Hodge theory: Lecture 2

We continue from the last time.

Theorem. Let $X$ be an irreducible $\mathbb{C}$-variety and $U \subset X$ a nonempty open subset. Then $U^{\mathrm{an}}$ is dense in $X^{\mathrm{an}}.$

We also recall from the last time that the theorem had an immediate corollary, which we will use later:

Corollary. Let $X$ be any $\mathbb{C}$-variety. If $U \subset X$ is Zariski-open and Zariski-dense, then $U(\mathbb{C}) \subset X(\mathbb{C})$ is dense in the classical topology.

Reference. The lecturer kindly notified us that we are following Mumford's Red Book (Theorem 1 on p.58).

Proof. It is a convoluted proof (in my opinion), so we have several steps.

Step 1. Reduction to the case where $X$ is affine.

Recall that our definition of variety includes finite type over $\mathbb{C},$ which includes quasi-compactness. Thus, we can choose a finite open cover $X = \bigcup_{i=1}^{r} U_{i}$ with every $U_{i} \neq \emptyset.$ Since $X$ is irreducible, we have $U \cap U_{i} \neq \emptyset$ for all $1 \leq i \leq r.$ Thus, if the theorem is true for affine case, then $(U \cap U_{i})^{\mathrm{an}} = U(\mathbb{C}) \cap U_{i}(\mathbb{C})$ would be dense in $U_{i}^{\mathrm{an}} = U_{i}(\mathbb{C}).$ Since $U = \bigcup_{i=1}^{r} (U \cap U_{i}),$ this implies that $$\overline{U(\mathbb{C})} = \bigcup_{i=1}^{r} (\overline{U(\mathbb{C})} \cap U_{i}(\mathbb{C})) = \bigcup_{i=1}^{r} U_{i}(\mathbb{C}) = X(\mathbb{C}),$$ with respect to the classical topology. Hence, for the rest of the proof, we may assume that $X$ is affine.

Step 2. Apply Nother normalization.

Recall (e.g., Vakil 11.2.4) the following lemma:

Lemma (Nother normalization). Given any integral domain finitely generated over a field $k,$ if the transcendental degree of $A$ over $k$ is $n$ (which is necessarily finite), then there exist $x_{1}, \dots, x_{n} \in A,$ algebraically independent over $k,$ such that $k[x_{1}, \dots, x_{n}] \hookrightarrow A$ is a finite extension of rings (which is necessarily a $k$-algebra map). In particular, by Lying Over, we have a surjective $k$-scheme map $\mathrm{Spec}(A) \twoheadrightarrow \mathbb{A}^{n}_{k}.$

Remark. We also recall that for any such domain $A,$ the transcendental degree of $A$ over $k$ is precisely the dimension of $\mathrm{Spec}(A),$ which can be proved using Noether normalization (e.g., Vakil 11.2.7). Without this approach, it does not seem clear (at least to me) how to show that $\dim(A)$ is finite (cf. Vakil 11.1.K).

Now, we have $X = \mathrm{Spec}(A)$ and writing $\dim(X) = n,$ we have a finite $\mathbb{C}$-algebra extension $\mathbb{C}[x_{1}, \dots, x_{n}] \hookrightarrow A.$ Denote by $\pi : X \twoheadrightarrow \mathbb{A}^{n}$ the surjective $k$-scheme map induced by this finite extension. Given any $\mathbb{C}$-point $q \in \mathbb{A}^{n}(\mathbb{C}),$ the fiber $\pi^{-1}(q) \subset X$ is a (nonempty) closed subset of dimension $0$ (e.g., Vakil 11.1.D), so all of its elements are closed points of $X,$ which are $\mathbb{C}$-points by Nullstellensatz. Thus, the induced map $\pi^{\mathrm{an}} : X(\mathbb{C}) \rightarrow \mathbb{A}^{n}(\mathbb{C}) = \mathbb{C}^{n}$ is surjective as well.

Let $Z := X \setminus U.$ We know from the last time that $Z(\mathbb{C}) = X(\mathbb{C}) \setminus U(\mathbb{C})$ not only as sets but as topological spaces with the classical topologies. Our goal is to show $$\overline{U(\mathbb{C})} = X(\mathbb{C}),$$ so it is enough to show that given $p \in Z(\mathbb{C}),$ we can find a sequence $(y_{m})$ in $U(\mathbb{C})$ such that $$\lim_{m \rightarrow \infty}y_{m} = p,$$ with respect to the Euclidean topology of $X(\mathbb{C}).$ (Note that $\pi(p) \in \mathbb{A}^{n}(\mathbb{C}) = \mathbb{C}^{n}$.)

