# Why are projective morphisms closed?

It is a well-known fact that if $X$ is a projective curve and $p \in X$ a smooth point, then any rational map $X \to Y$, $Y$ a projective variety, extends to a rational map $X \to Y$ regular at $p$. This is proposition I.6.8 in Hartshorne (in the case of $X$ an abstract non-singular curve), for example. However, the two proofs I have seen both assume that it suffices to consider the case $Y = \mathbb{P}^n$. As I understand it, this is because morphisms of projective varieties are proper, and in particular the image is closed. Where I can find a proof of this, in the case of varieties only? I found a proof here by Akhil Mathew, but I got lost when he started talking about base change.

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Does it help when I tell you that base change is simply another word for pull-back? In Wikipedia's notation $p_2$ is the base change of $f$ along $g$. (Think of $X$ as a bundle with base $Z$ and $P = X \times_Z Y$ as a bundle with base $Y$ and $f$ and $p_2$ the respective bundle projections). – t.b. Jun 7 '11 at 18:28
Dear Zhen, A projective variety, by definition, is something that is closed in projective space. So if you prove that a rational map $X \dashrightarrow Y$ extends to a map $X \to \mathbb{P}^n$, then the image must lie inside $Y$ (because $Y$ is closed). Now since $X$ is integral this means it scheme-theoretically factors through $Y$ as well. – Akhil Mathew Jun 7 '11 at 18:51
Dear @Akhil: Thanks for pointing the obvious out. My point set topology is too rusty to tell when intuition from metric spaces applies in non-metrisable contexts... – Zhen Lin Jun 7 '11 at 19:13
The "well-known fact" you mention in your first sentence is surely misstated: there is no relation betweemn $p$ and the rational map... When quoting things from Hartshorne, it is good to include the chapter number (each chapter has its own Proposition 6.8) – Mariano Suárez-Alvarez Jun 7 '11 at 19:47
You probably really mean "rational map regular at $p$" instead of "morphism" at the end of the 1st sentence, no? – Mariano Suárez-Alvarez Jun 7 '11 at 20:04

A) If you want to extend your rational map $f: X\to Y$ defined outside of $p$ across $p$, it is true that you can asssume that $Y=\mathbb P^n$ by embedding $Y$ into $\mathbb P^n$ as a closed set. This however has nothing to do with the completeness of $\mathbb P^n$ but follows from topology. Indeed, if $f$ sends $X\setminus \{p\}$ into $Y$, it will send the point $p$, which is in the closure of $X\setminus \{p\}$, into the closure of $f(X\setminus \{p\})$ and so in particular into $Y$.

B) Still completeness of $\mathbb P^n$ is a fundamental result that is proved here. This is a hand-out from a course given in 2009 by Mike Artin: I can't think of anybody more competent than him but I don't know if he wrote the note himself or if it was scribed by a student. Anyway, here is a link to the whole course, with many hand-outs, notes and exercises.

C) Addendum Not only don't we need completeness of $\mathbb P^n$ but the extension result is quite intuitive. In the complex case we would say that locally around $p=0$ the rational map can be written $z\mapsto F(z)=(f_0(z):...:f_n(z))$ where each $f_i(z)$ is identically zero or meromorphic of the form $f_i(z)=z^{n_i}u_i(z)$ for some $i\in \mathbb Z$ and $u_i(0)\neq 0$. If $n_0$ , say, is the smallest of the $n_i$'s ($n_0 \leq n_i$) then by multiplying all the homogeneous coordinates of $F(z)$ by $f_0 ^{-1}(z)=z^{-n_0} u_0^{-1}(z)$ we obtain $F(z)=(1:g_1(z):...:g_n(z))$ where the $g_i$'s are now holomorphic near $0$ so that the conclusion follows: $F$ can clearly be extended regularly across $p=0$.

In the purely algebraic case we adapt this idea by replacing $z$ by a uniformizing parameter near $p$, the $u_i$'s by units in $\mathcal O_p^\ast$ and the exponents $n_i$ of $z$ by the valuations of $f_i\in \mathcal O_p$, which is a discrete valuation ring thanks to the assumed regularity of $p$.

I have written this sketch because the technicalities of some proofs in the literature might obscure the actual naturality of this extension result.

