Now that we know local rings of the structure sheaves are Noetherian, we now want to establish that the local map given by the (un)analytification map is faithfully flat. Just like last time, we will only discuss the smooth case.
Proposition. Let $X$ be a smooth complex variety. Then the local map $\mathscr{O}_{X, x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$ induced by the analytification is faithfully flat for all $x \in X(\mathbb{C}).$
Proof. We consider the affine case first as the first step.
Step 1. Let $X = \mathbb{A}^{n}.$
Then consider the $\mathbb{C}$-algebra $R := \mathbb{C}\{z_{1}, \dots, z_{n}\}.$ It is a local ring with its unique maximal ideal being $$\mathfrak{m} := \{f \in R : f(0) = 0\} = (z_{1}, \dots, z_{n}).$$ Moreover, we have $$R/\mathfrak{m}^{N} \simeq \mathbb{C}[z_{1}, \dots, z_{n}]/(z_{1}, \dots, z_{n})^{N}$$ given by $z_{j} \mapsto z_{j},$ and this gives us $$\begin{align*}\hat{R} &= \lim_{N \geq 0} R/\mathfrak{m}^{N} \\ &\simeq \lim_{N \geq 0}\mathbb{C}[z_{1}, \dots, z_{n}]/(z_{1}, \dots, z_{n})^{N} \\ &\simeq \mathbb{C}[[z_{1}, \dots, z_{n}]]. \end{align*}$$ Recall that for any Noetherian local ring $(A, \mathfrak{m}),$ its completion $A \rightarrow \hat{A}$ is faitufully flat (e.g., Theorem on p.161 on Hochster's notes; he says "local" to mean Noetherian local). Thus, we deduce that $R \rightarrow \hat{R}$ is faithfully flat. We want to show that $$S := \mathbb{C}[z_{1}, \dots, z_{n}]_{(z_{1}, \dots, z_{n})} \rightarrow \mathbb{C}\{z_{1}, \dots, z_{n}\} = R$$ is faithfully flat. Fix any map $$M \rightarrow M'$$ of modules over $S.$ Then the following are equivalent:
- $M \rightarrow M'$ is injective;
- $M \otimes_{S} \hat{S} \rightarrow M' \otimes_{S} \hat{S}$ is injective;
- $M \otimes_{S} R \otimes_{R} \hat{R} \rightarrow M' \otimes_{S} R \otimes_{R} \hat{R}$ is injective;
- $M \otimes_{S} R \rightarrow M' \otimes_{S} R$ is injective,
because $S \rightarrow \hat{S}$ is faithfully flat and $\hat{S} \simeq \hat{R}.$ It follows that $S \rightarrow R$ is faithfully flat.
Remark. The fact that $R = \mathbb{C}\{z_{1}, \dots, z_{n}\}$ is a Noetherian local ring whose completion is $\mathbb{C}[[z_{1}, \dots, z_{n}]]$ also implies that $R$ is a regular local ring of dimension $n$ (see the paragraph following 29.2.D in Vakil).
To tackle the general case, we need the following statement.
Lemma 1. Let $X \hookrightarrow Y$ be a smooth algebraic varieties, where $X$ is a closed subvariety of $Y$ with this inclusion, and let $\mathscr{I} \subset \mathscr{O}_{Y}$ be the ideal sheaf associated to $X.$ Then the ideal sheaf of $X^{\mathrm{an}}$ is given by $\mathscr{I}^{\mathrm{an}} \subset \mathscr{O}_{Y^{\mathrm{an}}}.$
Step 2. Let $X$ be a (smooth) closed subvariety of $\mathbb{A}^{n}.$
Let $I$ be the (radical) ideal of $\mathbb{C}[z_{1}, \dots, z_{n}]$ defining $X.$ Fix $x \in X(\mathbb{C}).$ We have $$\begin{align*}\mathscr{O}_{X,x} &= (\mathscr{O}_{\mathbb{A}^{n}}/\tilde{I})_{x} \\ &= \mathscr{O}_{\mathbb{A}^{n}, x}/\tilde{I}_{x}.\end{align*}$$ Using Lemma 1, we have $$\tilde{I}^{\mathrm{an}}_{x} = \tilde{I}_{x}\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}$$ and $$\begin{align*}\mathscr{O}_{X^{\mathrm{an}},x} &= (\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}}/\tilde{I}^{\mathrm{an}})_{x} \\ &= \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}/\tilde{I}^{\mathrm{an}}_{x}.\end{align*}$$ Lemma also ensures that the map $\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$ is given by taking the quotient by $\tilde{I}_{x}$ of the map $$\mathscr{O}_{\mathbb{A}^{n}, x} \rightarrow \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x},$$ which we already showed to be faithfully flat. We want to show that $$\mathscr{O}_{\mathbb{A}^{n}, x}/\tilde{I}_{x} \rightarrow \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}/\tilde{I}_{x}\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}$$ is faithfully flat. Note that taking the completion of this map would give us an isomorphism. Hence, it is enough to show the following:
Lemma 2. Let $A \rightarrow B$ be a local map of Notherian local rings. If their completion is flat, then the map is flat.
