
## 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. Let $R$ be a domain, $I \subset R$ an ideal, and $a \in I$ a nonzero element. Then the affine blowup algebra $R[\frac{I}{a}]$ is a domain.

Proof. 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 $(R, \mathfrak m)$ 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

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. $\square$

Comment #1817 by Joseph Gunther on

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''.

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 toolbar).