The Stacks project

Proof. The first statement is clear from the construction of the relative Proj via glueing, see Constructions, Section 27.15. For $a \in I$ denote $a^{(1)}$ the element $a$ seen as an element of degree $1$ in the Rees algebra $\bigoplus _{n \geq 0} I^ n$. Since these elements generate the Rees algebra over $A$ we see that $\text{Proj}(\bigoplus _{d \geq 0} I^ d)$ is covered by the affine opens $D_{+}(a^{(1)})$. The affine scheme $D_{+}(a^{(1)})$ is the spectrum of the affine blowup algebra $A' = A[\frac{I}{a}]$, see Algebra, Definition 10.70.1. This finishes the proof. $\square$

Proof. Let $\mathcal{I}_2$ be the ideal sheaf of $Z_2$ in $X_2$. Denote $g : X_1 \to X_2$ the given morphism. Then the ideal sheaf $\mathcal{I}_1$ of $Z_1$ is the image of $g^*\mathcal{I}_2 \to \mathcal{O}_{X_1}$ (by definition of the inverse image, see Schemes, Definition 26.17.7). By Constructions, Lemma 27.16.10 we see that $X_1 \times _{X_2} X_2'$ is the relative Proj of $\bigoplus _{n \geq 0} g^*\mathcal{I}_2^ n$. Because $g$ is flat the map $g^*\mathcal{I}_2^ n \to \mathcal{O}_{X_1}$ is injective with image $\mathcal{I}_1^ n$. Thus we see that $X_1 \times _{X_2} X_2' = X_1'$. $\square$

Proof. As blowing up commutes with restrictions to open subschemes (Lemma 31.32.3) the first statement just means that $X' = X$ if $Z = \emptyset $. In this case we are blowing up in the ideal sheaf $\mathcal{I} = \mathcal{O}_ X$ and the result follows from Constructions, Example 27.8.14.

The second statement is local on $X$, hence we may assume $X$ affine. Say $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(A/I)$. By Lemma 31.32.2 we see that $X'$ is covered by the spectra of the affine blowup algebras $A' = A[\frac{I}{a}]$. Then $IA' = aA'$ and $a$ maps to a nonzerodivisor in $A'$ according to Algebra, Lemma 10.70.2. This proves the lemma as the inverse image of $Z$ in $\mathop{\mathrm{Spec}}(A')$ corresponds to $\mathop{\mathrm{Spec}}(A'/IA') \subset \mathop{\mathrm{Spec}}(A')$.

Consider the canonical map $\psi _{univ, 1} : b^*\mathcal{I} \to \mathcal{O}_{X'}(1)$, see discussion following Constructions, Definition 27.16.7. We claim that this factors through an isomorphism $\mathcal{I}_ E \to \mathcal{O}_{X'}(1)$ (which proves the final assertion). Namely, on the affine open corresponding to the blowup algebra $A' = A[\frac{I}{a}]$ mentioned above $\psi _{univ, 1}$ corresponds to the $A'$-module map

where $a^{(1)}$ is as in Algebra, Definition 10.70.1. We omit the verification that this is the map $I \otimes _ A A' \to IA' = aA'$. $\square$

Proof. We see that $b : X' \to X$ is an object of $\mathcal{C}$ according to Lemma 31.32.4. Let $f : Y \to X$ be an object of $\mathcal{C}$. We have to show there exists a unique morphism $Y \to X'$ over $X$. Let $D = f^{-1}(Z)$. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal sheaf of $Z$ and let $\mathcal{I}_ D$ be the ideal sheaf of $D$. Then $f^*\mathcal{I} \to \mathcal{I}_ D$ is a surjection to an invertible $\mathcal{O}_ Y$-module. This extends to a map $\psi : \bigoplus f^*\mathcal{I}^ d \to \bigoplus \mathcal{I}_ D^ d$ of graded $\mathcal{O}_ Y$-algebras. (We observe that $\mathcal{I}_ D^ d = \mathcal{I}_ D^{\otimes d}$ as $D$ is an effective Cartier divisor.) By the material in Constructions, Section 27.16 the triple $(1, f : Y \to X, \psi )$ defines a morphism $Y \to X'$ over $X$. The restriction

is unique. The open $Y \setminus D$ is scheme theoretically dense in $Y$ according to Lemma 31.13.4. Thus the morphism $Y \to X'$ is unique by Morphisms, Lemma 29.7.10 (also $b$ is separated by Constructions, Lemma 27.16.9). $\square$

