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The author of the first answer in this thread of mathoverflow concluded that a module $K'$ was finitely generated because it was squeezed between two finitely generated modules. In and of itself, this statement, as I interpret, it is false, because there are examples of non-finitely generated submodules of a f.g. module, and such non-f.g. submodules are squeezed between the latter and the zero module.

But given a little more context, his statement actually goes like this: "$K$ is a f.g. $R$-module and we have maps $K\to K'$ and $R^m \to R^n$ and an isomorphism $K'/\operatorname{im}(K) \cong R^m/\operatorname{im}(R^n)$. Therefore, we squeezed $K'$ between two f.g. modules and $K'$ must be f.g."

Why? By the way, is it also obvious that $R^m/\operatorname{im}(R^n)$ is f.g.?

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Rodrigo
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1 Answers1

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Any quotient (i.e. homomorphic image) of a f.g. module is again f.g. - just take the images of a generating set. Applying this to the case at hand:

Let $f : K \to K'$. Then $K$ f.g. $\implies f(K)$ f.g., and there is an exact sequence $0 \to f(K) \to K' \to K'/f(K) \to 0$, where $K'/f(K) \cong R^m/\text{im}(R^n)$ is f.g. If $\overline{x_1}, \ldots, \overline{x_n}$ is a generating set for $K'/f(K)$, then $K' = (x_1, \ldots, x_n)R + f(K)$ is finitely generated.

Probably what the author meant by "squeezed" was that it is the middle term in a short exact sequence. The reasoning just given shows that if the outer $2$ terms are f.g., then the middle is f.g. as well.

zcn
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