Let $R$ be a Noetherian commutative unital ring. It is generally speaking not true that the tensor product of two Noetherian $R$-algebras is Noetherian (e.g. take $R$ to be a field, and consider the tensor square of some crazy field extension).

What is true is that a tensor product of a finite type $R$-algebra and a Noetherian $R$-algebra is Noetherian. It is not true, however, that an $R$-algebra whose tensor product with any Noetherian $R$-algebra is Noetherian has to be of finite type (consider the power series ring, for example). What is the name for the class of $R$-algebras that have this property?

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    $\begingroup$ I don't think power series algebras have the property you claim. On the other hand, all localizations of finitely generated algebras do. $\endgroup$ May 1, 2019 at 12:25
  • $\begingroup$ I am not aware of even one example of a noetherian ring R, and two noetherian R-algebras A,B such that $A\otimes_R B$ is noetherian, in which not at least one of A,B is essentially of finite type over R. $\endgroup$
    – the L
    May 2, 2019 at 8:43
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    $\begingroup$ Meta discussion here: meta.mathoverflow.net/questions/4200/flood-of-new-users $\endgroup$ May 2, 2019 at 15:00

1 Answer 1


A (not necessarily commutative) algebra $A$ over a commutative noetherian ring $R$ is called strongly noetherian if for every noetherian $R$-algebra $R'$ the extension $A \otimes_R R'$ is noetherian. See this paper of Artin, Small, and Zhang for a reference. This property has also been discussed in this MO question.

This is an issue of importance in noncommutative algebra, where there exist finitely generated noetherian algebras that do not remain noetherian after extending the base field. (There is a well-known example of this phenomenon due to Resco and Small.) But as your question indicates, it is already an interesting property to consider for commutative algebras.


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