# Algebras such that the tensor product with any Noetherian algebra is Noetherian

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?

• I don't think power series algebras have the property you claim. On the other hand, all localizations of finitely generated algebras do. – Laurent Moret-Bailly May 1 '19 at 12:25
• 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. – the L May 2 '19 at 8:43
• Meta discussion here: meta.mathoverflow.net/questions/4200/flood-of-new-users – Steven Landsburg May 2 '19 at 15:00

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.