Proof. Since the exceptional divisor over $U'$ is cut out by the image of $a$ in $A' = A[\frac{I}{a}]$ one direction is clear. Conversely, assume that the image of $a$ in $\mathcal{O}_{X', x'}$ cuts out $E$. Since every element of $I$ maps to an element of the ideal defining $E$ over $b^{-1}(U)$ we see that elements of $I$ become divisible by $a$ in $\mathcal{O}_{X', x'}$. Thus for $f \in I^ n$ we can write $f = \psi (f) a^ n$ for some $\psi (f) \in \mathcal{O}_{X', x'}$. Observe that since $a$ maps to a nonzerodivisor of $\mathcal{O}_{X', x'}$ the element $\psi (f)$ is uniquely characterized by this. Then we define

Here we use the description of blowup algebras given following Algebra, Definition 31.32.1. The uniqueness mentioned above shows that this is an $A$-algebra homomorphism. This gives a morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{X', x"}) \to \mathop{\mathrm{Spec}}(A') = U'$. By the universal property of blowing up (Lemma 31.32.5) this is a morphism over $X'$, which of course implies that $x' \in U'$. $\square$

Proof. Immediate from the universal property of blowups (Lemma 31.32.5). $\square$

Proof. Combine Lemma 31.32.2 with Algebra, Lemma 10.70.9. $\square$

Proof. Combine Lemma 31.32.2 with Algebra, Lemma 10.70.10. $\square$

Proof. The exceptional divisor $E \subset X'$ is an effective Cartier divisor and $X' \setminus E \to X \setminus Z$ is an isomorphism, see Lemma 31.32.4. Thus it suffices to show the following: given an effective Cartier divisor $D \subset S$ of a scheme $S$ none of the generic points of irreducible components of $S$ are contained in $D$. To see this, we may replace $S$ by the members of an affine open covering. Hence by Lemma 31.13.2 we may assume $S = \mathop{\mathrm{Spec}}(A)$ and $D = V(f)$ where $f \in A$ is a nonzerodivisor. Then we have to show $f$ is not contained in any minimal prime ideal $\mathfrak p \subset A$. If so, then $f$ would map to a nonzerodivisor contained in the maximal ideal of $R_\mathfrak p$ which is a contradiction with Algebra, Lemma 10.25.1. $\square$

Proof. By Lemmas 31.32.2 and 31.13.2 this reduces to the following algebra fact: Let $A$ be a ring, $I \subset A$ an ideal, $a \in I$, and $x \in A$ a nonzerodivisor. Then the image of $x$ in $A[\frac{I}{a}]$ is a nonzerodivisor. Namely, suppose that $x (y/a^ n) = 0$ in $A[\frac{I}{a}]$. Then $a^ mxy = 0$ in $A$ for some $m$. Hence $a^ my = 0$ as $x$ is a nonzerodivisor. Whence $y/a^ n$ is zero in $A[\frac{I}{a}]$ as desired. $\square$

Proof. Let $E \subset X'$ be the exceptional divisor of $b$ which is an effective Cartier divisor by Lemma 31.32.4. Then $(b')^{-1}E$ is an effective Cartier divisor on $X''$ by Lemma 31.32.11. Let $E' \subset X''$ be the exceptional divisor of $b'$ (also an effective Cartier divisor). Consider the effective Cartier divisor $E'' = E' + (b')^{-1}E$. By construction the ideal of $E''$ is $(b \circ b')^{-1}\mathcal{I} (b \circ b')^{-1}\mathcal{J} \mathcal{O}_{X''}$. Hence according to Lemma 31.32.5 there is a canonical morphism from $X''$ to the blowup $c : Y \to X$ of $X$ in $\mathcal{I}\mathcal{J}$. Conversely, as $\mathcal{I}\mathcal{J}$ pulls back to an invertible ideal we see that $c^{-1}\mathcal{I}\mathcal{O}_ Y$ defines an effective Cartier divisor, see Lemma 31.13.9. Thus a morphism $c' : Y \to X'$ over $X$ by Lemma 31.32.5. Then $(c')^{-1}b^{-1}\mathcal{J}\mathcal{O}_ Y = c^{-1}\mathcal{J}\mathcal{O}_ Y$ which also defines an effective Cartier divisor. Thus a morphism $c'' : Y \to X''$ over $X'$. We omit the verification that this morphism is inverse to the morphism $X'' \to Y$ constructed earlier. $\square$

