## Tag `052P`

## 10.69. Blow up algebras

In this section we make some elementary observations about blowing up.

Definition 10.69.1. Let $R$ be a ring. Let $I \subset R$ be an ideal.

- The
blowup algebra, or theRees algebra, associated to the pair $(R, I)$ is the graded $R$-algebra $$ \text{Bl}_I(R) = \bigoplus\nolimits_{n \geq 0} I^n = R \oplus I \oplus I^2 \oplus \ldots $$ where the summand $I^n$ is placed in degree $n$.- Let $a \in I$ be an element. Denote $a^{(1)}$ the element $a$ seen as an element of degree $1$ in the Rees algebra. Then the
affine blowup algebra$R[\frac{I}{a}]$ is the algebra $(\text{Bl}_I(R))_{(a^{(1)})}$ constructed in Section 10.56.

In other words, an element of $R[\frac{I}{a}]$ is represented by an expression of the form $x/a^n$ with $x \in I^n$. Two representatives $x/a^n$ and $y/a^m$ define the same element if and only if $a^k(a^mx - a^ny) = 0$ for some $k \geq 0$.

Lemma 10.69.2. Let $R$ be a ring, $I \subset R$ an ideal, and $a \in I$. Let $R' = R[\frac{I}{a}]$ be the affine blowup algebra. Then

- the image of $a$ in $R'$ is a nonzerodivisor,
- $IR' = aR'$, and
- $(R')_a = R_a$.

Proof.Immediate from the description of $R[\frac{I}{a}]$ above. $\square$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 $S'$. $\square$Lemma 10.69.4. Let $R$ be a ring, $I \subset R$ an ideal, and $a \in I$. Set $R' = R[\frac{I}{a}]$. If $f \in R$ is such that $V(f) = V(I)$, then $f$ maps to a nonzerodivisor in $R'$ and $R'_f = R'_a = R_a$.

Proof.We will use the results of Lemma 10.69.2 without further mention. The assumption $V(f) = V(I)$ implies $V(fR') = V(IR') = V(aR')$. Hence $a^n = fb$ and $f^m = ac$ for some $b, c \in R'$. The lemma follows. $\square$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. $\square$Lemma 10.69.6. If $R$ is reduced then every (affine) blowup algebra of $R$ is reduced.

Proof.Let $I \subset R$ be an ideal and $a \in I$. Suppose $x/a^n$ with $x \in I^n$ is a nilpotent element of $R[\frac{I}{a}]$. Then $(x/a^n)^m = 0$. Hence $a^N x^m = 0$ in $R$ for some $N \geq 0$. After increasing $N$ if necessary we may assume $N = me$ for some $e \geq 0$. Then $(a^e x)^m = 0$ and since $R$ is reduced we find $a^e x = 0$. This means that $x/a^n = 0$ in $R[\frac{I}{a}]$. $\square$Lemma 10.69.7. If $R$ is a domain then every (affine) blowup algebra of $R$ is a domain.

Proof.Let $I \subset R$ be an ideal and $a \in I$ nonzero. Suppose $x/a^n$, $y/a^m$ with $x \in I^n$, $y \in I^m$ are elements of $R[\frac{I}{a}]$ whose product is zero. Then $a^N x y = 0$ in $R$. Since $R$ is a domain we conclude that either $x = 0$ or $y = 0$. $\square$Lemma 10.69.8. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $a \in I$. If $a$ is not contained in any minimal prime of $R$, then $\mathop{\mathrm{Spec}}(R[\frac{I}{a}]) \to \mathop{\mathrm{Spec}}(R)$ has dense image.

Proof.If $a^k x = 0$ for $x \in R$, then $x$ is contained in all the minimal primes of $R$ and hence nilpotent, see Lemma 10.16.2. Thus the kernel of $R \to R[\frac{I}{a}]$ consists of nilpotent elements. Hence the result follows from Lemma 10.29.6. $\square$Lemma 10.69.9. Let $R$ be a Noetherian ring. Let $a, a_2, \ldots, a_r$ be a regular sequence in $R$. With $I = (a, a_2, \ldots, a_r)$ the blowup algebra $R' = R[\frac{I}{a}]$ is isomorphic to $R'' = R[y_2, \ldots, y_r]/(a y_i - a_i)$.

