Proof of Blow Up Closure Lemma (22.2.G Vakil)

Question

To set up the problem, let
$$\require{AMScd} \begin{CD} W @>>> Z\\ @VVV @VVV \\ X @>>> Y \end{CD}$$

be a commutative square with the two horizontal maps closed embeddings corresponding to ideal sheafs of finite type. Then, take the fiber product with the blow up $\beta: Bl_X Y\rightarrow Y$ and arrive at a fibered diagram

$$\begin{CD} E_{\overline{Z}} @>>{Cartier}> \overline{Z}\\ @VV{cl. em.}V @VV{cl. em}V \\ W\times_Y E_X Y @>>{loc. principle}> Z\times_Y Bl_X Y\\ @VVV @VVV \\ E_X Y @>>{Cartier}> Bl_X Y \end{CD}$$,

where $\overline{Z}$ is the scheme theoretic closure of $(Z\times_Y Bl_XY)\backslash(W\times_Y Bl_X Y)$ in $Z\times_Y Bl_XY$, and $E_{\overline{Z}}$ is the pullback of $E_X Y$.

Question: Prove the Blow Up Closure Lemma 22.2.6: Assuming $Bl_X Y$ exists, $(\overline{Z},E_{\overline{Z}})$ is the blow-up of $Z$ along $W$.

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Notes: Because ($E_{\overline{Z}},\overline{Z})$ maps to $(W\times_Y E_X Y,Z\times_Y Bl_X Y)$ which maps to $(W,Z)$ we know by the universal property of the blowup that ($E_{\overline{Z}},\overline{Z})$ factors through $(E_W Z, Bl_W Z)$. How does one construct the inverse map to show isomorphism?

Answer 1

Let's do this in the case where $Z\hookrightarrow Y$ is a closed embedding (the more general case is just more tedious). Then we have three commutative diagrams arising from the universal property of the blow up. For simplicity I will be writing cartesian (fiber product) diagrams in the notation $(A,B)\rightarrow (C,D)$.

We have that $(E_W{Z},Bl_{W} Z)\rightarrow (W,Z)\hookrightarrow (X,Y)$ factors through $(E_W{Z},Bl_{W} Z)\rightarrow (E_X Y, Bl_X{Y})\rightarrow_\beta (X,Y)$, by the universal property of $Bl_X Y$. But since $(W,Z)\hookrightarrow (X,Y)$ is a closed embedding, we have that

(1) $$(E_W{Z},Bl_{W} Z)\rightarrow_{\beta_2} (W,Z)$$ factors through

$$(E_W{Z},Bl_{W} Z)\rightarrow_\delta (\beta^{-1}(W),\beta^{-1}(Z))\rightarrow_\beta (W,Z).$$

Secondly, we have that

(2)$$(E_{\bar{Z}},\bar{Z})\hookrightarrow_\alpha (\beta^{-1}(W),\beta^{-1}(Z))\rightarrow_{\beta} (W,Z)$$

factors through

$$(E_{\bar{Z}},\bar{Z})\rightarrow_{\gamma} (E_W Z,Bl_W Z)\rightarrow_{\beta_2} (W,Z),$$

by the universal property of $(E_W X,Bl_W Z)$. Thirdly, we have that

(3)

$$(E_W Z, Bl_W Z)\rightarrow_\delta (\beta^{-1}(W),\beta^{-1}(Z))$$

factors through

$$(E_W Z, Bl_W Z)\rightarrow_{\gamma'} (E_\bar{Z},\bar{Z})\rightarrow_\alpha (\beta^{-1}(W),\beta^{-1}(Z)),$$

by the universal property of the blowup $(E_{\bar{Z}},\bar{Z})$, which is the blow up of $(\beta^{-1}(W),\beta^{-1}(Z))$.

We will show that $\gamma$ and $\gamma'$ are the desired isomorphisms: Putting this together we have from (2) that $\beta\circ \alpha=\beta_2\circ \gamma$. Noting that $\beta_2$, a blowup morphism, is unique, we then have from (1) that $\beta_2\circ\gamma=\beta\circ\delta\circ \gamma$. Since the $\delta$ morphisms we chose to use are the same, we have from (3) that $\alpha\circ \gamma'=\delta$. Thus $\beta\circ \alpha=\beta\circ\alpha \circ \gamma' \circ\gamma$, which implies that $\gamma'\circ \gamma=id$. This argument can be carried out in (basically) reverse as well, showing that $\gamma'$ and $\gamma$ are isomorphisms between $(E_{\bar{Z}}, \bar{Z})$ and $(E_W Z,Bl_W {Z})$.