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!


1 Answer 1


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.

  • $\begingroup$ Thank you very much, MooS! I really like your argument! Should it be $\text{Ext}^1_R(R/I,R)$ instead of $\text{Ext}^1_R(I,R)$ in your answer? We want to show that every $f: I \rightarrow R$ can be extended to $R \rightarrow R$, i.e. that the inclusion in $0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$ induces a surjection in the corresponding hom-sets in $0 \rightarrow \text{Hom}_R(R/I,R) \rightarrow \text{Hom}_R(R,R) \rightarrow \text{Hom}_R(I,R) \rightarrow \text{Ext}^1_R(R/I,R) \rightarrow \cdots$. I believe that for this we would need $\text{Ext}^1_R(R/I,R)$ to vanish. $\endgroup$
    – Martin
    Apr 22, 2017 at 6:56
  • 1
    $\begingroup$ Yes, you are of course right. $\endgroup$
    – MooS
    Apr 22, 2017 at 14:25

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