Before proving this lemma, note that this is enough to ensure faithful flatness because any flat local ring map is faithfully flat: fix any flat local map $(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})$. Say $M \rightarrow M'$ is an $A$-module map, denote by $K$ its kernel. We assume that $K \otimes_{A} B = 0$ and want to show $K = 0.$ Fix any finitely generated $A$-submodule $Q \subset K,$ and it is enough to show that $Q = 0.$ Since $A \rightarrow B$ is flat, we have $Q \otimes_{A} B \hookrightarrow K \otimes_{A} B = 0.$ Hence $Q \otimes_{A} B = 0.$ Because $A \rightarrow B$ is a local map, this implies that $$Q \otimes_{A} (A/\mathfrak{m}) = 0.$$ Since $Q$ is finitely generated, we may apply Nakayama's lemma to conclude that $Q = 0,$ as desired.
Proof of Lemma 2. Denote by $C = \hat{A} = \hat{B}.$ Fix any injective $A$-module map $M \hookrightarrow M'.$ Denote by $Q$ the kernel of $M \otimes_{A} B \rightarrow M' \otimes_{A} B.$ We want to show $Q = 0.$ Since $B \rightarrow C$ is flat, we have the following exact sequence $$0 \rightarrow Q \otimes_{B} C \rightarrow M \otimes_{A} C \rightarrow M' \otimes_{A} C.$$ Since $A \rightarrow \hat{A} = C$ is flat, we have $Q \otimes_{B} C = 0.$ Since $B \rightarrow \hat{B} = C$ is a flat local map, the same trick we have used above implies that $Q = 0,$ as desired. $\Box$
Step 3. Let $X$ be any smooth complex variety.
Since the statement is only about the stalks, Step 2 takes care of this general case. This finishes the proof (given Lemma 1) of Proposition. $\Box$
A reference for proof of Lemma 1. This seems a bit more convoluted than I initially thought. I decided to skip this for now, but a proof is given for Proposition 4 in this link, which seems to be an English translation of Serre's GAGA. The hardest part seems to be an analytic version of Nullstellensatz, but I did not look into the details.
Let $\iota : X \hookrightarrow Y$ be a closed embedding of smooth complex varieties and $\mathscr{F}$ an $\mathscr{O}_{X}$-module. We have
Let $V \subset Y$ be any affine open. We have $$(\iota_{*}\mathscr{F})(V) = \mathscr{F}(V \cap X) \rightarrow (u_{X}^{*}\mathscr{F})(V^{\mathrm{an}} \cap X^{\mathrm{an}}) = ((u_{Y})_{*}\iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F})(V),$$ which gives a map $$\iota_{*}\mathscr{F} \rightarrow (u_{Y})_{*}\iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F}$$ of $\mathscr{O}_{Y}$-modules. Hence, we get a map $$u_{Y}^{*}\iota_{*}\mathscr{F} \rightarrow \iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F}$$ of $\mathscr{O}_{Y^{\mathrm{an}}}$-modules.
Fact. The map above is an isomoprhism (at least when $\mathscr{F}$ is coherent).
Personal remark. The above fact will be used soon. However, I must have missed how it is proven. This is significantly stronger than Lemma 1, which is the statement for the case $\mathscr{F} = \mathscr{O}_{X}.$ Given Lemma 1 being convoluted, this fact should not be so easy, or perhaps one can deduce this from Lemma 1, although I cannot see this right away.
