I have a reasonably precise question which I hope is clear enough to get a nice answer. Let R be a Noetherian non-commutative ring which is finite as a module (and flat/free if it helps) over it's center Z(R) which we can assume has finite Krull dimension. One can also assume R integral over Z(R). By a two sided prime ideal, I mean a two sided ideal P where if $xRy$ is contained in P, then either x and/or y are in P. Consider now the abelian category of left modules and suppose we know that $Ext^m(R/P,M)=0$ for any M and $m>n$ for some n (that is independent of P). Now there aren't enough (two-sided) prime ideals in a general non-commutative ring, but there are a fair number in rings such as the one I describe above.

*Question: Does it follow that for any finitely generated left module N, $Ext^m(N,M)=0$ for $m>n'$ which can depend on N *. The issue I am having is that we don't have quite as effective a filtration sequence of any N, we only have a filtration such that $N_i/N_{i-1}=I/P$ for some left ideal and a prime P. That might be the end of the story, but there might be other tricks I am not aware of. Either way, I haven't been able to sort it out or find a good reference. If there is an extra hypothesis that helps the situation, I'd like to know about it.


Yes. If we know that $\mathrm{Ext^{m}(R/P,M)}=0$ for all $m\gneq n$, then that tells us, in this situation, that the injective dimension of $M$ is less than or equal to $n$.

To see this, take an minimal injective resolution $I$ of $M$ (This is an injective resolution such that $ker(\partial_{i})\leq_{e}I^{i}$ for all $i$, i.e. you are doing the obvious thing and taking injective hulls at each step while constructing it). Let us define $E_{P}$ to be the injective hull of a uniform left ideal of $R/P$ as an $R$ module. This construction does not depend on the choice of uniform left ideal if you take your ring to be left noetherian.

A result, which appears as Lemma 2.3 in

K.A. Brown, Fully bounded noetherian rings of finite injective dimension, Quart. J. Math. Oxford (2), 41, (1990) 1-13

tells us that $E_{P}$ appears as a summand in the $i^{th}$ term of $I$ if and only if $\mathrm{Ext^{i}(R/P,M)}=0$ is not torsion as a left $R/P$ module. (Note that the bimodule structure of $R/P$ gives these $\mathrm{Ext}$ groups a left $R/P$ module structure.) Thus, if they all vanish for $m\geq n$, we have that for any $m\geq n$, $I^{m}$ contains no summands of the form $E_{P}$ for any prime.

However, your assumptions on $R$ tell us that $R$ is a left fully bounded noetherian ring. In such a ring, we have that every indecomposable injective is of the form $E_{P}$ for some prime $P$, and that every injective is a direct sum of indecomposable injectives. Thus $I^{m}=0$ for all $m\gneq n$, and $M$ has injective dimension less than or equal to $n$. In particular $\mathrm{Ext^{m}(N,M)}=0$ for any module $N$ and any $m\gneq n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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