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I am trying to prove:

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

In this link: Finite injective dimension of the residue field implies that the ring is regular, there is a proof using induction on $\dim R$, and there is a point which I can't understand. Since the answer given in the link is quite old, I've made a new post.

My question is:

  1. The answer says that use Theorem 3.1.17 to deduce $\operatorname{inj dim} \mathfrak{m}=0$. But to use Theorem 3.1.17 we need that $\mathfrak{m}$ has finite injective dimension. How do we know that $\mathfrak{m}$ has finite injective dimension?

  2. How does the result that $\mathfrak{m}$ is a direct summand of $R$ imply that $\mathfrak{m}=0$?

  3. As given in the answer, I am able to show that $\operatorname{inj dim}\mathfrak{m}/(x)<\infty$. But how should we apply this result to use the induction hypothesis?

user302934
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    For 1. I am not sure exactly what the intent was, but Thm. 3.1.17 implies $k$ is injective and then the injective hull of $\mathfrak m$ is a direct sum of copies of $k$ (using Thm. 3.2.8), which means $\mathfrak m$ is itself isomorphic to a direct sum of copies of $k$. For 2. The map $R\to k$ is split, so that $k$ is isomorphic an ideal of $R$ that contains invertible elements. For 3. If $R/xR$ is regular, then $R$ is regular by Prop. 2.2.4. – Karl Kroningfeld May 31 '22 at 17:19
  • @KarlKroningfeld Thanks! Since the question is exercise in section 3.1 I am trying to use materials in 3.1, and in the linked answer it seems that Thm 3.1.17 applied to $M=\mathfrak{m}$ implies that $\operatorname{inj dim} \mathfrak{m}=0$, but for this we need the assumption that $\mathfrak{m}$ is of finite injective dimension. Is this always true? – user302934 Jun 02 '22 at 04:46
  • @KarlKroningfeld Doesn't Prop. 2.2.4 only show that $R/xR$ is regular? – user302934 Jun 02 '22 at 04:49
  • I don't see how to apply that result to $\mathfrak m$ directly. How would we distinguish between $k[\varepsilon]$, the dual numbers, and $k$? In $k[\varepsilon]$ the maximal ideal is not of finite injective dimension (by the same argument as above). Re Prop. 2.2.4: Okay, yeah I thought I could just cite that proposition. Here's a different way. We have $R/xR$ is regular of dimension $d-1$, so $m/xR$ is generated by that many elements. Thus, $\mathfrak m$ has $d$ generators. – Karl Kroningfeld Jun 02 '22 at 13:22
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    Okay, here's a different idea for 1, that I think doesn't use anything beyond $\S$3.1. Again, Thm. 3.1.17 implies $k$ is injective, which means $\hom(\mathfrak m,k)=0$ via the ses $0\to \mathfrak m\to R \to k\to 0$. We have $\hom(\mathfrak m/\mathfrak m^2,k)\hookrightarrow\hom(\mathfrak m,k)$, so the vector space $\mathfrak m/\mathfrak m^2$ is $0$. – Karl Kroningfeld Jun 02 '22 at 13:42
  • @KarlKroningfeld Thanks. Can we prove the induction step as follows: if $x\in \mathfrak{m}-\mathfrak{m}^2$ is an element which is a non-zerodivisor of $R$ then by induction hypothesis $R/xR$ is regular of dimension $d-1$, so $\mathfrak{m}/xR$ is generated by $d-1$ elements; thus $\mathfrak{m}$ is generated by $d$ elements – user302934 Jun 02 '22 at 16:25
  • Yes (that's the "different way" from my earlier comment :)) In fact, if you want, you can write all this up in an answer to your question. Or, I will do so when I find the time. – Karl Kroningfeld Jun 02 '22 at 16:30

1 Answers1

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While I cannot speak for a different user's intent, I have some suggestions for how to fill in the details.

1 and 2. We want to show that $\mathfrak m=0$ in case $\dim R=0$ and $k$ has finite injective dimension. Note that $k$ is an injective module over $R$, by Theorem 3.1.17. Applying the exact functor $\hom_R(-,k)$ to the short exact sequence $0\to \mathfrak m\to R \to k \to 0$, the following is an exact sequence $$0\to\hom_R(k,k)\to\hom_R(R,k)\to\hom_R(\mathfrak m,k)\to 0$$ The map $\hom_R(k,k)\to\hom_R(R,k)$ is nonzero. Also, both $\hom_R(k,k)$ and $\hom_R(R,k)$ are isomorphic to $k$. It follows that $$\hom_R(\mathfrak m/\mathfrak m^2,k)\cong\hom_R(\mathfrak m,k)=0$$Applying Nakayama's lemma, we deduce $\mathfrak m=0$.

(Note that one can prove $\mathfrak m$ is injective by taking an injective hull and using Theorem 3.2.8, then $\mathfrak m=0$ follows from the fact that $0\to \mathfrak m\to R \to k\to 0$ is split exact.)

${}$3. This follows a suggestion in the comments as it seems to be clearer than the approach in the linked answer. Let $d=\dim R$ and $x\in\mathfrak m/\mathfrak m^2$ a non-zero divisor. Applying Lemma 3.1.16 (and its proof), we have isomorphisms$$\operatorname{Ext}_{R/xR}^i(k,\mathfrak m/xR)\cong \operatorname{Ext}_R^{i+1}(k,\mathfrak m)\quad\text{and}\quad \operatorname{Ext}_{R/xR}^i(k,R/xR)\cong\operatorname{Ext}_R^{i+1}(k,R)$$for all $i\ge 0$. Moreover, Proposition 3.1.14 and the fact that $\operatorname{inj dim}_Rk<\infty$ implies $$\operatorname{Ext}_R^i(k,\mathfrak m)\cong\operatorname{Ext}_R^i(k,R)\qquad(i\gg 0)$$Thus, in the long exact sequence associated to $0\to \mathfrak m/xR\to R/xR\to k\to 0$ we have isomorphisms$$\operatorname{Ext}_{R/xR}^i(k,\mathfrak m/xR)\xrightarrow{\sim}\operatorname{Ext}_{R/xR}^i(k,R/xR)\qquad(i\gg 0)$$and this forces $$\operatorname{Ext}_{R/xR}^i(k,k)=0\qquad(i\gg0)$$which means $k$ has a finite injective dimension over $R/xR$. The induction hypothesis implies $R/xR$ is regular. Hence, there exist $x_1,\dots,x_{d-1}\in\mathfrak m$ such that their images $\overline x_1,\dots,\overline x_{d-1}$ in $\mathfrak m/xR$ comprise a system of generators for $\mathfrak m/xR$. So, $\mathfrak m$ is generated by $x_1,\dots,x_{d-1},x$ and thus $R$ is regular.

Karl Kroningfeld
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    Thanks. I think the residue field of the local ring $R/xR$ is $(R/xR)/(\mathfrak{m}/xR)=R/\mathfrak{m}=k$, so doesn't the induction hypothesis apply directly to $R/xR$ without showing $\mathfrak{m}/xR$ has finite injective dimension over $R/xR$? – user302934 Jun 06 '22 at 09:38
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    @user302934 I have tried to write something up for part 3. I think your idea makes sense, though one has to be careful about the difference between the injective dimension for $k$ over $R$ versus the injective dimension for $k$ over $R/xR$. (Note that there are quotients $R/I$ that are not regular rings, so the connection is by no means direct.) – Karl Kroningfeld Jun 07 '22 at 05:21