Why $\ker \varphi|_N$ is finitely generated?

Question

Let $R$ Noetherian and $M$ finitely generated over $R$, prove $M$ is Noetherian.

Proof

We can suppose WLOG that $M\cong R^{\oplus s}$. The proof goes by induction. For $s=1$ it's obvious. Suppose $R^{\oplus s-1}$ is finitely generated. Consider $\varphi:R^{\oplus s}\to R^{\oplus s-1}$ the projection on the $s-1$ first coordinate. Let $N$ a submodule of $M$. We know that $K:=\ker \varphi|_N$ is a submodule of $R^{\oplus s}$ and since $\varphi|_N\neq 0$, we have that $K\neq R^{\oplus s}$. Now as I asked here, a submodule of a finitely generated module is not necessarily finitely generated. But in my couse it's written that $K$ is finitely generated, and since a submodule of a finitely generated module has no reason to be finitely generated, I don't understand why $K$ is finitely generated.

Answer 1

It is easy to see the equality $$\mathrm{ker}(\phi\vert_N) = (0^{\oplus(s-1)} \oplus R) \cap N.$$

Therefore, $\mathrm{ker}(\phi\vert_N)$ can be identified with a submodule of $R$, which is noetherian, so that $\mathrm{ker}(\phi\vert_N)$ is finitely generated as $R$-module.

Answer 2

What is $\ker\varphi$? It is isomorphic to $R$. Therefore $\ker\varphi|_N$ is an $R$-submodule of $R$, which is finitely generated because $R$ is a Noetherian ring. Note this is precisely the point in the proof where we need the fact that $R$ is Noetherian.