$M/x_nM$ finitely generated over $ k[x_1,..., x_{n-1}] $ as graded module

Question

I'm trying to figure out the following:

Let $M$ be a finitely generated graded module over $S=k[x_1,..., x_n]$ with standard grading. Let $K$ be the kernel of multiplication by $x_n$ in $M$. Then a text I am reading states that $K$ and $M/x_nM$ are finitely generated as $k [x_1,..., x_{n-1}]$-modules.

While I easily see them being finitely generated over $S$ by noetherianity of $S$ and hence $M$, I would love to see an elegant argument why it's even finitely generated without $x_n$. $ M $ of course doesn't have to be!

Thanks

Answer 1

$K=\{z\in M:x_nz=0\}$, so $x_nK=0$. We also have $x_n(M/x_nM)=0$. It's well known and easy to prove that if $I$ is an ideal in a ring $R$ and $N$ an $R$-module (finitely generated) such that $IN=0$ then $N$ is an $R/I$-module (finitely generated). In the end use that $k[x_1,\dots,x_n]/(x_n)\simeq k[x_1,\dots,x_{n-1}]$.