Since $\pi$ is a finite map, it sends closed subsets to closed subsets (e.g., Vakil 7.3.M), so $\pi(Z) \subset \mathbb{A}^{n}$ is Zariski-closed. Note that $\pi(Z) \subsetneq \mathbb{A}^{n}$ because if it were that $\pi(Z) = \mathbb{A}^{n},$ then the generic fiber must be contained $Z,$ but then in this case (because the ring map is injective) the generic fiber contains the generic point of $X,$ which would imply that $Z = X.$ This does not happen because $U = X \setminus Z$ is nonempty, so $\pi(Z)$ is a proper (closed) subset of $\mathbb{A}^{n}.$ Hence, we can write $$\pi(Z) = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/I) \simeq V(I) \subset \mathbb{A}^{n}$$ for some nonzero (radical) ideal $I \subset \mathbb{C}[x_{1}, \dots, x_{n}].$ Take any nonzero polynomial $g \in I$ so that we have $$\pi(Z) = V(I) \subset V(g) \subsetneq \mathbb{A}^{n}.$$ In other words, we have $Z \subset \pi^{-1}(V(g)).$ Next, we go analytic.

Consider $\varphi : \mathbb{R} \rightarrow \mathbb{C} = \mathbb{R}^{2}$ given by $$\varphi(t) := g(t\pi(p) + (1-t)w),$$ where we fix $w \in \mathbb{C}^{n}$ such that $g(w) \neq 0.$ (Note that $g(\pi(p)) = 0$ ) We have $\varphi(0) = g(w) \neq 0,$ and $\varphi$ is a polynomial in $t$ with complex coefficients, so it only vanishes at finitely many elements in $\mathbb{R},$ and in particular in the segment $[0, 1].$

Take any sequence $(t_{m})$ in $[0, 1]$ such that $\varphi(t_{m}) \neq 0$ for all $m,$ but $t_{m} \rightarrow 0,$ as $m \rightarrow \infty.$ Writing $u_{m} = t_{m}\pi(p) + (1 - t_{m})w \in \mathbb{C}^{n},$ we see that $g(u_{m}) \neq 0$ for all $m$ and $$\lim_{m \rightarrow \infty}u_{m} = \pi(p).$$ What do we need to do now?

Goal. We will replace $(u_{m})$ with a subsequence of it, and then find $y_{m} \in \pi^{-1}(u_{m})$ for each $m$ such that $y_{m} \rightarrow p$ as $m \rightarrow \infty.$ Once we are done with this, we have $\pi(y_{m}) = u_{m} \in D(g)$ so that $y_{m} \in \pi^{-1}(D(g)).$ Since $Z \subset \pi^{-1}(V(g)),$ which is disjoint to $\pi^{-1}(D(g)),$ we must have $y_{m} \in U = X \setminus Z$ so that $y_{m} \in U(\mathbb{C}).$

Let's take a break and think about what's been going on. After formally reducing the statement to the case where $X$ is affine. Our goal is to show that any $p \in Z(\mathbb{C}) = X(\mathbb{C}) \setminus U(\mathbb{C})$ can be analytically approached by points in $U(\mathbb{C}).$ To do so, we constructed a nice finite surjection $X \twoheadrightarrow \mathbb{A}^{n}$ via Noether normalization, and this gave us an analytic surjection $X(\mathbb{C}) \twoheadrightarrow \mathbb{C}^{n}.$ We took a nonzero polynomial $g \in \mathbb{C}[x_{1}, \dots, x_{n}]$ such that $\pi(Z) \subset V(g) \subsetneq \mathbb{A}^{n}$ and constructed a sequence $u_{m} \in D(g)(\mathbb{C}) \subset \mathbb{C}^{m}$ such that $u_{m} \rightarrow \pi(p)$ as $m \rightarrow \infty.$ We are now hoping to lift this convergence in $$\pi^{-1}(D(g))(\mathbb{C}) \subset U(\mathbb{C}) \subset X(\mathbb{C})$$ to achieve a convergence to $p,$ by possibly throwing away many $m$'s.

OK. Let's get back to the proof.

Step 3. Construction of such a subsequence.