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Ah, I see that Akhil made the point in A) in his comment while I was composing my answer. – Georges Elencwajg Jun 7 '11 at 19:34
I was in that class; it was scribed by a student. (I found the class somewhat confusing. It was supposed to be an undergraduate introduction, but I blink and he starts talking about sheaf cohomology...) – Qiaochu Yuan Jun 7 '11 at 19:34
Wow, undergraduate introduction : they don't lose time at MIT ! – Georges Elencwajg Jun 7 '11 at 19:42
There are also proofs (for varieties only) in Mumford's Red Book (I.9, Theorem 1) and in Shafarevich, Basic Algebraic Geometry I, section I.5.2 (p. 57 in the second edition). – Charles Staats Jun 7 '11 at 20:56
There's also a proof sketched in Exercise 14.1 (p. 318) of Eisenbud, Commutative Algebra. (To see where this fits in context, it is best to read at least the first 2-3 pages of Chapter 14.) – Charles Staats Jun 7 '11 at 21:09

We use the theory of specializations developed in Weil's Foundations of Algebraic Geometry, Chapter II. The main result of the theory of specializations(Lemma 7) below can be proved by "valuation theory + the axiom of choice". However, our proof(which is essentially the same as Weil's) does not use the axiom of choice.

Notation Let $X$ be a topological space. Let $E$ be a subset of $X$. We denote by $cl(E)$ the closure of $E$ in $X$.

Lemma 1 Let $X$ be a topological space. Let $x \in X$. Then $cl(\{x\})$ is irreducible.

Proof: Clear.

Definition 1 Let $X$ be a topological space. Suppose there exists $x \in X$ such that $X = cl(\{x\})$. We say $x$ is a generic point of $X$.

Definition 2 Let $X$ be a topological space. Let $x, y \in X$. Suppose $y \in cl(\{x\})$. Then we say $y$ is a specialization of $x$ and write $x \rightarrow y$.

We fix an algebraically closed field $\Omega$ which has infinite trancendence dimension over the prime subfield. Let $k$ be a subfield of $\Omega$ such that tr.dim $\Omega/k = \infty$. Let $E$ be a subset of a polynomial ring $k[X_1,\dots,X_n]$. We denote by $V(E)$ the common zeros of $E$ in $\Omega^n$. It is easy to see that by taking subsets of the form $V(E)$ as closed sets, we can define a topology on $\Omega^n$. We call this topology $k$-topology. A closed(resp. open) subset of $\Omega^n$ with respect to $k$-topology is called $k$-closed(resp. $k$-open) subset.

Let $E$ be a subset of $\Omega^n$. We denote by $I(E)$ the set $\{f \in k[X_1,\dots,X_n] |\ f(x) = 0$ for all $x \in E\}$. $I(E)$ is an ideal of $k[X_1,\dots,X_n]$. It is easy to see that $cl(E) = V(I(E))$. In particular, $cl(\{x\}) = V(I(\{x\}))$ for $x \in \Omega^n$. Let $x, y \in \Omega^n$. If $y$ is a specialization of $x$ with respect $k$-topology, we write $x \rightarrow_k y$ or simply $x \rightarrow y$ if there is no ambiguity. Hence $x \rightarrow_k y$ if and only if $f(y) =0$ for any $f \in k[X_1,\dots,X_n]$ such that $f(x) = 0$.

Lemma 2 Let $V$ be a $k$-irreducible subset of $\Omega^n$. Then $V$ has a generic point(this is one of the reasons why we need tr.dim $\Omega/k = \infty$).

Proof: Let $\mathfrak{p} = I(V)$. $\mathfrak{p}$ is a prime ideal. Let $K$ be the field of fractions of $k[X_1,\dots,X_n]/\mathfrak{p}$. Since tr.dim $K/k \le n$, there exists an $k$-embedding $\sigma\colon K \rightarrow \Omega$. Let $\bar x_i$ be the image of $x_i$ by the canonical homomorphism $k[X_1,\dots,X_n] \rightarrow k[X_1,\dots,X_n]/\mathfrak{p}$. Let $y = (\sigma(\bar x_1),\dots \sigma(\bar x_n))$. Then $\mathfrak{p} = I(\{y\})$. Hence $cl(\{y\}) = V(\mathfrak{p}) = V$. QED

We consider the set $\Omega_{\infty} = \Omega \cup \{\infty\}$, where $\infty$ is an element which does not belong to $\Omega$. Let $\rho\colon \Omega_{\infty} \rightarrow \Omega_{\infty}$ be the map defined as follows.