We now show the 1st GAGA for smooth projective varieties over $\mathbb{C}.$
Theorem. Let $X$ be a smooth projective variety over $\mathbb{C}.$ Then for any coherent $\mathscr{O}_{X}$-modules $\mathscr{F}, \mathscr{G},$ we have $$\mathrm{Ext}_{\mathscr{O}_{X}}^{i}(\mathscr{F}, \mathscr{G}) \simeq \mathrm{Ext}_{\mathscr{O}_{X^{\mathrm{an}}}}^{i}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$$ Before we prove this, we will organize some facts we will use without proofs. Fix any closed embedding $\iota : X \hookrightarrow \mathbb{P}^{n}.$
Fact 1. We repeat the fact we have said above: $$(\iota_{*}\mathscr{G})^{\mathrm{an}} \simeq \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}}.$$ The next fact should be true, but I cannot find any reference just yet.
Fact 2. We have $$H^{i}(X^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}}).$$ Personal remark. We can probably track down what the map actually is, but let's not think about it now. The proof will probably require using Čech complex, but again, let's skip this detail for now.
Fact 3. We have $h^{0}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}) = 1$ and $h^{i}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}) = 0$ for all $i > 0.$
Remark. We were told that we will later deduce Fact 3 from the singular Betti numbers of $\mathbb{P}^{n}(\mathbb{C})$ and Hodge theory.
Proof of Theorem. We have several steps.
Step 1. Consider the case $\mathscr{F} = \mathscr{O}_{X}.$ Then our statement to show reduces to $$H^{i}(X, \mathscr{G}) \simeq H^{i}(X^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$$ By Fact 2, we may show $$H^{i}(\mathbb{P}^{n}, \iota_{*}\mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}})$$ instead. By Fact 1, this means that we only need to show $$H^{i}(\mathbb{P}^{n}, \iota_{*}\mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), (\iota_{*}\mathscr{G})^{\mathrm{an}}),$$ so all in all, we reduce to the case where $X = \mathbb{P}^{n}.$
First, assume that $\mathscr{G}$ is a line bundle on $\mathbb{P}^{n}.$ More specifically, we assume $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m)$ for some $m \in \mathbb{Z}.$ We then argue by induction on $n \geq 0.$ When $n = 0,$ we have $\mathbb{P}^{0} = \{*\},$ so either side has $1$-dimensional $H^{0},$ as the space of functions from a point consist of constant functions, and all the higher degree cohomology groups vanish. Hence, we get the base case $n = 0$ when $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m).$
Remark. The vanishing uses Čech cohomology on either side. A statement (not a proof) we need here can be found as (b) and (c) on p.64 in Wells.
Assume that the statement works for any complex projective spaces of dimension $0, 1, \dots, n-1.$ Assuming $n \geq 1,$ consider the exact sequence $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}} \rightarrow i_{*}\mathscr{O}_{\mathbb{P}^{n-1}} \rightarrow 0,$$ where $i : \mathbb{P}^{n-1} \hookrightarrow \mathbb{P}^{n}$ is the inclusion map for the closed subscheme (e.g., 14.3.B in Vakil). Tensoring with $\mathscr{O}_{\mathbb{P}^{n}}(m),$ we get $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow i_{*}\mathscr{O}_{\mathbb{P}^{n-1}} \otimes_{\mathscr{O}_{\mathbb{P}^{n}}} \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow 0.$$ But then note that $$\begin{align*} (i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}) \otimes_{\mathscr{O}_{\mathbb{P}^{n}}} \mathscr{O}_{\mathbb{P}^{n}}(m) &\simeq i_{*}(\mathscr{O}_{\mathbb{P}^{n-1}} \otimes_{\mathscr{O}_{\mathbb{P}^{n-1}}} i^{*}\mathscr{O}_{\mathbb{P}^{n}}(m)) \\ &\simeq i_{*}i^{*}\mathscr{O}_{\mathbb{P}^{n}}(m) \\ &= i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m),\end{align*}$$ where the first isomorphism is given by 16.3.H (b) in Vakil. Thus, we have the following exact sequence: $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m) \rightarrow 0.$$ Applying the analytification functor with Fact 1, we also have an exact sequence $$0 \rightarrow (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}} \rightarrow \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m) \rightarrow i^{\mathrm{an}}_{*}(\mathscr{O}_{\mathbb{P}^{n-1}}(m))^{\mathrm{an}} \rightarrow 0.