# The Stacks Project

## Tag 00QB

Lemma 10.122.15. Let $R \to S$ be a finite type ring map. Suppose that $S$ is quasi-finite over $R$. Let $S' \subset S$ be the integral closure of $R$ in $S$. Then

1. $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(S')$ is a homeomorphism onto an open subset,
2. if $g \in S'$ and $D(g)$ is contained in the image of the map, then $S'_g \cong S_g$, and
3. there exists a finite $R$-algebra $S'' \subset S'$ such that (1) and (2) hold for the ring map $S'' \to S$.

Proof. Because $S/R$ is quasi-finite we may apply Theorem 10.122.13 to each point $\mathfrak q$ of $\mathop{\rm Spec}(S)$. Since $\mathop{\rm Spec}(S)$ is quasi-compact, see Lemma 10.16.10, we may choose a finite number of $g_i \in S'$, $i = 1, \ldots, n$ such that $S'_{g_i} = S_{g_i}$, and such that $g_1, \ldots, g_n$ generate the unit ideal in $S$ (in other words the standard opens of $\mathop{\rm Spec}(S)$ associated to $g_1, \ldots, g_n$ cover all of $\mathop{\rm Spec}(S)$).

Suppose that $D(g) \subset \mathop{\rm Spec}(S')$ is contained in the image. Then $D(g) \subset \bigcup D(g_i)$. In other words, $g_1, \ldots, g_n$ generate the unit ideal of $S'_g$. Note that $S'_{gg_i} \cong S_{gg_i}$ by our choice of $g_i$. Hence $S'_g \cong S_g$ by Lemma 10.23.2.

We construct a finite algebra $S'' \subset S'$ as in (3). To do this note that each $S'_{g_i} \cong S_{g_i}$ is a finite type $R$-algebra. For each $i$ pick some elements $y_{ij} \in S'$ such that each $S'_{g_i}$ is generated as $R$-algebra by $1/g_i$ and the elements $y_{ij}$. Then set $S''$ equal to the sub $R$-algebra of $S'$ generated by all $g_i$ and all the $y_{ij}$. Details omitted. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 29551–29565 (see updates for more information).

\begin{lemma}
\label{lemma-quasi-finite-open-integral-closure}
Let $R \to S$ be a finite type ring map.
Suppose that $S$ is quasi-finite over $R$.
Let $S' \subset S$ be the integral closure of $R$ in $S$. Then
\begin{enumerate}
\item $\Spec(S) \to \Spec(S')$ is a homeomorphism
onto an open subset,
\item if $g \in S'$ and $D(g)$ is contained in the image
of the map, then $S'_g \cong S_g$, and
\item there exists a finite $R$-algebra $S'' \subset S'$
such that (1) and (2) hold for the ring map
$S'' \to S$.
\end{enumerate}
\end{lemma}

\begin{proof}
Because $S/R$ is quasi-finite we may apply
Theorem \ref{theorem-main-theorem} to
each point $\mathfrak q$ of $\Spec(S)$.
Since $\Spec(S)$ is quasi-compact, see
Lemma \ref{lemma-quasi-compact}, we may choose
a finite number of $g_i \in S'$, $i = 1, \ldots, n$
such that $S'_{g_i} = S_{g_i}$, and such that
$g_1, \ldots, g_n$ generate the unit ideal in $S$
(in other words the standard opens of $\Spec(S)$ associated
to $g_1, \ldots, g_n$ cover all of $\Spec(S)$).

\medskip\noindent
Suppose that $D(g) \subset \Spec(S')$
is contained in the image. Then $D(g) \subset \bigcup D(g_i)$.
In other words, $g_1, \ldots, g_n$ generate the unit ideal of
$S'_g$. Note that $S'_{gg_i} \cong S_{gg_i}$ by our choice
of $g_i$. Hence $S'_g \cong S_g$ by Lemma \ref{lemma-cover}.

\medskip\noindent
We construct a finite algebra $S'' \subset S'$ as
in (3). To do this note that each $S'_{g_i} \cong S_{g_i}$
is a finite type $R$-algebra. For each $i$ pick
some elements $y_{ij} \in S'$ such that each
$S'_{g_i}$ is generated as $R$-algebra by $1/g_i$
and the elements $y_{ij}$. Then set $S''$
equal to the sub $R$-algebra of $S'$ generated by all $g_i$
and all the $y_{ij}$. Details omitted.
\end{proof}

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