Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You were told correctly. – Mariano Suárez-Alvarez 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
thanks for your input. – Heidi Jan 29 '11 at 23:44
Maybe one or more of these comments should be made into an answer? – Pete L. Clark Jan 30 '11 at 1:36
up vote 2 down vote accepted

Even for fields this fails 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 $F$.
share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.