$$ The long exact sequences of cohomology groups give the following commuting diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m-1)) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m)) @>{}>> H^{i}(\mathbb{P}^{n}, i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m)) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m-1)) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m)) \\ @V{}VV @V{}VV @V{\simeq}VV @V{}VV @V{}VV \\ H^{i}(\mathbb{P}^{n}, (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m)) @>{}>> H^{i}(\mathbb{P}^{n}, i^{\mathrm{an}}_{*}(\mathscr{O}_{\mathbb{P}^{n-1}}(m))^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m)),\end{CD}$$ where the isomorphism in the middle is given by the induction hypothesis. In the lower row, the notations $\mathbb{P}^{n}$ should be replaced with $\mathbb{P}^{n}(\mathbb{C}),$ but we kept them for the sake of margin. Applying five lemma, we see that showing the statement for the case $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m)$ is equivalent to showing the statement for the case $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m+1).$ Thus, working with induction on $|m|,$ we may reduce to the case $m = 0,$ which holds by Fact 3. This finishes the case where $\mathscr{G}$ is a line bundle on $\mathbb{P}^{n}.$
For general coherent $\mathscr{G},$ we want to get $$H^{i}(\mathbb{P}^{n}, \mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{G}^{\mathrm{an}})$$ for any $i \geq 0.$ Using Čech complex, both sides vanishes for $i \gg 0,$ so we may proceed with downward induction on $i$ (with all general coherent sheaves).
We may find an exact sequence $$0 \rightarrow \mathscr{K} \rightarrow \mathscr{E} \rightarrow \mathscr{G} \rightarrow 0,$$ where $\mathscr{E}$ is a finite direct sum of sheaves of the form $\mathscr{O}_{\mathbb{P}^{n}}(m)$ with finitely many various $m \in \mathbb{Z}.$ (For example, see Theorem 15.3.1 Vakil.) Taking direct sum of sheaves commutes with taking cohomology, so the statement must be true for $\mathscr{E}$ as well. Now, we have the following commutative diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} H^{i}(\mathbb{P}^{n}, \mathscr{K}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{E}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{G}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{K}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{E}) \\ @V{\alpha}VV @V{\simeq}VV @V{\beta}VV @V{\simeq}VV @V{\simeq}VV \\ H^{i}(\mathbb{P}^{n}, \mathscr{K}^{\mathrm{an}}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{E}^{\mathrm{an}}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{G}^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{K}^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{E}^{\mathrm{an}}),\end{CD}$$ where the fourth vertical isomoprhism is given by induction hypothesis. By five lemma, the map $\beta$ is surjective. Again by five lemma, this implies that $\alpha$ is injective. Since $\mathscr{G}$ is an arbitrary coherent sheaf, this implies that $\alpha$ is also surjective. By five lemma, this implies that $\beta$ is bijective, as desired. This finishes the case where $\mathscr{F} = \mathscr{O}_{X}.$
Step 2. Let $\mathscr{F}$ be locally free.
Then $\mathscr{F}^{\mathrm{an}}$ is locally free. We have $$\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \simeq H^{i}(X, \mathscr{F}^{\vee} \otimes \mathscr{G})$$ functorially in $\mathscr{F}$ (e.g., Vakil 30.2.I). Hence, by Step 1, we may resolve this situation.
Personal remark. Presumably, a similar fact is available for manifolds, and the map $$\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \rightarrow \mathrm{Ext}^{i}_{\mathscr{O}_{X^{\mathrm{an}}}}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}})$$ is given by $$H^{i}(X, \mathscr{F}^{\vee} \otimes \mathscr{G}) \rightarrow H^{i}(X^{\mathrm{an}}, (\mathscr{F}^{\mathrm{an}})^{\vee} \otimes \mathscr{G}^{\mathrm{an}}).$$ Moreover, we need to check $$(\mathscr{F}^{\mathrm{an}})^{\vee} \otimes \mathscr{G}^{\mathrm{an}} \simeq (\mathscr{F}^{\vee} \otimes \mathscr{G})^{\mathrm{an}},$$ which is probably true due to the fact that the analytification functor gives a flat base change. Still, I will leave these untouched for now.
