In order to prove the Hilbert base theorem ($A$ noetherian $\Longrightarrow$ $A[x]$ noetherian) consider an ideal $I\subset A[x]$, and $(f_1,\dots,f_k)$ such that the leading coefficients of the $f_i$ generate the ideal $J$ of the leading coefficients of all polynomials in $I$. So one can decompose each $f\in I$ as $f=h+F$, where $F\in(f_1,\dots,f_k)$ and $\deg h<s$, where $s$ is the maximum degree of the $f_i$.
(This requires a short proof by iterating a simple 'division' process.)
So, one can express $I$ as $(f_1,\dots, f_k)+M\cap I$, where $M=\left<1,x,\dots, x^{s-1}\right>$ is the submodule (not the ideal!) generated by the first $s$ powers of $x$. Therefore $I$ is finitely generated, since both terms are. But to say that $M\cap I$ is f.g. one needs to know that $M$ is noetherian. But now we mean as $A$-module, correct? And this is true why?
Thank you in advance.