The Stacks Project


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.

  1. The blowup algebra, or the 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 $n$.
  2. 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

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

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 $\mathop{\rm Spec}(R[\frac{I}{a}]) \to \mathop{\rm Spec}(R)$ 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 $$ A = \mathop{\rm colim}\nolimits R[\textstyle{\frac{I}{a}}] $$ is a directed colimit of affine blowups $R \to R[\frac{I}{a}]$ with the following properties

  1. $a \in I \subset \mathfrak m$,
  2. $I$ is finitely generated, and
  3. 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 16300–16526 (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}

    Comments (1)

    Comment #1817 by Joseph Gunther on February 4, 2016 a 7:06 am UTC

    Minor typos: I think that, in the proofs of Lemmas 10.69.6 and 10.69.9, the instances of A, A', and A'' should be R, R', and R''.

    Add a comment on tag 052P

    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.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?