$\rho(x) = 1/x$ if $x \in \Omega - \{0\}$.

$\rho(0) = \infty$.

$\rho(\infty) = 0$.

Clearly $\rho$ is a bijection.

Let $I = \{1,\dots,n\}$. For $i \in I$, let $\rho_i\colon \Omega_{\infty}^n \rightarrow \Omega_{\infty}^n$ be the map defined by $\rho_i(x_1,\dots,x_i,\dots,x_n) = (x_1,\dots, \rho(x_i), \dots,x_n)$. Clearly $\rho_i$ is a bijection and $\rho_i\rho_j = \rho_j\rho_i$ for any $i, j \in I$. Let $G$ be the subgroup of the symmetric group on $\Omega_{\infty}^n$ generated by the set $\{\rho_1,\dots,\rho_n\}$. Let $J$ be a subset of $I$. We define $\rho_J = \prod_{i\in J} \rho_i$ Clearly every element of $G$ can be uniquely written as $\rho_J$ for some $J \subset I$. Hence $|G| = 2^n$.

Lemma 3 Let $x, x' \in \Omega^n$. Let $\sigma \in G$. Let $y = \sigma(x)$, $y' = \sigma(x')$. Suppose $y$ and $y'$ are contained in $\Omega^n$. Then $x \rightarrow_k x'$ if and only if $y \rightarrow_k y'$.

Proof: Without loss of generality, we can assume that $\sigma = \rho_J$, where $J = \{1,\dots,m\}$. Suppose $x \rightarrow_k x'$. It suffices to prove that $y \rightarrow_k y'$. Note that none of $x_1,\dots,x_m$ and $x_1',\dots, x_m'$ are $0$, since $y$ and $y'$ are contained in $\Omega^n$.

Suppose $f(y) = 0$ for $f \in k[X_1,\dots,X_n]$. Then $g = (X_1\dots X_m)^k f(1/X_1,\dots,1/X_m, X_{m+1},\dots,X_n) \in k[X_1,\dots,X_n]$ for some integer $k > 0$. Since $f(y) = 0$, $g(x) = 0$. Since $x \rightarrow_k x'$, $g(x') = 0$. Hence $f(y') = 0$. Hence $y \rightarrow y'$. QED

Definition 3 Let $x, x' \in \Omega_{\infty}^n$. Suppose there exists $\sigma \in G$ such that $\sigma(x), \sigma(x') \in \Omega^n$ and $\sigma(x) \rightarrow_k \sigma(x')$. Then we say $x'$ is a specialization of $x$ over $k$ and write $x \rightarrow_k x'$. By Lemma 3, this does not depend on the choice of $\sigma$. Lemma 1 also shows that, if $x, x' \in \Omega^n$ and $x \rightarrow_k x'$ in the new definition, $x \rightarrow_k x'$ in the old definition(Definition 2).

Lemma 4 $0$ has only one specialization over $k$, namely $0$. Hence $\infty$ has only one specialization over $k$, namely $\infty$.

Proof: $I(\{0\}) = (X)$. Hence $V((X)) = 0$. Hence $cl(\{0\}) = \{0\}$. QED

Lemma 5 Let $x =(x_1,\dots,x_n), x' = (x_1',\dots,x_n') \in \Omega_\infty^n$. Suppose $x \rightarrow_k x'$. If $x_i = \infty$ for some $i$, then $x_i' = \infty$.

Proof. Since $x \rightarrow_k x'$, $x_i \rightarrow_k x_i'$. Hence, by Lemma 4, $x_i' = \infty$. QED

Lemma 6 Let $x, x' \in \Omega^n$, $y \in \Omega$. Suppose $x \rightarrow_k x'$. Then there exists $y' \in \Omega_{\infty}$ such that $(x, y) \rightarrow_k (x', y')$.

