The Stacks project

Lemma 10.123.9. Suppose $R\subset S$ is an inclusion of domains and let $x \in S$. Assume $x$ is (strongly) transcendental over $R$ and that $S$ is finite over $R[x]$. Then $R \to S$ is not quasi-finite at any prime of $S$.

Proof. As a first case, assume that $R$ is normal, see Definition 10.37.11. By Lemma 10.37.14 we see that $R[x]$ is normal. Take a prime $\mathfrak q \subset S$, and set $\mathfrak p = R \cap \mathfrak q$. Assume that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite. This would be the case if $R \to S$ is quasi-finite at $\mathfrak q$. Let $\mathfrak r = R[x] \cap \mathfrak q$. Then since $\kappa (\mathfrak p) \subset \kappa (\mathfrak r) \subset \kappa (\mathfrak q)$ we see that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak r)$ is finite too. Thus the inclusion $\mathfrak r \supset \mathfrak p R[x]$ is strict. By going down for $R[x] \subset S$, see Proposition 10.38.7, we find a prime $\mathfrak q' \subset \mathfrak q$, lying over the prime $\mathfrak pR[x]$. Hence the fibre $\mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p))$ contains a point not equal to $\mathfrak q$, namely $\mathfrak q'$, whose closure contains $\mathfrak q$ and hence $\mathfrak q$ is not isolated in its fibre.

If $R$ is not normal, let $R \subset R' \subset K$ be the integral closure $R'$ of $R$ in its field of fractions $K$. Let $S \subset S' \subset L$ be the subring $S'$ of the field of fractions $L$ of $S$ generated by $R'$ and $S$. Note that by construction the map $S \otimes _ R R' \to S'$ is surjective. This implies that $R'[x] \subset S'$ is finite. Also, the map $S \subset S'$ induces a surjection on $\mathop{\mathrm{Spec}}$, see Lemma 10.36.17. We conclude by Lemma 10.122.6 and the normal case we just discussed. $\square$

Comments (0)

There are also:

  • 3 comment(s) on Section 10.123: Zariski's Main Theorem

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Your email address will not be published.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 00Q1. Beware of the difference between the letter 'O' and the digit '0'.