## 10.70 Blow up algebras

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

Definition 10.70.1. Let $R$ be a ring. Let $I \subset R$ be an ideal.

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 $n$.

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 10.57.

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 10.70.2. 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 10.70.3. 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 $S'$.
$\square$

Example 10.70.4. Let $R$ be a ring. Let $P = R[t_1, \ldots , t_ n]$ be the polynomial algebra. Let $I = (t_1, \ldots , t_ n) \subset P$. With notation as in Definition 10.70.1 there is an isomorphism

\[ P[T_1, \ldots , T_ n]/(t_ iT_ j - t_ jT_ i) \longrightarrow \text{Bl}_ I(P) \]

sending $T_ i$ to $t_ i^{(1)}$. We leave it to the reader to show that this map is well defined. Since $I$ is generated by $t_1, \ldots , t_ n$ we see that our map is surjective. To see that our map is injective one has to show: for each $e \geq 1$ the $P$-module $I^ e$ is generated by the monomials $t^ E = t_1^{e_1} \ldots x_ n^{e_ n}$ for multiindices $E = (e_1, \ldots , e_ n)$ of degree $|E| = e$ subject only to the relations $t_ i t^ E = t_ j t^{E'}$ when $|E| = |E'| = e$ and $e_ a + \delta _{a i} = e'_ a + \delta _{a j},\ a = 1, \ldots , n$ (Kronecker delta). We omit the details.

Example 10.70.5. Let $R$ be a ring. Let $P = R[t_1, \ldots , t_ n]$ be the polynomial algebra. Let $I = (t_1, \ldots , t_ n) \subset P$. Let $a = t_1$. With notation as in Definition 10.70.1 there is an isomorphism

\[ P[x_2, \ldots , x_ n]/(t_1x_2 - t_2, \ldots , t_1x_ n - t_ n) \longrightarrow \textstyle {P[\frac{I}{a}] = P[\frac{I}{t_1}]} \]

sending $x_ i$ to $t_ i/t_1$. We leave it to the reader to show that this map is well defined. Since $I$ is generated by $t_1, \ldots , t_ n$ we see that our map is surjective. To see that our map is injective, the reader can argue that the source and target of our map are $t_1$-torsion free and that the map is an isomorphism after inverting $t_1$, see Lemma 10.70.2. Alternatively, the reader can use the description of the Rees algebra in Example 10.70.4. We omit the details.

Lemma 10.70.6. Let $R$ be a ring. Let $I = (a_1, \ldots , a_ n)$ be an ideal of $R$. Let $a = a_1$. Then there is a surjection

\[ R[x_2, \ldots , x_ n]/(a x_2 - a_2, \ldots , a x_ n - a_ n) \longrightarrow \textstyle {R[\frac{I}{a}]} \]

whose kernel is the $a$-power torsion in the source.

**Proof.**
Consider the ring map $P = \mathbf{Z}[t_1, \ldots , t_ n] \to R$ sending $t_ i$ to $a_ i$. Set $J = (t_1, \ldots , t_ n)$. By Example 10.70.5 we have $P[\frac{J}{t_1}] = P[x_2, \ldots , x_ n]/(t_1 x_2 - t_2, \ldots , t_1 x_ n - t_ n)$. Apply Lemma 10.70.3 to the map $P \to A$ to conclude.
$\square$

Lemma 10.70.7. 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 10.70.2 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 10.70.8. 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 10.70.2) 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 10.70.1.
$\square$

slogan
Lemma 10.70.9. 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 10.70.10. 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 10.70.11. 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 10.17.2. Thus the kernel of $R \to R[\frac{I}{a}]$ consists of nilpotent elements. Hence the result follows from Lemma 10.30.6.
$\square$

Lemma 10.70.12. 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.**
Any blowup algebra $R[\frac{I}{a}]$ is a domain contained in $K$ see Lemma 10.70.10. The lemma simply says that $A$ is the directed union of the ones where $a \in I$ have properties (1), (2), (3). If $R[\frac{I}{a}] \subset A$ and $R[\frac{J}{b}] \subset A$, then we have

\[ R[\textstyle {\frac{I}{a}}] \cup R[\textstyle {\frac{J}{b}}] \subset R[\textstyle {\frac{IJ}{ab}}] \subset A \]

The first inclusion because $x/a^ n = b^ nx/(ab)^ n$ and the second one because if $z \in (IJ)^ n$, then $z = \sum x_ iy_ i$ with $x_ i \in I^ n$ and $y_ i \in J^ n$ and hence $z/(ab)^ n = \sum (x_ i/a^ n)(y_ i/b^ n)$ is contained in $A$.

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.
$\square$

## Comments (1)

Comment #1817 by Joseph Gunther on