Proof: If there exists $y'$ in an finite extension of $k(x')$ such that $(x, y) \rightarrow_k (x', y')$, we are done. Suppose there exist no $y'$ in any finite extensuon of $k(x')$ such that $(x, y) \rightarrow_k (x', y')$. Then, by Lemma 4 of my answer to this question, $y \neq 0$ and $(x, 1/y) \rightarrow_k (x', 0)$. Hence $(x, y) \rightarrow_k (x', \infty)$. QED

Lemma 7(Fundamental theorem of the theory of specializations) Let $x, x' \in \Omega_\infty^n$, $y \in \Omega_\infty^m$. Suppose $x \rightarrow_k x'$. Then there exists $y' \in \Omega_{\infty}^m$ such that $(x, y) \rightarrow_k (x', y')$.

Proof: By using induction on $m$, it suffices to prove the lemma when $m = 1$. Since $x \rightarrow_k x'$, there exists $\sigma \in G$ such that $\sigma(x), \sigma(x') \in \Omega^n$ and $\sigma(x) \rightarrow_k \sigma(x')$. Since $(\sigma(x), \infty) \rightarrow_k (\sigma(x'), \infty)$, we may assume that $y \neq \infty$. Hence the assertion follows from Lemma 6. QED

Lemma 8(Weil, Foundations of algebraic geometry, Prop. 10, Sec. 3, Chap. II) Suppose $x \rightarrow_k x'$, where $x, x' \in \Omega_\infty^n$ Let $y = (y_1,\dots,y_m) \in \Omega^m$. Suppose $y \neq 0$. Let $I = {1,\dots,m}$. Then there exists $i_0 \in I$ such that $(x, z) \rightarrow_k (x, z')$, where $z = (y_1/y_{i_0},\dots,y_m/y_{i_0})$ and $z' \in \Omega^m$.

Proof: Without loss of generality, we may assume that $y_i \neq 0$ for all $i \in I$. Suppose no such $i_0$ exists. Let $u_{i,j} = y_i/y_j$ for all $(i, j) \in I\times I$. Let $u = (u_{i,j}) \in \Omega^{m^2}$. By Lemma 7, there exists $v = (v_{i,j}) \in \Omega_{\infty}^{m^2}$ such that $(x, u) \rightarrow_k (x, v)$. By the assumption, for each $j_0 \in I$ there exists $i \in I$ such that $v_{i,j_0} = \infty$. Since $u_{j_0,i} = 1/u_{i,j_0}, v_{j_0,i} = 0$. Hence we have a map $\phi\colon I \rightarrow I$ such that $v_{j, \phi(j)} = 0$. Let $j_0 = 1$ and $j_{k+1} = \phi(j_k)$ for $k = 0,1,2,\dots$ Since $I$ is finite, there exists $p, r$ such that $j_{p+r} = j_p$. Then $v_{j_p,j_{p+1}} = \cdots = v_{j_{p+r-1}, j_p} = 0$. But this is a contradiction, because $u_{j_p,j_{p+1}}\cdots u_{j_{p+r-1}, j_p} = 1$. QED

Let $\mathbf{P}^n(\Omega)$ be the $n$-projective space over $\Omega$. Let $E$ be a set of homogeneous polynomials in $k[X_0,\dots,X_n]$. We denote by $V_+(E)$ the common zeros of $E$ in $\mathbf{P}^n(\Omega)$. It is easy to see that by taking subsets of the form $V_+(E)$ as closed sets, we can define a topology on $\mathbf{P}^n(\Omega)$. We call this topology $k$-topology. A closed(resp. open) subset of $\mathbf{P}^n(\Omega)$ with respect to $k$-topology is called $k$-closed(resp. $k$-open) subset. It is easy to see that $\mathbf{P}^n(\Omega)$ is a $k$-variety defined in this question.

Lemma 9 Let $X$ be a topological space. Let $U$ be an open subset of $X$. Let $x, y \in U$. Then $x \rightarrow y$ in $X$ if and only if $x \rightarrow y$ in $U$.

Proof: Suppose $x \rightarrow y$ in $X$. Let $V'$ be an open neighborhood of $y$ in $U$. Then there exists an open subset $V$ of $X$ such that $V' = V \cap U$. Since $x \in V$, $x \in V$. Hence $x \in V'$. Hence $x \rightarrow y$ in $U$.