Since $\pi$ is a finite map, any of its fiber has finite size (e.g., Vakil 7.3.K), so we may write $$\pi^{-1}(\pi(p)) = \{\mathfrak{p}_{1}, \mathfrak{p}_{2}, \dots, \mathfrak{p}_{r}\}.$$ Without loss of generality, we may assume $\mathfrak{p}_{1} = p.$ Recall that we have $X = \mathrm{Spec}(A)$ with a finite $\mathbb{C}$-algebra extension $\mathbb{C}[x_{1}, \dots, x_{n}] \hookrightarrow A.$ Since $\mathfrak{p}_{i}$ are maximal ideals of $A,$ by the Chinese Remainder, we get a surjection $$A \twoheadrightarrow (A/\mathfrak{p}_{1}) \times \cdots \times (A/\mathfrak{p}_{r})$$ given by the ring projections. Thus, we may pick $f(\boldsymbol{x}) = f \in \mathfrak{p}_{1} = p$ such that $f \notin \mathfrak{p}_{i}$ for all $i \geq 2.$ Since the extension is finite, it is also integral. Hence, we may find $$F \in \mathbb{C}[x_{1}, \dots, x_{n}][t] = \mathbb{C}[x_{1}, \dots, x_{n}, t]$$ such that $$F = F(x_{1}, \dots, x_{n}, t)  = t^{d} + a_{1}(\boldsymbol{x})t^{d-1} + \cdots + a_{d-1}(\boldsymbol{x})t + a_{d}(\boldsymbol{x})$$ and $$F(x_{1}, \dots, x_{n}, f) = 0 \in A.$$ Since $\mathbb{C}[x_{1}, \dots, x_{n}, t]$ is a UFD, we may write $$F(\boldsymbol{x}, t) = F_{1}(\boldsymbol{x}, t) \cdots F_{s}(\boldsymbol{x}, t)$$ in it, where each $F_{i}(\boldsymbol{x}, t) \in \mathbb{C}[\boldsymbol{x}, t]$ is a monic irreducible polynomial. We have $$F(\boldsymbol{x}, f) = F_{1}(\boldsymbol{x}, f) \cdots F_{s}(\boldsymbol{x}, f) = 0 \in A$$ and since $A$ is a domain, there must be at least one $i$ such that $F_{i}(\boldsymbol{x}, f) = 0 \in A$ by replacing $F(\boldsymbol{x}, t)$ with $F_{i}(\boldsymbol{x}, t)$ if necessary, we may assume that $F(\boldsymbol{x}, t)$ is an irreducible polynomial.

This means that $(F)$ is a prime ideal of $\mathbb{C}[x_{1}, \dots, x_{n}, t],$ so $V(F) \subset \mathbb{A}^{n+1}$ is an irreducible Zariski-closed subset. Since $F(x_{1}, \dots, x_{n}, f) = 0$ in $A,$ we can define a $\mathbb{C}$-algebra map $$\mathbb{C}[x_{1}, \dots, x_{n}, t]/(F(x_{1}, \dots, x_{n}, t)) \rightarrow A$$ given by $t \mapsto f.$ Since the extension $\mathbb{C}[x_{1}, \dots, x_{n}] \hookrightarrow A$ factors as $$\mathbb{C}[x_{1}, \dots, x_{n}] \rightarrow \frac{\mathbb{C}[x_{1}, \dots, x_{n}, t]}{(F(x_{1}, \dots, x_{n}, t))} \rightarrow A,$$ where the second one is the map given by $t \mapsto f,$ the second map must be a finite map. Now, consider the induced maps of $\mathbb{C}$-schemes $$X = \mathrm{Spec}(A) \overset{\pi_{2}}{\longrightarrow} \mathrm{Spec}\left( \frac{\mathbb{C}[x_{1}, \dots, x_{n}, t]}{(F(x_{1}, \dots, x_{n}, t))} \right) \overset{\pi_{1}}{\longrightarrow} \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]) = \mathbb{A}^{n}$$ such that $\pi = \pi_{1} \circ \pi_{2}.$ We claim that $\pi_{2}$ is surjective.

