# Strongly Noetherian property. When is the tensor $A\otimes_{k}B$ Noetherian for Noetherian rings $A$ and $B$?

Let $k$ be a field. It is well-known that $A\otimes_{k}B$ is not necessarily Noetherian even if $k$-algebras $A$ and $B$ are Noetherian. For example $\mathbb{R}\otimes_{\mathbb{Q}}\mathbb{R}$.

1. When is the tensor $A\otimes_{k}B$ Noetherian for Noetherian "commutative" $k$-algebras $A$ and $B$?

2. What if $A$ is noncommutative? Are there any good criteria for $A\otimes_{k}B$ to be Noetherian? If $B$ is a finitely generated $k$-algebra, Hilbert's basis theorem implies that $A\otimes_{k}B$ is again Noetherian. So we need to check this with quite nasty $B$.

My primary motivation to ask these questions is the second question. Such ring $A$ is called a "strongly Noethrian ring" and has a lot of good properties, but I don't know many examples. Moreover I realized that things are not very clear even in commutative case and I need to understand commutative case first. I would appreciate it if experts on MO could let me know good criteria for this property and provide me with examples.

Rings I have in my mind are weakly noncommutative in the sense that they are commutative up to scalar multiplication such as quantum planes and their $good$ hypersurfaces.

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Just a quick question, is this a relative notion? When you say that $A$ is a strongly Noetherian ring, is there a fixed $k$ that comes along with it? Or is $A$ is a strongly Noetherian ring, does that mean the tensor product is Noetherian for all $k \to A$ and $k \to B$? –  Karl Schwede Sep 26 '12 at 18:42
We fix a $k$-algebra structure on $A$. Only $B$ varies. Thanks for clarifying the question. –  Fermion Sep 26 '12 at 19:17

You could try having a look at Yekutieli and Zhang's paper Homological Transcendence Degree (http://arxiv.org/abs/math/04120130). They call a $k$-algebra $A$ "doubly Noetherian" if $A \otimes_k A^{op}$ is Noetherian, and "rationally Noetherian" if $A \otimes_k U$ is Noetherian for every division ring $U$.

There's a lot of other stuff in the paper too, but the first couple of sections look at several conditions for when rings (usually simple Artinian or division) are doubly or rationally noetherian, and there's a nice example of a field in section 7 which is not doubly noetherian.

It's not quite what you asked (it certainly won't help with the $q$-plane, but maybe the $q$-torus or the $q$-division ring?) but it might help with intuition and provide some examples.

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Thank you for the reference, eithil. I was not aware of this paper. It may not be exactly what I am looking for, but I will take a look at it. –  Fermion Sep 27 '12 at 18:31

Try looking at section 4 of "Generic flatness for strongly noetherian algebras" by Artin, Small and Zhang. In fact, they work with a stronger property called universally noetherian (UN); this just means that you no longer require $B$ above to be commutative, merely noetherian. It turns out that most times when your ring is strongly noetherian it is possible to show it is universally noetherian too. In particular some nice results they prove are:

• [Propn 4.1] UN is preserved when taking Ore extensions, finite module extensions, localisations by denominator sets.

• [Propn 4.13] Certain twisted homogeneous coordinate rings have UN.

• [Propn. 4.24] Connected graded noetherian domains of GK dimension 2 are UN (alg. closed field hypothesis).

There are several other results in this section and indeed the whole paper is very nice and worth reading.

EDIT (by Nazih Nahlus): In this very nice paper (Thanks to Andrew), you also find that:

• [Prop. 4.1] (c) UN is also preserved under almost normalizing extensions.

• [Prop. 4.10] Suppose A is an N-filtered R-algebra. If the associated graded ring gr(A) is universally right noetherian, then so is A.

• [Cor. 4.11] Weyl algebras and universal enveloping algebras of finite-dimensional Lie algebras are universally noetherian.

I think one should rather say universally noetherian R-algebra (because we are tensoring over R).

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I found the paper a few days ago and am now reading it. Thank you for the information! –  Fermion Oct 1 '12 at 8:41

Since we have a natural identification $$A/I\otimes_{k} B \cong (A\otimes_{k}B)/(I\otimes_{k}B),$$ any quotients of UN rings Andrew mentioned are again UN.

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Let $k\subset K$ be a field extension. N. Bourbaki in Algebre. Chapitre 8, Modules et anneaux semi-simples (edition 1958), exercise 22, page 99, gives the following criterion: if the extension $k\subset K$ is algebraic, then $K\otimes_kK$ is Noetherian iff $[K:k]<\infty$.
Later on, in 1978, P. Vamos in the paper On the minimal prime ideals of a tensor product of two fields proves a more general result: for a field extension $k\subset K$, $K\otimes_kK$ is Noetherian iff $k\subset K$ is finitely generated.
Inspired by Vamos' result, Resco, Small and Wadsworth in the paper Tensor products of division rings and finite generation of subfields prove a (partially) noncommutative result: let $D$ be a division algebra over a field $k$ and $k\subset K$ a commutative subfield of $D$. Then $D\otimes_kL$ is Noetherian iff the extension $k\subset K$ is finitely generated. As a by-product they get that $D\otimes_kD^0$ Noetherian implies $k\subset K$ finitely generated for every commutative subfield $K$ of $D$ containing $k$.