This is taken from a forthcoming paper by János Kollár entitled “Variants of normality for Noetherian schemes”.

Lemma 10.119.2 (Kollár). Let $(R, \mathfrak m)$ be a local Noetherian ring. Then exactly one of the following holds:

1. $(R, \mathfrak m)$ is Artinian,

2. $(R, \mathfrak m)$ is regular of dimension $1$,

3. $\text{depth}(R) \geq 2$, or

4. there exists a finite ring map $R \to R'$ which is not an isomorphism whose kernel and cokernel are annihilated by a power of $\mathfrak m$ such that $\mathfrak m$ is not an associated prime of $R'$ and $R' \not= 0$.

Proof. Observe that $(R, \mathfrak m)$ is not Artinian if and only if $V(\mathfrak m) \subset \mathop{\mathrm{Spec}}(R)$ is nowhere dense. See Proposition 10.60.7. We assume this from now on.

Let $J \subset R$ be the largest ideal killed by a power of $\mathfrak m$. If $J \not= 0$ then $R \to R/J$ shows that $(R, \mathfrak m)$ is as in (4).

Otherwise $J = 0$. In particular $\mathfrak m$ is not an associated prime of $R$ and we see that there is a nonzerodivisor $x \in \mathfrak m$ by Lemma 10.63.18. If $\mathfrak m$ is not an associated prime of $R/xR$ then $\text{depth}(R) \geq 2$ by the same lemma. Thus we are left with the case when there is a $y \in R$, $y \not\in xR$ such that $y \mathfrak m \subset xR$.

If $y \mathfrak m \subset x \mathfrak m$ then we can consider the map $\varphi : \mathfrak m \to \mathfrak m$, $f \mapsto yf/x$ (well defined as $x$ is a nonzerodivisor). By the determinantal trick of Lemma 10.16.2 there exists a monic polynomial $P$ with coefficients in $R$ such that $P(\varphi ) = 0$. We conclude that $P(y/x) = 0$ in $R_ x$. Let $R' \subset R_ x$ be the ring generated by $R$ and $y/x$. Then $R \subset R'$ and $R'/R$ is a finite $R$-module annihilated by a power of $\mathfrak m$. Thus $R$ is as in (4).

Otherwise there is a $t \in \mathfrak m$ such that $y t = u x$ for some unit $u$ of $R$. After replacing $t$ by $u^{-1}t$ we get $yt = x$. In particular $y$ is a nonzerodivisor. For any $t' \in \mathfrak m$ we have $y t' = x s$ for some $s \in R$. Thus $y (t' - s t ) = x s - x s = 0$. Since $y$ is not a zero-divisor this implies that $t' = ts$ and so $\mathfrak m = (t)$. Thus $(R, \mathfrak m)$ is regular of dimension 1. $\square$

Comment #2227 by David Savitt on

I guess you need a little argument to see that (2) and (3) each exclude (4). [For (2), $R'$ is maximal CM hence free of rank one. For (3), use the long exact sequence of $\text{Ext}^1(R/\mathfrak{m},-)$ associated to $R \to R' \to R'/R$.]

Comment #2230 by on

Yes, this proof is really a bit too terse. I shall work on it when I go through the comments in a few weeks.

Comment #3818 by Sandor Kovacs on

At the end of the fourth paragraph (starting with "If $y\mathfrak m\subset x\mathfrak m$ then...") it should be saying that "R′/R is a finite R-module annihilated by a power of $\mathfrak m$" not that "R′/R is a finite R-module annihilated by a power of x".

Comment #6243 by Brendan Seamas Murphy on

I think the lemma is slightly wrong as stated, because condition 1 implies condition 4, taking $R' = 0$. If $(R, \mathfrak{m})$ is a local Artinian ring then $\mathfrak{m}^r = 0$ for some $r$, so the conditions that the kernel and cokernel of the map are automatic. The map $R \to R'$ isn't an isomorphism since $0$ isn't a local ring, it's a finite map, and as $R'$ is zero there are no associated primes (thus $\mathfrak{m}$ isn't an associated prime of $R'$).

Comment #6244 by Brendan Seamas Murphy on

*that the kernel and cokernel of the map are annihilated are automatic

Comment #6245 by on

Yes, I agree. So the fix is to say in part (4) that $R'$ is nonzero, right? In other words, this is the usual mistake of forgetting to check what happens with $0$.

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