Proof.There is a canonical map $R'' \to R'$ sending $y_i$ to the class of $a_i/a$. Since every element $x$ of $I$ can be written as $ra + \sum r_i a_i$ we see that $x/a = r + \sum r_i a_i/a$ is in the image of the map. Hence our map is surjective. Suppose that $z = \sum r_E y^E \in R''$ maps to zero in $R'$. Here we use the multi-index notation $E = (e_2, \ldots, e_r)$ and $y^E = y_2^{e_2} \ldots y_r^{e_r}$. Let $d$ be the maximum of the degrees $|E| = \sum e_i$ of the multi-indices which occur with a nonzero coefficient $r_E$ in $z$. Then we see that $$ a^d z = \sum r_E a^{d - |E|} a_2^{e_2} \ldots a_r^{e_r} $$ is zero in $R$; here we use that $a$ is a nonzerodivisor on $R$. Since a regular sequence is quasi-regular by Lemma 10.68.2 we conclude that $r_E \in I$ for all $E$. This means that $z$ is divisible by $a$ in $R''$. Say $z = az'$. Then $z'$ is in the kernel of $R'' \to R'$ and we see that $z'$ is divisible by $a$ and so on. In other words, $z$ is an element of $\bigcap a^n R''$. Since $R''$ is Noetherian by Krull's intersection theorem $z$ maps to zero in $R''_\mathfrak p$ for every prime ideal $\mathfrak p$ containing $aR''$, see Remark 10.50.6. On the other hand, if $\mathfrak p \subset R''$ does not contain $a$, then $R''_a \cong R_a \cong R'_a$ and we find that $z$ maps to zero in $R''_\mathfrak p$ as well. We conclude that $z$ is zero by Lemma 10.23.1. $\square$Lemma 10.69.10. Let $(R, \mathfrak m)$ be a local domain with fraction field $K$. Let $R \subset A \subset K$ be a valuation ring which dominates $R$. Then $$ A = \mathop{\mathrm{colim}}\nolimits R[\textstyle{\frac{I}{a}}] $$ is a directed colimit of affine blowups $R \to R[\frac{I}{a}]$ with the following properties

- $a \in I \subset \mathfrak m$,
- $I$ is finitely generated, and
- the fibre ring of $R \to R[\frac{I}{a}]$ at $\mathfrak m$ is not zero.