Proof for surjectivity of $\pi_{2}$. First note that $\pi_{2}$ is given by a finite $\mathbb{C}$-algebra extension to its image, which is a closed subset of the scheme in the middle. By Krull's principal ideal theorem that $$\dim\left(\mathrm{Spec}\left( \frac{\mathbb{C}[x_{1}, \dots, x_{n}, t]}{(F(x_{1}, \dots, x_{n}, t))} \right)\right) = n.$$ Hence, if $\pi_{2}$ is not surjective, then $\pi_{2}(X)$ is a proper closed subset of this $n$-dimensional irreducible scheme (which is where we use our smart choice of $F$ being irreducible), so $\dim(\pi_{2}(X)) < n.$ However, denoting $B := \mathbb{C}[\boldsymbol{x}, t]/(F(\boldsymbol{x}, t)),$ the finite $\mathbb{C}$-algebra map $B \rightarrow A$ inducing $\pi_{2}$ factors as $$B \twoheadrightarrow B/J \hookrightarrow A,$$ where the latter one is a finite extension. This gives $\pi_{2},$ which breaks into $$X = \mathrm{Spec}(A) \twoheadrightarrow \mathrm{Spec}(B/J) \hookrightarrow \mathrm{Spec}(B),$$ so $\mathrm{Spec}(B/J) = \pi_{2}(X).$ Since $B/J \hookrightarrow A$ is a finite extension, we have $\dim(\pi_{2}(X)) = \dim(B/J) = \dim(A) = n$ (e.g., Vakil 11.1.E), so it would contradict what we had earlier: $\dim(\pi_{2}(X)) < n.$ This means that $\pi_{2}$ must be surjective.

There is more to the argument above. That is, we see that $$V(J) = V(0) = \mathrm{Spec}(B),$$ so $\sqrt{J} = \sqrt{(0)}$ in $B.$ Since $B$ is a domain, this implies that $\sqrt{J} = (0),$ so $J = (0).$ Thus, after all $\pi_{2}$ is induced by a finite extension $B \hookrightarrow A$ of $\mathbb{C}$-algebras.

Now, note that for any closed point $\mathfrak{m} \in X = \mathrm{Spec}(A),$ we have $$\pi_{2}(\mathfrak{m}) = (\pi(\mathfrak{m}), f(\mathfrak{m})) \in V(F)(\mathbb{C}) \subset \mathbb{C}^{n+1},$$ where $f(\mathfrak{m})$ is the image of $f$ in $A/\mathfrak{m} \simeq \mathbb{C}.$ This gives us $F(\pi(\mathfrak{m}), f(\mathfrak{m})) = 0.$ Since $f(p) = 0,$ we have $F(\pi(p), 0) = 0.$ Recall that $$F(\boldsymbol{x}, t) = t^{d} + a_{1}(\boldsymbol{x})t^{d-1} + \cdots + a_{d-1}(\boldsymbol{x})t + a_{d}(\boldsymbol{x}),$$ which lets us have $a_{d}(\pi(p)) = 0.$ For any $q \in \mathbb{C}^{n},$ denote by $w_{1}(q), \dots, w_{d}(q) \in \mathbb{C}$ the roots of $F(q, t)$ in $t$ allowing multiplicity. We have $$|a_{d}(q)| = |w_{1}(q)| \cdots |w_{d}(q)|.$$ Since $u_{m} \rightarrow \pi(p),$ we have $$a_{d}(u_{m}) \rightarrow a_{d}(\pi(p)) = 0,$$ as $m \rightarrow \infty.$ In other words, we have $$\lim_{m \rightarrow \infty} \left( |w_{1}(u_{m})| \cdots |w_{d}(u_{m})| \right) = 0.$$ For each $m,$ pick $t_{m} \in \{w_{1}(u_{m}), \dots, w_{d}(u_{m})\}$ so that $$|t_{i}| = \min\{|w_{1}(u_{m})|, \dots, |w_{d}(u_{m})|\}.$$ This means that $$|t_{m}|^{d} \leq |w_{1}(u_{m})| \cdots |w_{d}(u_{m})|,$$ so $t_{m} \rightarrow 0$ as $m \rightarrow \infty.$ Since $t_{m} = w_{i}(u_{m})$ for some $i,$ we have $F(u_{m}, t_{m}) = 0$ as well. Let's rewrite these two properties:

• $\lim_{m \rightarrow 0}t_{m} = 0$ in $\mathbb{C}$ and
• $F(u_{m}, t_{m}) = 0$ for all $m \geq 0.$

