The Stacks project

Lemma 58.31.6. In the situation of Lemma 58.31.5 the normalization of $X$ in $Y$ is a finite locally free morphism $\pi : Y' \to X$ such that

  1. the restriction of $Y'$ to $X \setminus D$ is isomorphic to $Y$,

  2. $D' = \pi ^{-1}(D)_{red}$ is an effective Cartier divisor on $Y'$, and

  3. $D'$ is a regular scheme.

Moreover, étale locally on $X$ the morphism $Y' \to X$ is a finite disjoint union of morphisms

\[ \mathop{\mathrm{Spec}}(A[x]/(x^ e - f)) \to \mathop{\mathrm{Spec}}(A) \]

where $A$ is a Noetherian ring, $f \in A$ is a nonzerodivisor with $A/fA$ regular, and $e \geq 1$ is invertible in $A$.

Proof. This is just an addendum to Lemma 58.31.5 and in fact the truth of this lemma follows almost immediately if you've read the proof of that lemma. But we can also deduce the lemma from the result of Lemma 58.31.5. Namely, taking the normalization of $X$ in $Y$ commutes with étale base change, see More on Morphisms, Lemma 37.19.2. Hence we see that we may prove the statements on the local structure of $Y' \to X$ étale locally on $X$. Thus, by Lemma 58.31.5 we may assume that $X = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian ring, that we have a nonzerodivisor $f\in A$ such that $A/fA$ is regular, and that $Y$ is a finite disjoint union of spectra of rings $A_ f[x]/(x^ e - f)$ where $e$ is invertible in $A$. We omit the verification that the integral closure of $A$ in $A_ f[x]/(x^ e - f)$ is equal to $A' = A[x]/(x^ e - f)$. (To see this argue that the localizations of $A'$ at primes lying over $(f)$ are regular.) We omit the details. $\square$

Comments (0)

There are also:

  • 6 comment(s) on Section 58.31: Tame ramification

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EYH. Beware of the difference between the letter 'O' and the digit '0'.