I read the following question:
Let $M$ be a f.g. $A$-module and $\varphi:M\to A^n$ a surjective $A$-homomorphism. Show that $\operatorname{Ker}(\varphi)$ is finitely generated.
I was thinking about this, because seems to be obvious that $\operatorname{Ker}(\varphi)$ is finitely generated, once is a submodule of $M$ (f.g.).
Maybe I do not understand the question.
Thanks in advance.
On
This problem is true as stated, with no hypotheses on the ring $A$, because of the hypothesis that $\phi : M \to A^n$ is surjective. Let $K = \mathrm{Ker}(\phi)$, so we have a short exact sequence $0 \to K \to M \to A^n \to 0$.
The short version of what I want to say next is "$A^n$ is free, so $0 \to K \to M \to A^n \to 0$ splits, so $M \cong K \oplus A^n$." Here is the longer version. Write $\iota$ for the inclusion $K \to M$ and $\pi$ for the surjection $M \to A^n$. Let $e_1$, $e_2$, ..., $e_n$ be the generators of $A^n$. Lift them to $m_1$, $m_2$, ..., $m_n \in M$, meaning that $\pi(m_j) = e_j$. Let $\sigma : A^n \to M$ be the map $\sum a_j e_j \to \sum a_j m_j$. So $\pi(\sigma(x)) = x$. This map $\sigma$ is called a splitting (or sometimes, a right splitting) of the sequence $0 \to K \to M \to A^n \to 0$.
Whenever you have a splitting, you can do the following trick: For $m \in M$, let $\rho(m) = m - \sigma(\pi(m))$. Since $\pi(\rho(m)) = \pi(m) = \pi(\sigma(\pi(m))) = \pi(m) - \pi(m) = 0$, we have $\rho(M) \in K$. So $\rho$ gives a map $M \to K$. As an exercise, check that $\rho(\iota(k))=k$ for any $k \in K$ and $\rho(\sigma(x)) = 0$ for any $x \in A^n$. I claim that $M \cong K \oplus A^n$. The map in one direction is $m \mapsto (\rho(m), \pi(m))$; the inverse map is $(k, x) \mapsto \iota(x)+\sigma(x)$. Checking that these are inverse is left as an exercise.
So, the short way or the long way, we conclude that $M \cong K \oplus A^n$. In particular, $K$ is a quotient of $M$, and $M$ is finitely generated, so $K$ is as well.
See Wikipedia's article on split short exact sequences for more.
Under many conditions, the hypothesis that $\phi$ is surjective is not needed. For Noetherian rings, any submodule of a finitely generated module is finitely generated, so you are done immediately. More generally, if $R$ is a coherent ring, the kernel of any map $A^m \to A^n$ is a finitely generated $A$-module.
The simplest example I know of a non-coherent ring is $A=k[x_1, x_2, x_3, \ldots]/\langle x_i x_j \rangle_{i \leq j}$. In other words, take the polynomial ring in infinitely many variables and set all products of two variables equal to $0$. Then multiplication by $x_1$ is a map $A \to A$, whose kernel is the maximal ideal $\langle x_1, x_2, \ldots, \rangle$, but this maximal ideal is not finitely generated.
As stated in the title, it even is false for $K$-vector spaces ($K$ being the base field). Consider $n$ distinct scalars $a_1, a_2,\dots, a_n$, and the linear map: \begin{align} f:K[X]&\longrightarrow K^n \\ P&\longmapsto\bigl(P(a_1),P(a_2),\dots,P(a_n)\bigr) \end{align} This map is surjective because of the existence of the Lagrange polynomial.
As stated in the question itself (with $M$ finitely generated) it is true if $A$ is a coherent ring:
Indeed, $M$ is the quotient of a finitely generated free module, i.e. there exists some $m\in\mathbf N$ and a surjective map $\;p:A^m\longrightarrow M$.
Now, as $A^m$ and $A^n$ are coherent modules, $\operatorname{Ker}(\varphi\circ p)=p^{-1}(\operatorname{Ker}\varphi)$ is coherent, which implies it is finitely generated, hence $\operatorname{Ker}\varphi=p\bigl(\operatorname{Ker}(\varphi\circ p\bigr)\bigr)$ is finitely generated.