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. 



Finite Homological Dimension of R/P for all P for module finite non-commutative rings
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 $m\gneq n$.