Step 3. Let $\mathscr{F}$ be any coherent $\mathscr{O}_{X}$-module.
Choose any exact sequence $$0 \rightarrow \mathscr{K} \rightarrow \mathscr{E} \rightarrow \mathscr{F} \rightarrow 0$$ of coherent sheaves, where $\mathscr{E}$ is locally free. We have the following commutative diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} \mathrm{Ext}^{i-1}(\mathscr{F}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{E}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{K}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i}(\mathscr{E}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i}(\mathscr{F}, \mathscr{G}) \\ @V{\simeq}VV @V{\simeq}VV @V{\simeq}VV @V{\gamma}VV @V{\simeq}VV \\ \mathrm{Ext}^{i-1}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{E}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{K}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i}(\mathscr{E}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}),\end{CD}$$ where the isomorphisms on degree $i-1$ are given by induction hypothesis. Note that the base case is free because the negative degrees all vanish. By five lemma, we see $\gamma$ is injective (although the first vertical map is not necessary). We may apply five lemma again by introducing the next vertical map, which can now be assumed to be injective to conclude $\gamma$ is surjective. This finishes the proof (modulo a bunch of stuff I did not check). $\Box$
Remark. The fact that $R = \mathbb{C}\{z_{1}, \dots, z_{n}\}$ is a Noetherian local ring whose completion is $\mathbb{C}[[z_{1}, \dots, z_{n}]]$ also implies that $R$ is a regular local ring of dimension $n$ (see the paragraph following 29.2.D in Vakil).
To tackle the general case, we need the following statement.
Lemma 1. Let $X \hookrightarrow Y$ be a smooth algebraic varieties, where $X$ is a closed subvariety of $Y$ with this inclusion, and let $\mathscr{I} \subset \mathscr{O}_{Y}$ be the ideal sheaf associated to $X.$ Then the ideal sheaf of $X^{\mathrm{an}}$ is given by $\mathscr{I}^{\mathrm{an}} \subset \mathscr{O}_{Y^{\mathrm{an}}}.$
Step 2. Let $X$ be a (smooth) closed subvariety of $\mathbb{A}^{n}.$
Let $I$ be the (radical) ideal of $\mathbb{C}[z_{1}, \dots, z_{n}]$ defining $X.$ Fix $x \in X(\mathbb{C}).$ We have $$\begin{align*}\mathscr{O}_{X,x} &= (\mathscr{O}_{\mathbb{A}^{n}}/\tilde{I})_{x} \\ &= \mathscr{O}_{\mathbb{A}^{n}, x}/\tilde{I}_{x}.\end{align*}$$ Using Lemma 1, we have $$\tilde{I}^{\mathrm{an}}_{x} = \tilde{I}_{x}\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}$$ and $$\begin{align*}\mathscr{O}_{X^{\mathrm{an}},x} &= (\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}}/\tilde{I}^{\mathrm{an}})_{x} \\ &= \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}/\tilde{I}^{\mathrm{an}}_{x}.\end{align*}$$ Lemma also ensures that the map $\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$ is given by taking the quotient by $\tilde{I}_{x}$ of the map $$\mathscr{O}_{\mathbb{A}^{n}, x} \rightarrow \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x},$$ which we already showed to be faithfully flat. We want to show that $$\mathscr{O}_{\mathbb{A}^{n}, x}/\tilde{I}_{x} \rightarrow \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}/\tilde{I}_{x}\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}$$ is faithfully flat. Note that taking the completion of this map would give us an isomorphism. Hence, it is enough to show the following:
Lemma 2. Let $A \rightarrow B$ be a local map of Notherian local rings. If their completion is flat, then the map is flat.
Before proving this lemma, note that this is enough to ensure faithful flatness because any flat local ring map is faithfully flat: fix any flat local map $(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})$. Say $M \rightarrow M'$ is an $A$-module map, denote by $K$ its kernel. We assume that $K \otimes_{A} B = 0$ and want to show $K = 0.$ Fix any finitely generated $A$-submodule $Q \subset K,$ and it is enough to show that $Q = 0.$ Since $A \rightarrow B$ is flat, we have $Q \otimes_{A} B \hookrightarrow K \otimes_{A} B = 0.$ Hence $Q \otimes_{A} B = 0.$ Because $A \rightarrow B$ is a local map, this implies that $$Q \otimes_{A} (A/\mathfrak{m}) = 0.$$ Since $Q$ is finitely generated, we may apply Nakayama's lemma to conclude that $Q = 0,$ as desired.
