# Exercise on finitely generated $A$-modules

Here is the exercise I'm trying to solve:

Let $$M$$ be a finitely generated $$A$$-module (where $$A$$ is a commutative ring) and let $$g:M\rightarrow A^n$$ a surjective $$A$$-module morphism. Prove that $$\text{Ker}(g)$$ is finitely generated.

Here is what I'd do:

Since $$g$$ is surjective I'd choose $$\{x_1,x_2,\dots,x_n\}\subset M$$ s.t. $$g(x_i)=e_i$$ for all $$i=1\dots n$$, where $$e_i$$ is the i-th canonical generator of the direct product $$A^n$$.
Then I'd consider \begin{align}\varphi: A^n \oplus\text{Ker}(g) &\longrightarrow M\\ ((a_1,\dots,a_n),y) &\longmapsto a_1 x_1 + \dots + a_n x_n + y\end{align} which is an $$A$$-module isomorphism. Hence $$M \cong A^n\oplus \text{Ker}(g)$$ Since $$M, A^n$$ are both finitely generated as $$A$$-modules, $$\text{Ker}(g)$$ also needs to be finitely generated.

Is this acceptable? I have the feeling that it could be solved in a more concise way, without having to explicitely construct a morphism. Thanks!

• I doubt that there is any better way than noting that the kernel is a direct summand of $M$, hence, finitely generated too. You could circumvent this a little by noting that if $f\colon A^n\to M$ is the map $e_i\mapsto x_i$ (the first component of your $\varphi$), then $\mathrm{id}_M-fg$ maps $M$ surjectively onto $\ker g$. This may feel a bit more concise. – Ben Jun 24 '20 at 8:52
• You need to be careful with the argument that a submodule of a finitely generated module is also finitely generated. This may fail to be true if the ring $A$ is not Noetherian. However I think that the map $id_M - fg$ of Ben is an elegant way to prove finite generation of the kernel. – Louis Hainaut Jun 24 '20 at 9:48
• The point is not that it is a sub-module but a direct summand. This should be made clear, of course. – Ben Jun 24 '20 at 10:07

## 1 Answer

I doubt that there is any better way than noting that the kernel is a direct summand of $$M$$, hence, finitely generated too. You could circumvent this a little by noting that if $$f\colon A^n\to M$$ is the map $$e_i\mapsto x_i$$ (the first component of your $$\varphi$$), then $$\mathrm{id}_M-fg$$ maps $$M$$ surjectively onto $$\ker g$$. Perhaps this feels a bit more concise.

In case you just don't want to deal with elements, here is a closely related argument in a slightly more abstract phrasing. Consider the short exact sequence $$0\to \ker g\to M\to A^n\to 0$$ and the associated exact sequence after applying $$\hom(-,\ker g)$$: $$0\to\hom(A^n,\ker g)\to\hom(M,\ker g)\to\hom(\ker g,\ker g)\to \mathrm{ext}^1_A(A^n,\ker g).$$ Since $$A^n$$ is projective, $$\mathrm{ext}^1_A(A^n,M)=0$$ and so the map $$\hom(M,\ker g)\to\hom(\ker g,\ker g)$$, $$\varphi\mapsto \varphi|_{\ker g}$$, is surjective. Thus, there exists a morphism $$\varphi\colon M\to \ker g$$ such that $$\varphi|_{\ker g}=\mathrm{id}_{\ker g}$$. In particular, $$\varphi$$ is surjective and since $$M$$ is finitely generated, so is $$\ker g$$.