## Tag `02ML`

Chapter 10: Commutative Algebra > Section 10.121: Quasi-finite maps

Lemma 10.121.10. 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 10.24.1. Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.31.2. 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 10.60.3 and 10.114.4) $\kappa(\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ 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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 29072–29080 (see updates for more information).

```
\begin{lemma}
\label{lemma-generically-finite}
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.
\end{lemma}
\begin{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 \ref{lemma-minimal-prime-reduced-ring}.
Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see
Lemma \ref{lemma-locally-nilpotent}.
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 \ref{lemma-finite-type-algebra-finite-nr-primes} and
\ref{lemma-Noether-normalization})
$\kappa(\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$
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.
\end{proof}
```

## Comments (4)

## Add a comment on tag `02ML`

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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.