Proof of Lemma 2. Denote by $C = \hat{A} = \hat{B}.$ Fix any injective $A$-module map $M \hookrightarrow M'.$ Denote by $Q$ the kernel of $M \otimes_{A} B \rightarrow M' \otimes_{A} B.$ We want to show $Q = 0.$ Since $B \rightarrow C$ is flat, we have the following exact sequence $$0 \rightarrow Q \otimes_{B} C \rightarrow M \otimes_{A} C \rightarrow M' \otimes_{A} C.$$ Since $A \rightarrow \hat{A} = C$ is flat, we have $Q \otimes_{B} C = 0.$ Since $B \rightarrow \hat{B} = C$ is a flat local map, the same trick we have used above implies that $Q = 0,$ as desired. $\Box$
Step 3. Let $X$ be any smooth complex variety.
Since the statement is only about the stalks, Step 2 takes care of this general case. This finishes the proof (given Lemma 1) of Proposition. $\Box$
A reference for proof of Lemma 1. This seems a bit more convoluted than I initially thought. I decided to skip this for now, but a proof is given for Proposition 4 in this link, which seems to be an English translation of Serre's GAGA. The hardest part seems to be an analytic version of Nullstellensatz, but I did not look into the details.
Let $\iota : X \hookrightarrow Y$ be a closed embedding of smooth complex varieties and $\mathscr{F}$ an $\mathscr{O}_{X}$-module. We have
- the analytification $\iota^{\mathrm{an}} : X^{\mathrm{an}} \hookrightarrow Y^{\mathrm{an}}$ of the closed embedding,
- the unanalytification $u_{X} : X^{\mathrm{an}} \rightarrow X$ of $X,$ and
- the unanalytification $u_{Y} : Y^{\mathrm{an}} \rightarrow Y$ of $Y.$
Let $V \subset Y$ be any affine open. We have $$(\iota_{*}\mathscr{F})(V) = \mathscr{F}(V \cap X) \rightarrow (u_{X}^{*}\mathscr{F})(V^{\mathrm{an}} \cap X^{\mathrm{an}}) = ((u_{Y})_{*}\iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F})(V),$$ which gives a map $$\iota_{*}\mathscr{F} \rightarrow (u_{Y})_{*}\iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F}$$ of $\mathscr{O}_{Y}$-modules. Hence, we get a map $$u_{Y}^{*}\iota_{*}\mathscr{F} \rightarrow \iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F}$$ of $\mathscr{O}_{Y^{\mathrm{an}}}$-modules.
Fact. The map above is an isomoprhism (at least when $\mathscr{F}$ is coherent).
Personal remark. The above fact will be used soon. However, I must have missed how it is proven. This is significantly stronger than Lemma 1, which is the statement for the case $\mathscr{F} = \mathscr{O}_{X}.$ Given Lemma 1 being convoluted, this fact should not be so easy, or perhaps one can deduce this from Lemma 1, although I cannot see this right away.
We now show the 1st GAGA for smooth projective varieties over $\mathbb{C}.$
Theorem. Let $X$ be a smooth projective variety over $\mathbb{C}.$ Then for any coherent $\mathscr{O}_{X}$-modules $\mathscr{F}, \mathscr{G},$ we have $$\mathrm{Ext}_{\mathscr{O}_{X}}^{i}(\mathscr{F}, \mathscr{G}) \simeq \mathrm{Ext}_{\mathscr{O}_{X^{\mathrm{an}}}}^{i}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$$ Before we prove this, we will organize some facts we will use without proofs. Fix any closed embedding $\iota : X \hookrightarrow \mathbb{P}^{n}.$
Fact 1. We repeat the fact we have said above: $$(\iota_{*}\mathscr{G})^{\mathrm{an}} \simeq \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}}.$$ The next fact should be true, but I cannot find any reference just yet.
Fact 2. We have $$H^{i}(X^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}}).$$ Personal remark. We can probably track down what the map actually is, but let's not think about it now. The proof will probably require using Čech complex, but again, let's skip this detail for now.
