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

Let $(R\,,\, S\,,\, _RM_S\,,\, _SN_R\,,\, f\,,\, g)$ be a Morita context with $NM=S$ and $R$ right Noetherian. Show that $S$ is right Noetherian as well. If we further assume $R$ is an affine $\mathbb{k}$-algebra (for some commutative ring $\mathbb{k}$), show that $S$ is as well.

I've tried to mimic a bit of what I've seen in some texts I've skimmed through, but it hasn't helped so far. I've been trying to find an inclusion of the lattice of submodules of $S$ into the lattice of submodules of $R$; perhaps it's better to try and do it directly.... But I don't know how to relate the Morita context to a chain of ideals of $S$.

Any help is appreciated

share|cite|improve this question
When is a ring "k affine"? – rschwieb Apr 15 '13 at 21:18
Whoops! I meant to assume that $R$ is an affine $\mathbb{k}$-algebra. It has been edited – Bey Apr 16 '13 at 11:26
I have no idea what you mean by "affine algebra" either. Wikipedia seems to indicate you might mean "finitely generated algebra"? – rschwieb Apr 16 '13 at 13:15
Yes, "affine" means it is finitely generated as an algebra over $\mathbb{k}$. Sorry for not being clear; I thought it conventional terminology – Bey Apr 16 '13 at 18:53
So for instance, you consider $F[x]$ is an affine $F$ algebra, right? Hmm, that's interesting: I do not ever recall hearing that being an affine $k$ algebra is Morita invariant, although it wouldn't surprise me. Does this property have a module theoretic characterization? – rschwieb Apr 16 '13 at 19:30
up vote 1 down vote accepted

Probably the way to think of it is to establish the functor $F(-):Mod-R\to Mod-S$ isomorphic to $-\otimes_R M$ for a progenerator $_RM_S$ and then look at the properties it preserves.

First establish that if $N_R$ is finitely generated, then so is the left $S$ module $F(N)$. Then, identifying the submodules of $F(N)$ as images of submodules of $N$ ia $F$, you can easily say that all submodules of $F(N)$ are finitely generated.

Applying this to $N_R=R_R$ would finish the job.

Actually it is probably possible to argue directly with a correspondence of submodules using $F$ and a counterpart inverse functor $G$, but I lack the experience to state that confidently.

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.