Lemma 10.119.3. Let $R$ be a local ring with maximal ideal $\mathfrak m$. Assume $R$ is Noetherian, has dimension $1$, and that $\dim (\mathfrak m/\mathfrak m^2) > 1$. Then there exists a ring map $R \to R'$ such that

1. $R \to R'$ is finite,

2. $R \to R'$ is not an isomorphism,

3. the kernel and cokernel of $R \to R'$ are annihilated by a power of $\mathfrak m$, and

4. $\mathfrak m$ is not an associated prime of $R'$.

Proof. This follows from Lemma 10.119.2 and the fact that $R$ is not Artinian, not regular, and does not have depth $\geq 2$ (the last part because the depth does not exceed the dimension by Lemma 10.72.3). $\square$

Comment #1504 by Kollar on

Dear Johan,

2 comments on Lemma tag 00P9.

1. About R' you want to also claim that m is not an associated prime of R'. (To avoid eg R'=R+(R/m) with R/m being nilpotent.)

2. You may like the following variant of Serre's criterion of normality, which implies the Lemma (and other things as well).

Let $(R,m)$ be a local Noetherian ring. Then exactly one of the following holds.

1. $(R,m)$ is Artinian,
2. $(R,m)$ is regular of dimension $1$,
3. $\depth_mR\geq 2$ or
4. $(R,m)$ is a non-normal pair (= in your terminology: there is an R' as you want)

\end{lem}

Proof.
$(R,m)$ is not Artinian iff $V(m)\subset \spec R$ is nowhere dense; we assume this from now on.

Let $J\subset R$ be the largest ideal killed by a power of $m$. If $J\neq 0$ then $R\to R/J$ shows that $(R,m)$ is a non-normal pair.

Otherwise $J=0$ and there is an $r\in m$ that is not a zero-divisor. If $m$ is not an associated prime of $R/(r)$ then $\depth_mR\geq 2$. Thus we are left with the case when there is an $a\in R\setminus (r)$ such that $am\subset rR$.

If $am\subset rm$ then, by the determinantal trick proposition 2.4 of Atiyah--Macdonald, $a/r$ satisfies a monic polynomial hence $R\subset R[a/r]$ shows that $(R,m)$ is a non-normal pair.

Otherwise there is a $t_0\in m$ such that $at_0=r$; in particular $a$ is not a zero-divisor. For any $t\in m$ we have $at=rt'$ for some $t'\in R$. Thus $a(t-t't_0)=at-t'(at_0)=at-t'r=0$.
Since $a$ is not a zero-divisor this implies that $t=t't_0$ and so $m=(t_0)$. Thus $(R,m)$ is regular of dimension 1. $\qed$

Comment #1508 by on

OK, thank you very much. I added your lemma and I used it to prove this one. See this commit. Of course this lemma can be used to give a quick proof of Serre's criterion, but I haven't made those changes yet.

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