finitely generated modules over a principal ideal domain


generated by \PMlinkescapephrasegenerating set

Let R be a principal ideal domainMathworldPlanetmath and let M be a finitely generatedMathworldPlanetmathPlanetmath R- module.


Let M be a submoduleMathworldPlanetmath of the R-module Rn. Then M is free and finitely generated by sn elements.


For n=1 this is clear, since M is an ideal of R and is generated by some element aR. Now suppose that the statement is true for all submodules of Rm,1mn-1.

For a submodule M of Rn we define f:MR by (k1,,kn)k1. The image of f is an ideal in R. If ={0}, then Mker(f)=(0)×Rn-1. Otherwise, =(g),g0. In the first case, elements of ker(f) can be bijectively mapped to Rn-1 by the function ker(f)Rn-1 given by (0,k1,,kn-1)(k1,,kn-1); so the image of M under this mapping is a submodule of Rn-1, which by the induction hypothesis is finitely generated and free.

Now let xM such that f(x)=gh and yM with f(y)=g. Then f(x-hy)=f(x)-f(hy)=0, which is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to x-hyker(f)Rn:=N which is isomorphicPlanetmathPlanetmathPlanetmath to a submodule of Rn-1. This shows that Rx+N=M.

Let {g1,,gs} be a basis of N. By assumptionPlanetmathPlanetmath, sn-1. We’ll show that {x,g1,,gs} is linearly independentMathworldPlanetmath. So let rx+i=1srigi=0. The first componentMathworldPlanetmathPlanetmath of the gi are 0, so the first component of rx must also be 0. Since f(x) is a multiple of g0 and 0=rf(x), then r=0. Since {g1,,gs} are linearly independent, {x,g1,,gs} is a generating set of M with s+1n elements. ∎


If M is a finitely generated R-module over a PID generated by s elements and N is a submodule of M, then N can be generated by s or fewer elements.


Let {g1,,gs} be a generating set of M and f:RsM, (r1,,rs)i=1srigi. Then the inverse imagePlanetmathPlanetmath N of N is a submodule of Rs, and according to lemma Lemma. can be generated by s or fewer elements. Let n1,,nt be a generating set of n; then ts, and since f is surjectivePlanetmathPlanetmath, f(n1),,f(nt) is a generating set of N. ∎


Let M be a finitely generated module over a principal ideal domain R.


Note that M/tor(M) is torsion-free, that is, tor(M/tor(M))={0}. In particular, if M is torsion-free, then M is free.


Let tor(M) be a proper submodule of M. Then there exists a finitely generated free submodule F of M such that M=Ftor(M).

Proof of (I): Let T=tor(M). For mM, m¯ denotes the coset modulo T generated by m. Let m be a torsion element of M/T, so there exists αR{0} such that αm¯=0, which means αm¯T. But then αm is a member of T, and this implies that M/T has no non-zero torsion elements (which is obvious if M=tor(M)).

Now let M be a finitely generated torsion-free R-module. Choose a maximal linearly independent subset S of M, and let F be the submodule of M generated by S. Let {m1,,mn} be a set of generatorsPlanetmathPlanetmathPlanetmath of M. For each i=1,,n there is a non-zero riR such that rimiF. Put r=i=1nri. Then r is non-zero, and we have rmiF for each i=1,,n. As M is torsion-free, the multiplication by r is injectivePlanetmathPlanetmath, so MrMF. So M is isomorphic to a submodule of a free moduleMathworldPlanetmathPlanetmath, and is therefore free.

Proof of (II): Let π:MM/T be defined by aa+T. Then π is surjective, so m1,,mtM can be chosen such that π(mi)=ni, where the ni’s are a basis of M/T. If 0M=i=1taimi, then 0n=i=1taini. Since n1,,nt are linearly independent in N it follows 0=a1==at. So the submodule spanned by m1,,mt of M is free.

Now let m be some element of M and π(m)=i=1taini. This is equivalent to m-(i=1taini)ker(π)=T. Hence, any mM is a sum of the form f+t, for some fF and tT. Since F is torsion-free, FT={0}, and it follows that M=FT.

Title finitely generated modules over a principal ideal domain
Canonical name FinitelyGeneratedModulesOverAPrincipalIdealDomain
Date of creation 2013-03-22 13:55:22
Last modified on 2013-03-22 13:55:22
Owner yark (2760)
Last modified by yark (2760)
Numerical id 21
Author yark (2760)
Entry type Topic
Classification msc 13E15