The second property tells us that $(u_{m}, t_{m}) \in V(F)(\mathbb{C}).$ Since $\pi_{2}$ induced a surjective map $X(\mathbb{C}) \twoheadrightarrow V(F)(\mathbb{C}),$ there exists some $y_{m} \in \pi_{2}^{-1}((u_{m}, t_{m})) \subset X(\mathbb{C}).$ Note that this means two things:

• $\pi(y_{m}) = u_{m}$ and
• $f(y_{m}) = t_{m}.$

In particular, the second property gives us $\lim_{m \rightarrow \infty}f(y_{m}) = 0.$

Claim. The sequence $(y_{m})_{m_{\geq 0}}$ is sits inside a compact subset of $X(\mathbb{C}).$ In particular, it has a convergent subsequence (e.g., Munkres Theorem 28.2, because the topology on $X(\mathbb{C})$ is given by the Euclidean metric).

For now, suppose that we have proven the claim. Then we exchange $(y_{m})$ with its convergent subsequence provided by the claim to assume that $y_{m} \rightarrow y$ for some $y \in X(\mathbb{C}),$ as $m \rightarrow \infty.$ Since $\pi$ is continuous with respect to the classical topology on the sets of $\mathbb{C}$-points, we have $$\pi(p) = \lim_{m \rightarrow \infty}u_{m} = \lim_{m \rightarrow \infty}\pi(y_{m}) = \pi(y).$$ That is, we have $$y \in \pi^{-1}(\pi(p)) = \{p = \mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}\}.$$ What's surprising is that $y$ is necessarily equal to $p.$

To see this, recall that we have chosen $f \in A$ such that $f \in \mathfrak{p}_{1} = p$ and $f \notin \mathfrak{p}_{2}, \dots, \mathfrak{p}_{r}.$ The corresponding regular function $$f : X = \mathrm{Spec}(A) \rightarrow \mathbb{A}^{1}$$ induces a continuous function $f : X(\mathbb{C}) \rightarrow \mathbb{C}$ with respect to the classical topologies, and $f(p) = 0$ while $f(\mathfrak{p}_{i}) \neq 0$ for $i \geq 2.$ Therefore, we have $$f(y) = \lim_{m \rightarrow \infty}f(y_{m}) = 0,$$ but as $y \in \pi^{-1}(\pi(p))$ and the only point at which $f$ vanishes is $p,$ so we must have $y = p.$ Hence, it only remains to show the claim.

Step 4. Proof of the claim.

Now, since $A$ is finitely generated $\mathbb{C}$-algebra, we may choose finitely many generators $h_{1}, \dots, h_{N} \in A.$ Then we have the $\mathbb{C}$-algebra surjection $$\mathbb{C}[y_{1}, \dots, y_{N}] \twoheadrightarrow \mathbb{C}[h_{1}, \dots, h_{N}] = A,$$ where each indeterminate $y_{i}$ mapsto $h_{i},$ which gives us a closed embedding $$X = \mathrm{Spec}(A) \hookrightarrow \mathbb{A}^{N}.$$ More concretely, on the sets of $\mathbb{C}$-points, each $x \in X(\mathbb{C})$ maps to $(h_{1}(x), \dots, h_{N}(x)) \in \mathbb{C}^{n}$ under this map. This map is a closed homeomorphic embedding $X(\mathbb{C}) \hookrightarrow \mathbb{C}^{n},$ with respect to the classical topology. To show the claim, it is enough to show that the image $(h_{1}(y_{m}), \dots, h_{N}(y_{m}))_{m \geq 0}$ of the sequence $(y_{m})_{m \geq 0}$ under this embedding is bounded in $\mathbb{C}^{n}.$ (This will finish the proof because any closed subset of a compact set is compact.) For this, we are in $\mathbb{C}^{n},$ so it is enough to bound the sequence $(h_{i}(y_{m}))_{m \geq 0}$ in $\mathbb{C}$ for each $1 \leq i \leq N.$