Conversely suppose $x \rightarrow y$ in $U$. Let $V$ be an open subset of $X$ such that $y \in V$. Then $V \cap U$ is an open neighborhood of $y$ in $U$. Hence $x \in V \cap U \subset V$. Hence $x \rightarrow y$ in $X$. QED

Lemma 10 Let $X = \mathbf{P}^n(\Omega)$. We regard $X$ as a $k$-variety. Let $Y$ be a $k$-variety. Then $X\times Y$ is a $k$-variety as defined in this question. Let $(x, y) \in X\times Y$. Suppose $y \rightarrow y'$ in $Y$. Then there exists $x' \in X$ such that $(x, y) \rightarrow (x', y')$ in $X\times Y$.

Proof. Let $V$ be an affine open neighborhood of $y'$ in $Y$. Since $y \rightarrow y'$, $y \in V$. Hence, by Lemma 9, we may assume $Y$ is an affine $k$-variety. Then the assertion follows from Lemma 8. QED

Lemma 11 Let $X$ be an irrreducible topological space. Let $U$ be a non-empty open subset of $X$. Suppose $U$ has a generic point(Definition 1) $x$. Then $x$ is a generic point of $X$.

Proof: Let $y \in X$. Let $V$ be an open neighborhood of $y$ in $X$. Since $X$ is irreducible, $V \cap U = \emptyset$. Hence $x \in V \cap U \subset V$. Hence $x \rightarrow y$ in $X$. QED

Lemma 12 Let $f\colon X \rightarrow Y$ be a continuous map of topological spaces. Suppose $X$ is irreducible and has a generic point $x$. Then $f(x)$ is a generic point of $cl(f(X))$.

Proof: Let $y = f(x)$. Since $cl(\{y\}) \subset cl(f(X))$, it suffices to prove that $cl(f(X)) \subset cl(\{y\})$. Let $y' \in cl(f(X))$. It suffices to prove that $y \rightarrow y'$. Let $V$ be a neighborhood of $y'$ in $Y$. There exists $x' \in X$ such that $f(x') \in V$. Since $f$ is continuous, there exists a neighborhood $U$ of $x'$ such that $f(U) \subset V$. Since $x \in U$, $y = f(x) \in V$. Hence $y \rightarrow y'$ as desired. QED

Lemma 13 Let $X = \mathbf{P}^n(\Omega)$. Let $Y$ be a $k$-variety. Then the projection map $\pi\colon X\times Y \rightarrow Y$ is a closed map.

Proof: Let $Z$ be a closed subset of $X\times Y$. It suffices to prove that $\pi(Z)$ is closed. Since $X\times Y$ is a Noethrian space, $Z = Z_1\cup\cdots\cup Z_r$, where each $Z_i$ is a closed irreducible subset. Then $\pi(Z) = \pi(Z_1)\cup\cdots\cup \pi(Z_r)$. Hence $cl(\pi(Z)) = cl(\pi(Z_1)\cup\cdots\cup cl(\pi(Z_r))$. Hence we may assume that $Z$ is irreducible. By Lemma 2 and Lemma 11, $Z$ has a generic point $(x, y)$. By Lemma 11, $y$ is a generic point of $cl(\pi(Z))$. Let $y' \in cl(\pi(Z))$. Then $y \rightarrow y'$. By Lemma 10, there exists $x' \in X$ such that $(x, y) \rightarrow (x', y')$ in $X\times Y$. Since $(x',y') \in Z, y' \in \pi(Z)$. Hence $cl(\pi(Z)) = \pi(Z)$. QED

Lemma 14 Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties(see this question). Let $x \in X$ and $y = f(x)$. Let $y' \in Y$ be such that $cl(\{y\}) = cl(\{y'\})$. Then there exists $x' \in X$ such that $cl({x}) = cl({x'})$ and $f(x') = y'$.

Proof: Let $X'$ be the reduced scheme of finite type over $k$ corresponding to $X$ by this question. Then the underlying set of $X$ can be idenitified with $Hom_k(Spec(\Omega), X')$. Then the assertion follows from this question. QED

Definition 4 Let $X$ be a $k$-variety. Let $x \in X$. We denote by $k(x)$ the residue field of the local ring $\mathcal{O}_x$. We say $x$ is algebraic if $k(x)$ is algebraic over $k$.

Lemma 15 Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties. Let $X_0$(resp. $Y_0$) be the set of algebraic points of $X$(resp. $Y$). Then $f(X_0) \subset Y_0$.

