In particular, I'm told if $k$ is commutative (ring), $R$ and $S$ are commutative $k$-algebras such that $R$ is noetherian, and $S$ is a finitely generated $k$-algebra, then the tensor product $R\otimes_k S$ of $R$ and $S$ over $k$ is a noetherian ring.
Commutative Rings – When is a Tensor Product of Two Commutative Rings Noetherian?
abstract-algebracommutative-algebranoetherian
Best Answer
Even for algebras over finite fields, “tensor products of Noetherian rings are Noetherian” may fail dramatically. Assume for example that $K=F((x_i)_{i \in B})$ is a function field. When $B$ is finite, then $K \otimes_F K$ is a localization of $F[(x_i)_{i \in B}, (x'_i)_{i \in B}]$, thus noetherian. Now assume that $B$ is infinite. Then $\Omega^1_{K/F}$ has dimension $|B|$. Since it is isomorphic to $I/I^2$, where $I$ is the kernel of the multiplication map $K \otimes_F K \to K, x \otimes y \mapsto x \cdot y$, it follows that $I$ is not finitely generated, hence $K \otimes_F K$ is not noetherian.
The general case treated in the following paper:
Here is a selection of some results of that paper: Let $K,L$ be extensions of a field $F$.