Recall that we have a finite (and hence integral) extension $$\mathbb{C}[x_{1}, \dots, x_{n}] \hookrightarrow A$$ given by Nother normalization. Thus, for each $h_{i},$ we may find a monic polynomial $$H_{i}(\boldsymbol{x}, t) = t^{d_{i}} + a_{i,1}(\boldsymbol{x})t^{d_{i}-1} + \cdots + a_{i,d_{i}-1}(\boldsymbol{x})t + a_{i,d_{i}}(\boldsymbol{x}) \in \mathbb{C}[\boldsymbol{x}, t]$$ such that $H_{i}(\boldsymbol{x}, h_{i}) = 0$ in $A.$ This factors $\mathbb{C}[\boldsymbol{x}] \hookrightarrow A$ as $$\mathbb{C}[\boldsymbol{x}] \rightarrow \mathbb{C}[\boldsymbol{x}, t]/(H_{i}(\boldsymbol{x}, t)) \rightarrow A,$$ where the second map is given by $t \mapsto h_{i}.$ This induces $\mathbb{C}$-scheme maps $$X = \mathrm{Spec}(A) \rightarrow \mathbb{C}[\boldsymbol{x}, t]/(H_{i}(\boldsymbol{x}, t)) \rightarrow \mathbb{A}^{n},$$ which compose to $\pi,$ and the first map gives $$x \mapsto (\pi(x), h_{i}(x)) \in V(H_{i})(\mathbb{C}) \subset \mathbb{C}^{n+1},$$ for any $x \in X(\mathbb{C}).$ Therefore, taking $x = y_{m},$ we see that $h_{i}(y_{m}) \in \mathbb{C}$ is a root of $$H_{i}(\pi(y_{m}), t) = t^{d_{i}} + a_{i,1}(\pi(y_{m}))t^{d_{i}-1} + \cdots + a_{i,d_{i}-1}(\pi(y_{m}))t + a_{i,d_{i}}(\pi(y_{m})) \in \mathbb{C}[t].$$ Since each $a_{i,j}(\boldsymbol{x})$ is a polynomial in $\mathbb{C}[\boldsymbol{x}]$ we have continuity to help us conclude that $$\lim_{m \rightarrow \infty}a_{i,j}(\pi(y_{m})) = \lim_{m \rightarrow \infty}a_{i,j}(u_{m}) = \lim_{m \rightarrow \infty}a_{i,j}(\pi(p)) \in \mathbb{C}.$$ In particular, each $(a_{i,j}(\pi(y_{m})))_{m \geq 0}$ is a bounded sequence in $\mathbb{C}.$ Thus, the following lemma finishes the proof.

Lemma. Let $(a_{1,m})_{m \geq 0}, \dots, (a_{d,m})_{m \geq 0}$ be bounded sequences in $\mathbb{C}.$ Then the roots of $f(z) = z^{d} + a_{1,m}z^{d-1} + \cdots + a_{d-1,m}z + a_{d,m} \in \mathbb{C}[z]$ are bounded in $\mathbb{C}.$

Proof of Lemma. We will use Rouché's theorem. Let $R \gg 1$ such that $|a_{1,m}| + \dots + |a_{d,m}| < R/d$ for all $m \geq 0.$ Then we have $$\begin{align*}|a_{1,m}R^{d-1} + \cdots + a_{d-1,m}z + a_{d,m}| &\leq |a_{1,m}|R^{d-1} + \cdots + |a_{d-1,m}|R + |a_{d,m}| \\ & \leq |a_{1,m}|R^{d-1} + \cdots + |a_{d-1,m}|R^{d-1} + |a_{d,m}|R^{d-1} \\ &< R^{d}.\end{align*}$$ Thus, by Rouché's theorem, the number of roots of the polynomial $z^{d}$ is equal to the number of the roots of $f(z)$ in the open disk $D(0; R) = \{a \in \mathbb{C} : |a| < R\},$ counting with multiplicity. The first one has $d$ roots in $D(0; R),$ so all the roots of the polynomial $f(z)$ must be in $D(0; R).$ This finishes the proof. $\Box$

We are done! $\Box$

Constructable sets. Recall that constructible subsets of a Noetherian topological space $X$ are generated by open subsets of $X$ by taking finite intersections and complements. A constructible subset is precisely a finite disjoint union of locally closed subsets, each of which is the intersection of one open subset and one closed subset (Vakil 7.4.A).

Remark. A locally closed subscheme of a scheme $X$ seems to be a more subtle notion: $Z \overset{\text{closed}}{\hookrightarrow} U \overset{\text{open}}{\hookrightarrow} X$. Vakil 8.1.M discusses why it is subtle. He references Stacks Project for an example that prohibits us from changing the order of open and closed embeddings, but I have never taken a look at this.

What's important about constructible subsets is that their structures are preserved when we push them forward with reasonable morphisms.

Chevalley's theorem (Vakil 7.4.2). Let $\pi : X \rightarrow Y$ be finite type morphism of Noetherian schemes. For any constructible subset $W \subset X,$ the image $\pi(X) \subset Y$ is constructible.

