# Finite injective dimension of the residue field implies that the ring is regular

Let $(R,\mathfrak m,k)$ be a noetherian local ring. If $\operatorname{inj dim}_R k$ is finite, then $R$ is regular.

This is exercise 3.1.26 from Bruns and Herzog, Cohen-Macaulay Rings. I don't see how I can use the results from this chapter to solve it. I think we must use the Ext long exact sequence, but I don't see how.

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What results do you already have? – Alex Becker Feb 17 '12 at 22:29
These notes may help: math.uchicago.edu/~may/MISC/RegularLocal.pdf – Parsa Feb 18 '12 at 4:46
I don't see any new usefull info in your notes. I don't have any results yet. – Andrei Feb 19 '12 at 11:11

There are at least two different ways to prove that $\operatorname{inj dim}\ k<\infty$ implies $R$ regular.

The first one uses the fact that $\operatorname{gl dim} R=\operatorname{inj dim} k$ and the characterization of regular local rings via $\operatorname{gl dim}$.

The second approach is more elementary and proceeds by induction on $d=\text{dim}\ R$. If $d=0$, then use Theorem 3.1.17 (same book) to deduce that $\operatorname{inj dim}\mathfrak m=0$, so $\mathfrak m$ is a direct summand of $R$, that is, $\mathfrak m=(0)$. If $d>0$, then choose an element $x\in \mathfrak m-\mathfrak m^2$ which is not a zerodivisor on $R$. In order to use the induction hypothesis you have to prove now that $\operatorname{inj dim} \mathfrak m/(x)<\infty$. This is almost clear if you notice that $\mathfrak m/x\mathfrak m$ is isomorphic with the direct sum of $(x)/x\mathfrak m$ and $\mathfrak m/(x)$ and use Corollary 3.1.15.

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the definition says that inj dim R is finite, my question is why inj dim k is finite? – Andrei Jun 22 '12 at 20:37