Let $R$ be a Noetherian ring and $n \in \mathbb{Z}$ such that for any f.g. $R$-module $M$ and $k > n$ we know that $\text{Ext}^k_R(M,R) = 0$. Does it follow from this that $\text{Ext}^k_R(M,R) = 0$ for arbitrary $R$-modules $M$ as well? I am trying to show that the $R$-module $R$ has injective dimension $\leq n$ and in order to do so I have to show that $\text{Ext}^k_R(M,R) = 0$ for arbitrary $M$.
(I have tried to look at this question, where nobody found an answer. However, that question was more general than mine because I am just interested in the $R$-module $R$ and I'm allowing $R$ to be Noetherian.)
I thought that I could try to write $M=F/N$ as a quotient of a free module $F$. Then we know that $\text{Ext}^k_R(F,R) = \text{Ext}^k_R(\bigoplus R,R) \cong \prod \text{Ext}^k_R(R,R) = 0$, where $F = \oplus R$. But I am not sure if this is of any use because I don't know anything about $N$.
Thank you very much!
It is true by Baer's criterion:
The assumption implies $\operatorname{Ext}^1_R(I,R)=0$ for any ideal and by Baer's criterion this shows that $R$ is an injective $R$-module, thus $\operatorname{Ext}^k_R(-,R)=0$ for any $k > 0$.
This was the case $n=0$. For arbitrary $n$, just take any injective resolution of $R$ and consider the co-kernel at the $n$-th step. Then you do the same argument for this co-kernel to see that it is injective. Thus the resolution stops after $n$ steps.