ i+1 (n i ) + s λ i+1 (n i+1 ) = r x i s x i+1 = gcd(x i, x i+1 ) That is, Rg Rx i and Rg Rx i+1. The maximality property of Rx i requires that Rx i = Rg. Thus, Rx i+1 Rx i, as claimed. This proves existence of a decomposition as indicated. Proof of uniqueness is far better treated after introduction of a further idea, namely, exterior algebra. Thus, for the moment, we will not prove uniqueness, but will defer this until the later point when we treat exterior algebra. [14] At this point it is not clear that this maximal ideal is unique, but by the end of the proof we will see that it is. The fact that any ascending chain of proper ideals in a PID has a maximal element, that is, that a PID is Noetherian, is proven along with the proof that a PID is a unique factorization domain.

11 Garrett: Abstract Algebra Recovering the earlier structure theorem The above structure theorem on finitely-generated free modules M over PIDs R and submodules N M gives the structure theorem for finitely-generated modules as a corollary, as follows. Let F be a finitely-generated R-module with generators [15] the free R-module on generators i : S M. Let f 1,..., f n. Let S = {f 1,..., f n }, and let M be q : M F be the unique R-module homomorphism such that q(i(f k )) = f k for each generator f k. Since q(m) contains all the generators of F, the map q is surjective. [16] Let N = ker q, so by a basic isomorphism theorem F M/N By the theorem of the last section, M has a basis m 1,..., m t and there are uniquely determined [17] r 1 r 2... r t R such that r 1 m 1,..., r t m t is a basis for N. Then since by F M/N (Rm 1 /Rr 1 m 1 )... (Rm t /Rrm t ) R/ r 1... R/ r t Rm i /Rr i m i R/ r i rm i + Rr i m i r + Rr i This gives an expression for F of the sort desired. 6. Submodules of free modules Let R be a principal ideal domain. Let A be a well-ordered set, and M a free module on generators e α for α A. Let N be a submodule of M. For α A, let I α = {r R : there exist r β, β < α : r e α + β<α r β e β N} Since R is a PID, the ideal I α has a single generator ρ α (which may be 0). Let n α N be such that n α = ρ α e α + β<α r β e β for some r β R. This defines ρ α and n α for all α A by transfinite induction. [6.0.1] Theorem: N is free on the (non-zero elements among) n α. [15] It does not matter whether or not this set is minimal, only that it be finite. [16] We will have no further use for the generators f k of F after having constructed the finitely-generated free module M which surjects to F. [17] Uniquely determined up to units.

12 178 Finitely-generated modules Proof: It is clear that I α is an ideal in R, so at least one element n α exists, though it may be 0. For any element n N lying in the span of {e β : β α}, for some r R the difference n rn α lies in the span of {e β : β < α}. We claim that the n α span N. Suppose not, and let α A be the first index such that there is n N not in that span, with n expressible as n = β α r βe β. Then r α = r ρ α for some r R, and for suitable coefficients s β R n rn α = s β e β β<α This element must still fail to be in the span of the n γ s. Since that sum is finite, the supremum of the indices with non-zero coefficient is strictly less than α. This gives a contradiction to the minimality of α, proving that the n α span N. Now prove that the (non-zero) n α s are linearly independent. Indeed, if we have a non-trivial (finite) relation 0 = β r β n β let α be the highest index (among finitely-many) with r α 0 and n α 0. Since n α is non-zero, it must be that ρ α 0, and then the expression of n α in terms of the basis {e γ } includes e α with non-zero coefficient (namely, ρ α ). But no n β with β < α needs e α in its expression, so for suitable s β R 0 = β r β n β = r α ρ α e α + β<α s β e β contradicting the linear independence of the e α s. Thus, we conclude that the n β s are linearly independent.

13 Garrett: Abstract Algebra 179 Exercises 11.[6.0.1] Find two integer vectors x = (x 1, x 2 ) and y = (y 1, y 2 ) such that gcd(x 1, x 2 ) = 1 and gcd(y 1, y 2 ) = 1, but Z 2 /(Zx + Zy) has non-trivial torsion. 11.[6.0.2] Show that the Z-module Q is torsion-free, but is not free. 11.[6.0.3] Let G be the group of positive rational numbers under multiplication. Is G a free Z-module? Torsion-free? Finitely-generated? 11.[6.0.4] Let G be the quotient group Q/Z. Is G a free Z-module? Torsion-free? Finitely-generated? 11.[6.0.5] Let R = Z[ 5], and let M = R 2 + R (1 + 5) Q( 5). Show that M is not free over R, although it is torsion-free. 11.[6.0.6] Given an m-by-n matrix M with entries in a PID R, give an existential argument that there are matrices A (n-by-n) and B (m-by-m) with entries in R and with inverses with entries in R, such that AMB is diagonal. 11.[6.0.7] Describe an algorithm which, given a 2-by-3 integer matrix M, finds integer matrices A, B (with integer inverses) such that AM B is diagonal. 11.[6.0.8] Let A be a torsion abelian group, meaning that for every a A there is 1 n Z such that n a = 0. Let A(p) be the subgroup of A consisting of elements a such that p l a = 0 for some integer power p l of a prime p. Show that A is the direct sum of its subgroups A(p) over primes p. 11.[6.0.9] (*) Let A be a subgroup of R n such that in each ball there are finitely-many elements of A. Show that A is a free abelian group on at most n generators.

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