Now, let's come back to our discussion.

Lemma. Let $X$ be any $\mathbb{C}$-variety and $W \subset X$ a constructible subset. Then taking closure of $W(\mathbb{C})$ in the classical topology yields the same set as taking closure of it in the Zariski topology.

Proof. Denote by $\overline{W(\mathbb{C})}^{\mathrm{an}}$ the closure of $W(\mathbb{C})$ with respect to the classical topology and $\overline{W(\mathbb{C})}^{\mathrm{Zar}}$ using the Zariski topology. In any topology, the closure of a subset $S$ of a topological space $T$ means that the smallest closed subset of $T$ containing $S.$ Since the classical topology is finer (i.e., having more closed subsets) taking the closure in it can get only smaller, so we have $$\overline{W(\mathbb{C})}^{\mathrm{an}} \subset \overline{W(\mathbb{C})}^{\mathrm{Zar}}.$$ Note that so far $W$ could have been any subset of $X.$

For the opposite inclusion, we use that $W$ is constructible. This lets us write $$W = (U_{1} \cap Z_{1}) \sqcup \cdots \sqcup (U_{r} \cap Z_{r}),$$ where $U_{i}$ are open and $Z_{i}$ are closed in $X.$ Without loss of generality, we may assume each $U_{i} \cap Z_{i} \neq \emptyset.$ We have $$\overline{W} = \overline{(U_{1} \cap Z_{1})} \cup \cdots \cup \overline{(U_{r} \cap Z_{r})}$$ in $X$ with respect to the Zariski topology. Each $Z_{i}$ is a finite union of irreducible components say $Z_{i} = Z_{i,1} \cup \cdots \cup Z_{i, s_{i}},$ so we may collect all nonempty $U_{i} \cap Z_{i,j}$ and take the union and call it $D.$ Note that $D$ is a dense open subset of $\overline{W},$ so Corollary on top of this posting implies that $D(\mathbb{C})$ is dense in $\overline{W}(\mathbb{C}) = \overline{W(\mathbb{C})}^{\mathrm{Zar}}$ in the classical topology. Since we have $$D(\mathbb{C}) \subset W(\mathbb{C}) \subset \overline{W(\mathbb{C})}^{\mathrm{Zar}},$$ this implies that we have $$\overline{W(\mathbb{C})}^{\mathrm{an}} = \overline{\overline{W(\mathbb{C})}^{\mathrm{Zar}}}^{\mathrm{an}} \supset \overline{W(\mathbb{C})}^{\mathrm{Zar}},$$ as desired. $\Box$

Theorem. Let $X$ be any $\mathbb{C}$-variety.

(1) $X$ is separated if and only if $X^{\mathrm{an}}$ is Hausdorff.

(2) $X$ is complete if and only if $X^{\mathrm{an}}$ is compact and Hausdorff.

(3) Given a morphism $f : X \rightarrow Y$ of separated varieties, the following are equivalent:

• $f$ is proper (algebraically)
• $f^{\mathrm{an}}$ is proper (topologically)

Remark. We say a continuous map $\phi : Y \rightarrow Z$ is proper if any compact Hausdorff subset of $Z$ pulls back to a compact Hausdorf subset of $Y.$

Proof of (1). In general, the diagonal map $\Delta : X \rightarrow X \times X$ is a locally closed embedding (i.e., a closed embedding and then an open embedding; see Vakil 10.1.3). Hence $\Delta$ is a closed embedding if and only if $\Delta(X) \subset X \times X$ is Zariski-closed. (This is generally true; see Vakil 10.1.4). Since $\Delta(X)$ is a constructible subset of $X \times X,$ we have the following chain of logical equivalences:

• $X$ is separated;
• $\Delta$ is a closed embedding;
• $\Delta(X) \subset X \times X$ is Zariski-closed;
• $\Delta(X)^{\mathrm{an}} \subset (X \times X)^{\mathrm{an}} = X^{\mathrm{an}} \times X^{\mathrm{an}}$ is closed;
• $X$ is Hausdorff.

This finishes the proof (for (1)). $\Box$

Remark. A relevant result to the first theorem in this posting is as follows: if $X$ is an irreducible $\mathbb{C}$-variety, then $X^{\mathrm{an}}$ is connected.

My thought. The last remark will probably be proven by comparision of cohomologies.