# The Stacks Project

## Tag 0BBI

Lemma 10.69.5. Let $R$ be a ring, $I \subset R$ an ideal, $a \in I$, and $f \in R$. Set $R' = R[\frac{I}{a}]$ and $R'' = R[\frac{fI}{fa}]$. Then there is a surjective $R$-algebra map $R' \to R''$ whose kernel is the set of $f$-power torsion elements of $R'$.

Proof. The map is given by sending $x/a^n$ for $x \in I^n$ to $f^nx/(fa)^n$. It is straightforward to check this map is well defined and surjective. Since $af$ is a nonzero divisor in $R''$ (Lemma 10.69.2) we see that the set of $f$-power torsion elements are mapped to zero. Conversely, if $x \in R'$ and $f^n x \not = 0$ for all $n > 0$, then $(af)^n x \not = 0$ for all $n$ as $a$ is a nonzero divisor in $R'$. It follows that the image of $x$ in $R''$ is not zero by the description of $R''$ following Definition 10.69.1. $\square$

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\begin{lemma}
Let $R$ be a ring, $I \subset R$ an ideal, $a \in I$, and $f \in R$.
Set $R' = R[\frac{I}{a}]$ and $R'' = R[\frac{fI}{fa}]$. Then
there is a surjective $R$-algebra map $R' \to R''$ whose kernel
is the set of $f$-power torsion elements of $R'$.
\end{lemma}

\begin{proof}
The map is given by sending $x/a^n$ for $x \in I^n$ to $f^nx/(fa)^n$.
It is straightforward to check this map is well defined and surjective.
Since $af$ is a nonzero divisor in $R''$
(Lemma \ref{lemma-affine-blowup}) we see that the set of $f$-power
torsion elements are mapped to zero. Conversely, if $x \in R'$
and $f^n x \not = 0$ for all $n > 0$, then $(af)^n x \not = 0$
for all $n$ as $a$ is a nonzero divisor in $R'$. It follows
that the image of $x$ in $R''$ is not zero by the description of
$R''$ following Definition \ref{definition-blow-up}.
\end{proof}

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