Proof. The surjection of graded $\mathcal{O}_ X$-algebras

defines via Constructions, Lemma 27.18.5 a closed immersion

Hence $b$ is projective, see Morphisms, Definition 29.43.1. The second statement follows for example from the characterization of relatively ample invertible sheaves in Morphisms, Lemma 29.37.4. Some details omitted. $\square$

Proof. The condition that $Z \to X$ is of finite presentation means that $Z$ is cut out by a finite type quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$, see Morphisms, Lemma 29.21.7. Write $\mathcal{A} = \bigoplus _{n \geq 0} \mathcal{I}^ n$ so that $X' = \underline{\text{Proj}}(\mathcal{A})$. Note that $X \setminus Z$ is a quasi-compact open of $X$ by Properties, Lemma 28.24.1. Since $b^{-1}(X \setminus Z) \to X \setminus Z$ is an isomorphism (Lemma 31.32.4) the same result shows that $b^{-1}(X \setminus Z) \setminus Z'$ is quasi-compact open in $X'$. Hence $U = X \setminus (Z \cup b(Z'))$ is quasi-compact open in $X$. By Lemma 31.31.5 there exist a $d > 0$ and a finite type $\mathcal{O}_ X$-submodule $\mathcal{F} \subset \mathcal{I}^ d$ such that $Z' = \underline{\text{Proj}}(\mathcal{A}/\mathcal{F}\mathcal{A})$ and such that the support of $\mathcal{I}^ d/\mathcal{F}$ is contained in $X \setminus U$.

Since $\mathcal{F} \subset \mathcal{I}^ d$ is an $\mathcal{O}_ X$-submodule we may think of $\mathcal{F} \subset \mathcal{I}^ d \subset \mathcal{O}_ X$ as a finite type quasi-coherent sheaf of ideals on $X$. Let's denote this $\mathcal{J} \subset \mathcal{O}_ X$ to prevent confusion. Since $\mathcal{I}^ d / \mathcal{J}$ and $\mathcal{O}/\mathcal{I}^ d$ are supported on $X \setminus U$ we see that $V(\mathcal{J})$ is contained in $X \setminus U$. Conversely, as $\mathcal{J} \subset \mathcal{I}^ d$ we see that $Z \subset V(\mathcal{J})$. Over $X \setminus Z \cong X' \setminus b^{-1}(Z)$ the sheaf of ideals $\mathcal{J}$ cuts out $Z'$ (see displayed formula below). Hence $V(\mathcal{J})$ equals $Z \cup b(Z')$. It follows that also $V(\mathcal{I}\mathcal{J}) = Z \cup b(Z')$ set theoretically. Moreover, $\mathcal{I}\mathcal{J}$ is an ideal of finite type as a product of two such. We claim that $X'' \to X$ is isomorphic to the blowing up of $X$ in $\mathcal{I}\mathcal{J}$ which finishes the proof of the lemma by setting $Y = V(\mathcal{I}\mathcal{J})$.

First, recall that the blowup of $X$ in $\mathcal{I}\mathcal{J}$ is the same as the blowup of $X'$ in $b^{-1}\mathcal{J} \mathcal{O}_{X'}$, see Lemma 31.32.12. Hence it suffices to show that the blowup of $X'$ in $b^{-1}\mathcal{J} \mathcal{O}_{X'}$ agrees with the blowup of $X'$ in $Z'$. We will show that

as ideal sheaves on $X''$. This will prove what we want as $\mathcal{I}_ E^ d$ cuts out the effective Cartier divisor $dE$ and we can use Lemmas 31.32.7 and 31.32.12.

To see the displayed equality of the ideals we may work locally. With notation $A$, $I$, $a \in I$ as in Lemma 31.32.2 we see that $\mathcal{F}$ corresponds to an $R$-submodule $M \subset I^ d$ mapping isomorphically to an ideal $J \subset R$. The condition $Z' = \underline{\text{Proj}}(\mathcal{A}/\mathcal{F}\mathcal{A})$ means that $Z' \cap \mathop{\mathrm{Spec}}(A[\frac{I}{a}])$ is cut out by the ideal generated by the elements $m/a^ d$, $m \in M$. Say the element $m \in M$ corresponds to the function $f \in J$. Then in the affine blowup algebra $A' = A[\frac{I}{a}]$ we see that $f = (a^ dm)/a^ d = a^ d (m/a^ d)$. Thus the equality holds. $\square$