Fact 3. We have $h^{0}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}) = 1$ and $h^{i}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}) = 0$ for all $i > 0.$
Remark. We were told that we will later deduce Fact 3 from the singular Betti numbers of $\mathbb{P}^{n}(\mathbb{C})$ and Hodge theory.
Proof of Theorem. We have several steps.
Step 1. Consider the case $\mathscr{F} = \mathscr{O}_{X}.$ Then our statement to show reduces to $$H^{i}(X, \mathscr{G}) \simeq H^{i}(X^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$$ By Fact 2, we may show $$H^{i}(\mathbb{P}^{n}, \iota_{*}\mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}})$$ instead. By Fact 1, this means that we only need to show $$H^{i}(\mathbb{P}^{n}, \iota_{*}\mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), (\iota_{*}\mathscr{G})^{\mathrm{an}}),$$ so all in all, we reduce to the case where $X = \mathbb{P}^{n}.$
First, assume that $\mathscr{G}$ is a line bundle on $\mathbb{P}^{n}.$ More specifically, we assume $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m)$ for some $m \in \mathbb{Z}.$ We then argue by induction on $n \geq 0.$ When $n = 0,$ we have $\mathbb{P}^{0} = \{*\},$ so either side has $1$-dimensional $H^{0},$ as the space of functions from a point consist of constant functions, and all the higher degree cohomology groups vanish. Hence, we get the base case $n = 0$ when $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m).$
Remark. The vanishing uses Čech cohomology on either side. A statement (not a proof) we need here can be found as (b) and (c) on p.64 in Wells.
Assume that the statement works for any complex projective spaces of dimension $0, 1, \dots, n-1.$ Assuming $n \geq 1,$ consider the exact sequence $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}} \rightarrow i_{*}\mathscr{O}_{\mathbb{P}^{n-1}} \rightarrow 0,$$ where $i : \mathbb{P}^{n-1} \hookrightarrow \mathbb{P}^{n}$ is the inclusion map for the closed subscheme (e.g., 14.3.B in Vakil). Tensoring with $\mathscr{O}_{\mathbb{P}^{n}}(m),$ we get $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow i_{*}\mathscr{O}_{\mathbb{P}^{n-1}} \otimes_{\mathscr{O}_{\mathbb{P}^{n}}} \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow 0.$$ But then note that $$\begin{align*} (i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}) \otimes_{\mathscr{O}_{\mathbb{P}^{n}}} \mathscr{O}_{\mathbb{P}^{n}}(m) &\simeq i_{*}(\mathscr{O}_{\mathbb{P}^{n-1}} \otimes_{\mathscr{O}_{\mathbb{P}^{n-1}}} i^{*}\mathscr{O}_{\mathbb{P}^{n}}(m)) \\ &\simeq i_{*}i^{*}\mathscr{O}_{\mathbb{P}^{n}}(m) \\ &= i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m),\end{align*}$$ where the first isomorphism is given by 16.3.H (b) in Vakil. Thus, we have the following exact sequence: $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m) \rightarrow 0.$$ Applying the analytification functor with Fact 1, we also have an exact sequence $$0 \rightarrow (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}} \rightarrow \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m) \rightarrow i^{\mathrm{an}}_{*}(\mathscr{O}_{\mathbb{P}^{n-1}}(m))^{\mathrm{an}} \rightarrow 0.$$ The long exact sequences of cohomology groups give the following commuting diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m-1)) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m)) @>{}>> H^{i}(\mathbb{P}^{n}, i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m)) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m-1)) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m)) \\ @V{}VV @V{}VV @V{\simeq}VV @V{}VV @V{}VV \\ H^{i}(\mathbb{P}^{n}, (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m)) @>{}>> H^{i}(\mathbb{P}^{n}, i^{\mathrm{an}}_{*}(\mathscr{O}_{\mathbb{P}^{n-1}}(m))^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m)),\end{CD}$$ where the isomorphism in the middle is given by the induction hypothesis. In the lower row, the notations $\mathbb{P}^{n}$ should be replaced with $\mathbb{P}^{n}(\mathbb{C}),$ but we kept them for the sake of margin. Applying five lemma, we see that showing the statement for the case $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m)$ is equivalent to showing the statement for the case $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m+1).$ Thus, working with induction on $|m|,$ we may reduce to the case $m = 0,$ which holds by Fact 3. This finishes the case where $\mathscr{G}$ is a line bundle on $\mathbb{P}^{n}.$
For general coherent $\mathscr{G},$ we want to get $$H^{i}(\mathbb{P}^{n}, \mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{G}^{\mathrm{an}})$$ for any $i \geq 0.$ Using Čech complex, both sides vanishes for $i \gg 0,$ so we may proceed with downward induction on $i$ (with all general coherent sheaves).
