The Stacks project

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). $\square$

Comments (2)

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 be a local Noetherian ring. Then exactly one of the following holds.

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


is not Artinian iff is nowhere dense; we assume this from now on.

Let be the largest ideal killed by a power of . If then shows that is a non-normal pair.

Otherwise and there is an that is not a zero-divisor. If is not an associated prime of then . Thus we are left with the case when there is an such that .

If then, by the determinantal trick proposition 2.4 of Atiyah--Macdonald, satisfies a monic polynomial hence shows that is a non-normal pair.

Otherwise there is a such that ; in particular is not a zero-divisor. For any we have for some . Thus .
Since is not a zero-divisor this implies that and so . Thus is regular of dimension 1.

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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