# The Stacks Project

## Tag 0BIP

Lemma 10.69.3. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal and $a \in I$. Set $J = IS$ and let $b \in J$ be the image of $a$. Then $S[\frac{J}{b}]$ is the quotient of $S \otimes_R R[\frac{I}{a}]$ by the ideal of elements annihilated by some power of $b$.

Proof. Let $S'$ be the quotient of $S \otimes_R R[\frac{I}{a}]$ by its $b$-power torsion elements. The ring map $$S \otimes_R R[\textstyle{\frac{I}{a}}] \longrightarrow S[\textstyle{\frac{J}{b}}]$$ is surjective and annihilates $a$-power torsion as $b$ is a nonzerodivisor in $S[\frac{J}{b}]$. Hence we obtain a surjective map $S' \to S[\frac{J}{b}]$. To see that the kernel is trivial, we construct an inverse map. Namely, let $z = y/b^n$ be an element of $S[\frac{J}{b}]$, i.e., $y \in J^n$. Write $y = \sum x_is_i$ with $x_i \in I^n$ and $s_i \in S$. We map $z$ to the class of $\sum s_i \otimes x_i/a^n$ in $S'$. This is well defined because an element of the kernel of the map $S \otimes_R I^n \to J^n$ is annihilated by $a^n$, hence maps to zero in $S'$. $\square$

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\begin{lemma}
\label{lemma-blowup-base-change}
Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal
and $a \in I$. Set $J = IS$ and let $b \in J$ be the image of $a$.
Then $S[\frac{J}{b}]$ is the quotient of $S \otimes_R R[\frac{I}{a}]$
by the ideal of elements annihilated by some power of $b$.
\end{lemma}

\begin{proof}
Let $S'$ be the quotient of $S \otimes_R R[\frac{I}{a}]$ by its
$b$-power torsion elements. The ring map
$$S \otimes_R R[\textstyle{\frac{I}{a}}] \longrightarrow S[\textstyle{\frac{J}{b}}]$$
is surjective and annihilates $a$-power torsion as $b$ is a nonzerodivisor
in $S[\frac{J}{b}]$. Hence we obtain a surjective map $S' \to S[\frac{J}{b}]$.
To see that the kernel is trivial, we construct an inverse map. Namely, let
$z = y/b^n$ be an element of $S[\frac{J}{b}]$, i.e., $y \in J^n$.
Write $y = \sum x_is_i$ with $x_i \in I^n$ and $s_i \in S$.
We map $z$ to the class of $\sum s_i \otimes x_i/a^n$ in
$S'$. This is well defined because an element of the kernel of the map
$S \otimes_R I^n \to J^n$ is annihilated by $a^n$, hence maps to zero in $S'$.
\end{proof}

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