The Stacks project

Lemma 29.46.6. Let $A \subset B$ be a ring extension such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism. If $A \not= B$, then either there exists a $b \in B$, $b \not\in A$ with $b^2 \in A$ and $b^3 \in A$ or there exists a prime number $p$ and a $b \in B$, $b \not\in A$ with $pb \in A$ and $b^ p \in A$.

Proof. The argument is almost exactly the same as in the proof of Lemma 29.46.5 but we write everything out to make sure it works.

Recall that $A \subset B$ is integral (Lemma 29.44.7). By Lemma 29.46.1 we may assume that $B$ is generated by a single element over $A$. Hence $B$ is finite over $A$ (Algebra, Lemma 10.36.5). Hence the support of $B/A$ as an $A$-module is closed and not empty (Algebra, Lemmas 10.40.5 and 10.40.2). Let $\mathfrak p \subset A$ be a minimal prime of the support. After replacing $A \subset B$ by $A_\mathfrak p \subset B_\mathfrak p$ (permissible by Lemma 29.46.3) we may assume that $(A, \mathfrak m)$ is a local ring, that $B$ is finite over $A$, and that $B/A$ has support $\{ \mathfrak m\} $ as an $A$-module. Since $B/A$ is a finite module, we see that $I = \text{Ann}_ A(B/A)$ satisfies $\mathfrak m = \sqrt{I}$ (Algebra, Lemma 10.40.5). Let $\mathfrak m' \subset B$ be the unique prime ideal lying over $\mathfrak m$. Because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a homeomorphism, we find that $\mathfrak m' = \sqrt{IB}$. For $f \in \mathfrak m'$ pick $n \geq 1$ such that $f^ n \in IB$. Then also $f^{n + 1} \in IB$. Since $IB \subset A$ by our choice of $I$ we conclude that $f^ n, f^{n + 1} \in A$. Using Lemma 29.46.4 we conclude our lemma is true if $\mathfrak m' \not\subset A$. If $\mathfrak m' \subset A$, then $\mathfrak m' = \mathfrak m$. Since $A \not= B$ we conclude the map $\kappa = A/\mathfrak m \to B/\mathfrak m' = \kappa '$ of residue fields cannot be an isomorphism. By Lemma 29.10.2 we conclude that the characteristic of $\kappa $ is a prime number $p$ and that the extension $\kappa '/\kappa $ is purely inseparable. Pick $b \in B$ whose image in $\kappa '$ is an element not contained in $\kappa $ but whose $p$th power is in $\kappa $. Then $b \not\in A$, $b^ p \in A$, and $pb \in A$ (because $pb \in \mathfrak m' = \mathfrak m \subset A$) as desired. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 29.46: Universal homeomorphisms of affine schemes

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