Let $M$ be a Noetherian $A$-module. Show that $M[x]$ is a Noetherian $A[x]$-module.
Comments: I was able to show that $M[x] \cong M\otimes_A A[x]$. It is possible to construct an isomorphism $\phi$ so that $$\frac{M\oplus A[x]}{ker(\phi)} \cong M\otimes_A A[x].$$ Thus, it would suffice to show that $M$ and $A[x]$ are Noetherian $A$-modules. $M$ is Noetherian $A$-module by assumption. The problem lies in proving that $A[x]$ is too. I thought about using the Hilbert’s Basis Theorem, but I do not have the hypothesis of being Noetherian.
