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I'm revising Commutative Algebra and have a quick (possibly stupid) question - I can't get my head around the fact that a submodule of a finitely generated module is not necessarily finitely generated - so M is finitely generated if there is a finite set Y such that very element of M is a linear combination of elements of Y, but if A is a submodule of M any element in A is also an element of M and thus can be written as a linear combination of elements of Y as well?

Any help greatly appreciated, thanks!!

jemima
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  • See http://math.stackexchange.com/questions/125015/finitely-generated-module-with-a-submodule-thats-not-finitely-generated for an example of what can go wrong. The problem is that the generators you pick for your module might not be in the submodule. – jmracek Jan 01 '14 at 22:21
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    Well certainly any ring is a finitely generated module over itself. But what are the submodules of rings called? When are they finitely generated? – PVAL-inactive Jan 01 '14 at 22:22
  • @JennyFirman $10$ questions, $0$ accepted answers! Why? (Btw, your question is broad enough to be tagged as "abstract-algebra".) –  Jan 01 '14 at 22:46

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The problem is that the generators for $M$ may not lie in the submodule $A$. As an example, every ideal $I$ in a commutative ring $R$ is an $R$-submodule of the $R$-module $R$, but there are rings with ideals that are not finitely generated (i.e., non-Noetherian rings), although $R$ is always finitely generated as an $R$-module (by the element $1$). Explicitly, the ideal $(x_1, x_2, ...)$ in the polynomial ring $k[x_1, x_2, ...]$ in infinitely many variables requires infinitely many generators (note: the generator $1$ for $R$ is not contained in $(x_1, x_2, ...)$).

zcn
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You "proof" doesn't work since $Y$ is not a subset of $A$ (which is the minimal requirement for a generating system).

Rings over which (left) submodules of f.g. (left) modules are f.g. are precisely the (left) noetherian rings.

Martin Brandenburg
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