# When is a tensor product of two commutative rings noetherian?

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.

• You were told correctly. – Mariano Suárez-Álvarez Jan 29 '11 at 3:06
• 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. – Pete L. Clark Jan 29 '11 at 3:43
• 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.) – Akhil Mathew Jan 29 '11 at 13:11
• Maybe one or more of these comments should be made into an answer? – Pete L. Clark Jan 30 '11 at 1:36
• See also this awesome answer of François Brunault mathoverflow.net/a/323253/461 – 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.