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.

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    $\begingroup$ You were told correctly. $\endgroup$ – Mariano Suárez-Álvarez Jan 29 '11 at 3:06
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    $\begingroup$ Indeed your hypotheses imply that $R \otimes_k S$ is finitely generated as an algebra over the Noetherian ring $R$, hence Noetherian by the Hilbert Basis Theorem. $\endgroup$ – Pete L. Clark Jan 29 '11 at 3:43
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    $\begingroup$ Though in general, it is false that the tensor product of two noetherian rings is noetherian (take e.g. a non-perfect field $k$ of characteristic $p$ and consider $R=S$ to be the perfect closure. Then I claim that $R \otimes_k S$ is non-noetherian. Indeed, for each $n$, consider $\alpha \in R$ such that $\alpha^{p^n} \in k$ but $\alpha^{p^{n-1}} \notin k$. Then $(1 \otimes \alpha - \alpha \otimes 1)$ is such that the $p^{n}$th power is zero but the $p^{n-1}$th power is not. Hence the nilradical is not nilpotent, meaning the tensor product is nonnoetherian.) $\endgroup$ – Akhil Mathew Jan 29 '11 at 13:11
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    $\begingroup$ Maybe one or more of these comments should be made into an answer? $\endgroup$ – Pete L. Clark Jan 30 '11 at 1:36
  • $\begingroup$ See also this awesome answer of François Brunault mathoverflow.net/a/323253/461 $\endgroup$ – Pierre-Yves Gaillard Feb 15 '19 at 16:33

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:

P. Vámos, On the minimal prime ideals of a tensor product of two fields, Mathematical Proceedings of the Cambridge Philosophical Society, 84 (1978), pp. 25-35

Here is a selection of some results of that paper: Let $K,L$ be extensions of a field $F$.

  • If $K$ is a finitely generated field extension of $F$, then $K \otimes_F L$ is noetherian.
  • If $K,L \subseteq F^{\mathrm{alg}}$ are separable algebraic extensions of $F$, and $L$ is normal, then $K \otimes_F L$ is noetherian iff $K \otimes_F L$ is a finite product of fields iff $[K \cap L : F] < \infty$.
  • If there is an extension $M$ of $F$ which sits inside $K$ and $L$, which has a strictly ascending chain of intermediate fields, then $K \otimes_F L$ is not noetherian.
  • If $K \otimes_F L$ is noetherian, then $\min(\mathrm{tr.deg}_F(K),\mathrm{tr.deg}_F(L)) < \infty$.
  • $K \otimes_F K$ is noetherian iff the ascending chain condition holds for intermediate fields of $K/F$ iff $K$ is a finitely generated field extension of $F$.

If $S$ is finitely generated as a $k$-algebra, we can write $S\cong k[x_1,\ldots,x_n]/I$ for some $n\in\mathbb{N}$ and some ideal $I$. It follows that $$ R\otimes_kS\cong R\otimes_k(k[x_1,\ldots,x_n]/I)\cong R[x_1,\ldots,x_n]/I $$ Since $R$ is noetherian, it follows from Hilbert's basis theorem that $R[x_1,\ldots,x_n]$ is noetherian. Finally, homomorphic images of noetherian rings are noetherian, so that $R[x_1,\ldots,x_n]/I$ is noetherian.


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