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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10$\begingroup$ You were told correctly. $\endgroup$– Mariano Suárez-ÁlvarezJan 29, 2011 at 3:06
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13$\begingroup$ 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. ClarkJan 29, 2011 at 3:43
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7$\begingroup$ 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 MathewJan 29, 2011 at 13:11
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4$\begingroup$ Maybe one or more of these comments should be made into an answer? $\endgroup$– Pete L. ClarkJan 30, 2011 at 1:36
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$\begingroup$ See also this awesome answer of François Brunault mathoverflow.net/a/323253/461 $\endgroup$– Pierre-Yves GaillardFeb 15, 2019 at 16:33
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2 Answers
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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 $F$.
Peter LeFanu Lumsdaine
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answered Mar 23, 2013 at 1:47
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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.
