The Stacks project

Lemma 15.107.2. Let $A \to B$ be a ring map. Let $\mathfrak q \subset B$ be a prime ideal lying over the prime $\mathfrak p \subset A$. Assume

  1. $A$ is a domain,

  2. $A_\mathfrak p$ is geometrically unibranch,

  3. $A \to B$ is unramified at $\mathfrak q$, and

  4. $A_\mathfrak p \to B_\mathfrak q$ is injective.

Then there exists a $g \in B$, $g \not\in \mathfrak q$ such that $B_ g$ is étale over $A$.

Proof. By Algebra, Proposition 10.152.1 after replacing $B$ by a principal localization, we can find a standard étale ring map $A \to B'$ and a surjection $B' \to B$. Denote $\mathfrak q' \subset B'$ the inverse image of $\mathfrak q$. We will show that $B' \to B$ is injective after possibly replacing $B'$ by a principal localization.

In this paragraph we reduce to the case that $B'$ is a domain. Since $A$ is a domain, the ring $B'$ is reduced, see Algebra, Lemma 15.42.1. Let $K$ be the fraction field of $A$. Then $B' \otimes _ A K$ is étale over a field, hence is a finite product of fields, see Algebra, Lemma 10.143.4. Since $A \to B'$ is étale (hence flat) the minimal primes of $B'$ are lie over $(0) \subset A$ (by going down for flat ring maps). We conclude that $B'$ has finitely many minimal primes, say $\mathfrak r_1, \ldots , \mathfrak r_ r \subset B'$. Since $A_\mathfrak p$ is geometrically unibranch and $A \to B'$ étale, the ring $B'_{\mathfrak q'}$ is a domain, see Lemmas 15.106.8 and 15.106.7. Hence $\mathfrak q' \supset \mathfrak r_ i$ for exactly one $i = i_0$. Choose $g' \in B'$, $g' \not\in \mathfrak r_{i_0}$ but $g' \in \mathfrak r_ i$ for $i \not= i_0$, see Algebra, Lemma 10.15.2. After replacing $B'$ and $B$ by $B'_{g'}$ and $B_{g'}$ we obtain that $B'$ is a domain.

Assume $B'$ is a domain, in particular $B' \subset B'_{\mathfrak q'}$. If $B' \to B$ is not injective, then $J = \mathop{\mathrm{Ker}}(B'_{\mathfrak q'} \to B_\mathfrak q)$ is nonzero. By Lemma 15.107.1 applied to $A_\mathfrak p \to B'_{\mathfrak q'}$ we find a nonzero element $a \in A_\mathfrak p$ mapping to zero in $B_\mathfrak q$ contradicting assumption (4). This finishes the proof. $\square$

Comments (0)

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Your email address will not be published.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 0GSC. Beware of the difference between the letter 'O' and the digit '0'.