We may find an exact sequence $$0 \rightarrow \mathscr{K} \rightarrow \mathscr{E} \rightarrow \mathscr{G} \rightarrow 0,$$ where $\mathscr{E}$ is a finite direct sum of sheaves of the form $\mathscr{O}_{\mathbb{P}^{n}}(m)$ with finitely many various $m \in \mathbb{Z}.$ (For example, see Theorem 15.3.1 Vakil.) Taking direct sum of sheaves commutes with taking cohomology, so the statement must be true for $\mathscr{E}$ as well. Now, we have the following commutative diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} H^{i}(\mathbb{P}^{n}, \mathscr{K}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{E}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{G}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{K}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{E}) \\ @V{\alpha}VV @V{\simeq}VV @V{\beta}VV @V{\simeq}VV @V{\simeq}VV \\ H^{i}(\mathbb{P}^{n}, \mathscr{K}^{\mathrm{an}}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{E}^{\mathrm{an}}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{G}^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{K}^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{E}^{\mathrm{an}}),\end{CD}$$ where the fourth vertical isomoprhism is given by induction hypothesis. By five lemma, the map $\beta$ is surjective. Again by five lemma, this implies that $\alpha$ is injective. Since $\mathscr{G}$ is an arbitrary coherent sheaf, this implies that $\alpha$ is also surjective. By five lemma, this implies that $\beta$ is bijective, as desired. This finishes the case where $\mathscr{F} = \mathscr{O}_{X}.$
Step 2. Let $\mathscr{F}$ be locally free.
Then $\mathscr{F}^{\mathrm{an}}$ is locally free. We have $$\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \simeq H^{i}(X, \mathscr{F}^{\vee} \otimes \mathscr{G})$$ functorially in $\mathscr{F}$ (e.g., Vakil 30.2.I). Hence, by Step 1, we may resolve this situation.
Personal remark. Presumably, a similar fact is available for manifolds, and the map $$\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \rightarrow \mathrm{Ext}^{i}_{\mathscr{O}_{X^{\mathrm{an}}}}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}})$$ is given by $$H^{i}(X, \mathscr{F}^{\vee} \otimes \mathscr{G}) \rightarrow H^{i}(X^{\mathrm{an}}, (\mathscr{F}^{\mathrm{an}})^{\vee} \otimes \mathscr{G}^{\mathrm{an}}).$$ Moreover, we need to check $$(\mathscr{F}^{\mathrm{an}})^{\vee} \otimes \mathscr{G}^{\mathrm{an}} \simeq (\mathscr{F}^{\vee} \otimes \mathscr{G})^{\mathrm{an}},$$ which is probably true due to the fact that the analytification functor gives a flat base change. Still, I will leave these untouched for now.
Step 3. Let $\mathscr{F}$ be any coherent $\mathscr{O}_{X}$-module.
Choose any exact sequence $$0 \rightarrow \mathscr{K} \rightarrow \mathscr{E} \rightarrow \mathscr{F} \rightarrow 0$$ of coherent sheaves, where $\mathscr{E}$ is locally free. We have the following commutative diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} \mathrm{Ext}^{i-1}(\mathscr{F}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{E}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{K}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i}(\mathscr{E}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i}(\mathscr{F}, \mathscr{G}) \\ @V{\simeq}VV @V{\simeq}VV @V{\simeq}VV @V{\gamma}VV @V{\simeq}VV \\ \mathrm{Ext}^{i-1}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{E}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{K}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i}(\mathscr{E}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}),\end{CD}$$ where the isomorphisms on degree $i-1$ are given by induction hypothesis. Note that the base case is free because the negative degrees all vanish. By five lemma, we see $\gamma$ is injective (although the first vertical map is not necessary). We may apply five lemma again by introducing the next vertical map, which can now be assumed to be injective to conclude $\gamma$ is surjective. This finishes the proof (modulo a bunch of stuff I did not check). $\Box$
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