Proof.Consider a finite subset $E \subset A$. Say $E = \{e_1, \ldots, e_n\}$. Choose a nonzero $a \in R$ such that we can write $e_i = f_i/a$ for all $i = 1, \ldots, n$. Set $I = (f_1, \ldots, f_n, a)$. We claim that $R[\frac{I}{a}] \subset A$. This is clear as an element of $R[\frac{I}{a}]$ can be represented as a polynomial in the elements $e_i$. The lemma follows immediately from this observation. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 16334–16560 (see updates for more information).

```
\section{Blow up algebras}
\label{section-blow-up}
\noindent
In this section we make some elementary observations about blowing up.
\begin{definition}
\label{definition-blow-up}
Let $R$ be a ring.
Let $I \subset R$ be an ideal.
\begin{enumerate}
\item The {\it blowup algebra}, or the {\it Rees algebra}, associated to
the pair $(R, I)$ is the graded $R$-algebra
$$
\text{Bl}_I(R) =
\bigoplus\nolimits_{n \geq 0} I^n =
R \oplus I \oplus I^2 \oplus \ldots
$$
where the summand $I^n$ is placed in degree $n$.
\item Let $a \in I$ be an element. Denote $a^{(1)}$ the element $a$
seen as an element of degree $1$ in the Rees algebra. Then the
{\it affine blowup algebra} $R[\frac{I}{a}]$ is the algebra
$(\text{Bl}_I(R))_{(a^{(1)})}$ constructed in Section \ref{section-proj}.
\end{enumerate}
\end{definition}
\noindent
In other words, an element of $R[\frac{I}{a}]$ is represented by
an expression of the form $x/a^n$ with $x \in I^n$. Two representatives
$x/a^n$ and $y/a^m$ define the same element if and only if
$a^k(a^mx - a^ny) = 0$ for some $k \geq 0$.
\begin{lemma}
\label{lemma-affine-blowup}
Let $R$ be a ring, $I \subset R$ an ideal, and $a \in I$.
Let $R' = R[\frac{I}{a}]$ be the affine blowup algebra. Then
\begin{enumerate}
\item the image of $a$ in $R'$ is a nonzerodivisor,
\item $IR' = aR'$, and
\item $(R')_a = R_a$.
\end{enumerate}
\end{lemma}
\begin{proof}
Immediate from the description of $R[\frac{I}{a}]$ above.
\end{proof}
\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}
\begin{lemma}
\label{lemma-blowup-in-principal}
Let $R$ be a ring, $I \subset R$ an ideal, and $a \in I$.
Set $R' = R[\frac{I}{a}]$. If $f \in R$ is such that $V(f) = V(I)$,
then $f$ maps to a nonzerodivisor in $R'$ and $R'_f = R'_a = R_a$.
\end{lemma}
\begin{proof}
We will use the results of Lemma \ref{lemma-affine-blowup}
without further mention.
The assumption $V(f) = V(I)$ implies $V(fR') = V(IR') = V(aR')$.
Hence $a^n = fb$ and $f^m = ac$ for some $b, c \in R'$.
The lemma follows.
\end{proof}
\begin{lemma}
\label{lemma-blowup-add-principal}
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}
\begin{lemma}
\label{lemma-blowup-reduced}
If $R$ is reduced then every (affine) blowup algebra of $R$ is reduced.
\end{lemma}
\begin{proof}
Let $I \subset R$ be an ideal and $a \in I$. Suppose $x/a^n$ with
$x \in I^n$ is a nilpotent element of $R[\frac{I}{a}]$. Then
$(x/a^n)^m = 0$. Hence $a^N x^m = 0$ in $R$ for some $N \geq 0$.
After increasing $N$ if necessary we may assume $N = me$ for some
$e \geq 0$. Then $(a^e x)^m = 0$ and since $R$ is reduced we find
$a^e x = 0$. This means that $x/a^n = 0$ in $R[\frac{I}{a}]$.
\end{proof}
\begin{lemma}
\label{lemma-blowup-domain}
If $R$ is a domain then every (affine) blowup algebra of $R$ is a domain.
\end{lemma}
\begin{proof}
Let $I \subset R$ be an ideal and $a \in I$ nonzero.
Suppose $x/a^n$, $y/a^m$ with $x \in I^n$, $y \in I^m$
are elements of $R[\frac{I}{a}]$ whose product is zero.
Then $a^N x y = 0$ in $R$. Since $R$ is a domain we conclude
that either $x = 0$ or $y = 0$.
\end{proof}
\begin{lemma}
\label{lemma-blowup-dominant}
Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $a \in I$.
If $a$ is not contained in any minimal prime of $R$, then
$\Spec(R[\frac{I}{a}]) \to \Spec(R)$ has dense image.
\end{lemma}
\begin{proof}
If $a^k x = 0$ for $x \in R$, then $x$ is contained in all the
minimal primes of $R$ and hence nilpotent, see
Lemma \ref{lemma-Zariski-topology}.
Thus the kernel of $R \to R[\frac{I}{a}]$ consists of nilpotent
elements. Hence the result follows from
Lemma \ref{lemma-image-dense-generic-points}.
\end{proof}
\begin{lemma}
\label{lemma-blowup-regular-sequence}
Let $R$ be a Noetherian ring. Let $a, a_2, \ldots, a_r$ be a
regular sequence in $R$. With $I = (a, a_2, \ldots, a_r)$ the
blowup algebra $R' = R[\frac{I}{a}]$ is isomorphic to
$R'' = R[y_2, \ldots, y_r]/(a y_i - a_i)$.
\end{lemma}
\begin{proof}
There is a canonical map $R'' \to R'$ sending $y_i$ to the class
of $a_i/a$. Since every element $x$ of $I$ can be written
as $ra + \sum r_i a_i$ we see that
$x/a = r + \sum r_i a_i/a$ is in the image of the map. Hence
our map is surjective. Suppose that $z = \sum r_E y^E \in R''$
maps to zero in $R'$. Here we use the multi-index notation
$E = (e_2, \ldots, e_r)$ and $y^E = y_2^{e_2} \ldots y_r^{e_r}$.
Let $d$ be the maximum of the degrees $|E| = \sum e_i$ of the
multi-indices which occur with a nonzero coefficient $r_E$ in $z$.
Then we see that
$$
a^d z = \sum r_E a^{d - |E|} a_2^{e_2} \ldots a_r^{e_r}
$$
is zero in $R$; here we use that $a$ is a nonzerodivisor on $R$.
Since a regular sequence is quasi-regular
by Lemma \ref{lemma-regular-quasi-regular}
we conclude that $r_E \in I$ for all $E$.
This means that $z$ is divisible by $a$ in $R''$.
Say $z = az'$. Then $z'$ is in the kernel of $R'' \to R'$
and we see that $z'$ is divisible by $a$ and so on.
In other words, $z$ is an element of $\bigcap a^n R''$.
Since $R''$ is Noetherian by Krull's intersection theorem
$z$ maps to zero in $R''_\mathfrak p$ for every prime ideal
$\mathfrak p$ containing $aR''$, see
Remark \ref{remark-intersection-powers-ideal}.
On the other hand, if $\mathfrak p \subset R''$ does
not contain $a$, then $R''_a \cong R_a \cong R'_a$ and
we find that $z$ maps to zero in $R''_\mathfrak p$ as well.
We conclude that $z$ is zero by Lemma \ref{lemma-characterize-zero-local}.
\end{proof}
\begin{lemma}
\label{lemma-valuation-ring-colimit-affine-blowups}
Let $(R, \mathfrak m)$ be a local domain with fraction field $K$.
Let $R \subset A \subset K$ be a valuation ring which dominates $R$.
Then
$$
A = \colim R[\textstyle{\frac{I}{a}}]
$$
is a directed colimit of affine blowups $R \to R[\frac{I}{a}]$ with
the following properties
\begin{enumerate}
\item $a \in I \subset \mathfrak m$,
\item $I$ is finitely generated, and
\item the fibre ring of $R \to R[\frac{I}{a}]$ at $\mathfrak m$
is not zero.
\end{enumerate}
\end{lemma}
\begin{proof}
Consider a finite subset $E \subset A$. Say $E = \{e_1, \ldots, e_n\}$.
Choose a nonzero $a \in R$ such that we can write $e_i = f_i/a$ for
all $i = 1, \ldots, n$. Set $I = (f_1, \ldots, f_n, a)$.
We claim that $R[\frac{I}{a}] \subset A$. This is clear as an element
of $R[\frac{I}{a}]$ can be represented as a polynomial in the elements
$e_i$. The lemma follows immediately from this observation.
\end{proof}
```

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