Lemma 10.122.10. Let $R \to S$ be a ring map of finite type. Let $\mathfrak p \subset R$ be a minimal prime. Assume that there are at most finitely many primes of $S$ lying over $\mathfrak p$. Then there exists a $g \in R$, $g \not\in \mathfrak p$ such that the ring map $R_ g \to S_ g$ is finite.

**Proof.**
Let $x_1, \ldots , x_ n$ be generators of $S$ over $R$. Since $\mathfrak p$ is a minimal prime we have that $\mathfrak pR_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.25.1. Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.32.3. By assumption the finite type $\kappa (\mathfrak p)$-algebra $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ has finitely many primes. Hence (for example by Lemmas 10.61.3 and 10.115.4) $\kappa (\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ is a finite ring map. Thus we may find monic polynomials $P_ i \in R_{\mathfrak p}[X]$ such that $P_ i(x_ i)$ maps to zero in $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$. By what we said above there exist $e_ i \geq 1$ such that $P(x_ i)^{e_ i} = 0$ in $S_{\mathfrak p}$. Let $g_1 \in R$, $g_1 \not\in \mathfrak p$ be an element such that $P_ i$ has coefficients in $R[1/g_1]$ for all $i$. Next, let $g_2 \in R$, $g_2 \not\in \mathfrak p$ be an element such that $P(x_ i)^{e_ i} = 0$ in $S_{g_1g_2}$. Setting $g = g_1g_2$ we win.
$\square$

## 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Your email address will not be published.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.

## Comments (4)

Comment #954 by JuanPablo on

Comment #962 by Johan on

Comment #2451 by John Smith on

Comment #2493 by Johan on