# Tensor product of noetherian modules is also noetherian?

Here are my thoughts. I want to show that every submodule of this tensor product is finitely generated (I think that is easier than trying to show it satisfies the ascending chain condition) Also, because $A$ is a noetherian ring, $A[x]$ is also noetherian by Hilbert basis theorem. I want to take a submodule of $M\otimes A[x]$ and show that it projects to a submodule of $M$ and a submodule of $A[x]$ Then use the generators of each of these submodules, tensor them together to show that I now have generators for my original submodule again. Is this the right idea?

Also, how does the action of $A[x]$ work on $M\bigotimes_A A[x]$? How is it an $A[x]$ module? Would it be

$a(x)*(m\otimes p(x)) = m\otimes p(x)a(x)?$

Source: Old qual exam Spring 1992 5a.

First of all, you're working too hard -- $A[x]$ is a Noetherian ring, so all we need is $M\otimes A[x]$ to be finitely generated. We know that $M$ is finitely generated as an $A$-module because it is Noetherian, and $A[x]$ is a cyclic module -- that is, it's generated by exactly one element. All you need to do is transplant the generators of $M$ and $A[x]$ into a set of generators for $M\otimes A[x]$.
The fact that $M\otimes A[x]$ is an $A[x]$-module is the consequence of a more general property often referred to as extension of scalars. Note that this isn't some "arbitrary" tensor product: it's $M\otimes_A A[x]$. The action of $A[x]$ can be defined on basic tensors by $p(x).m\otimes q(x)=m\otimes p(x)q(x)$. The universal property of the tensor product makes this an actual action of $A[x]$ on the module.