Proof: Let $x \in X_0$ and $y = f(x)$. $f$ induces a $k$-homomorphism $\mathcal{O}_y \rightarrow \mathcal{O}_x$. Hence it induces a $k$-homomorphism $k(y) \rightarrow k(x)$. Since $k(x)$ is algebraic over $k$, $k(y)$ is algebraic over $k$. QED

Lemma 16 Let $x$ be a point of $k$-variety $X$. Then there exists an algebraic point $x_0 \in X$ such that $x \rightarrow x_0$.

Proof: By Lemma 9, we may assume that $X$ is affine. Let $A$ be the affine coordinate ring of $X$. Let $P = \{f \in A|\ f(x) = 0\}$. Let $\mathfrak{m}$ be a maximal ideal such that $P \subset \mathfrak{m}$. Let $K = A/\mathfrak{m}$. Since $K$ is algebraic over $k$, there exists a $k$-embedding $\phi\colon K \rightarrow \Omega$. $\phi$ induces a $k$-homomorphism $\alpha\colon A \rightarrow \Omega$. $\alpha$ can be idenitified with a point $x_0 \in X$. Clearly $x_0$ is algebraic. Since $Ker(\alpha) = \mathfrak{m}$, $x \rightarrow x_0$. QED

Lemma 17 Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties. Let $X_0$(resp. $Y_0$) be the set of algebraic points of $X$(resp. $Y$). By Lemma 15, $f$ induces a map $f_0\colon X_0 \rightarrow Y_0$. Suppose $f$ is a closed map. Then so is $f_0$.

Proof: Let $Z_0$ be a closed subset of $X_0$. There exists a closed subset $Z$ of $X$ such that $Z \cap X_0 = Z_0$. It suffices to prove that $f(Z) \cap Y_0 = f(Z_0)$. Clearly $f(Z_0) \subset f(Z) \cap Y_0$. Let $y_0 \in f(Z) \cap Y_0$. It suffices to prove that $y_0 \in f(Z_0)$. There exists $z \in Z$ such that $f(z) = y_0$. By Lemma 16, there exists $z_0 \in Z_0$ such that $z \rightarrow z_0$. Then $y_0 = f(z) \rightarrow f(z_0)$. Since $y_0$ is algebraic, $cl(\{y_0\}) = cl(\{f(z_0)\})$. Hence, by Lemma 14, there exists $w_0 \in X$ such that $cl({z_0}) = cl({w_0})$ and $f(w_0) = y_0$. Since $Z$ is closed, $w_0 \in Z$. Since $z_0$ is algebraic, so is $w_0$. Hence $w_0 \in Z_0$. Hence $y_0 \in f(Z_0)$. QED

Proposition Let $k$ be an algebraically closed field. Let $X = \mathbf{P}^n(k)$ be a projective $n$-space over $k$. Let $Y$ be a variety over $k$ in the sense of Serre's FAC. Then the porojection map $\pi\colon X\times Y \rightarrow Y$ is a closed map.

Proof: Let $Z$ be a closed subset of $X\times Y$. Let $(U_i)$ be a finite affine open cover of $Y$. Since $(X \times U_i)$ is an affine open cover of $X\times Y$ and $\pi(Z) \cap U_i = \pi(Z\cap (X\times U_i))$, we may assume that $Y$ is affine. Then $Y$ is identified with a Zariski closed subset of $k^m$. There exists an ideal $I$ of $k[X_1,\dots,X_m]$ such that $Y$ is the set of common zeros of $I$ in $k^m$. Let $Y'$ be the set of common zeros of $I$ in $\Omega^m$, where $\Omega$ is an algebraically closed field such that tr.dim $\Omega/k = \infty$.Since $Y'$ is a $k$-closed subset of $\Omega^m$, $Y'$ is an affine $k$-variety. Let $X' = \mathbf{P}^n(\Omega)$. Since $X\times Y$ is the set of algebraic points of a $k$-variety $X'\times Y'$, the assertion follows from Lemma 13 and Lemma 17. QED

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Let $k$ be an algebraically closed field. Let $X$ be a reduced separated scheme of finite type over $k$. Let $X_0$ be the set of closed points of $X$. By this question, $(X_0, \mathcal{O}_X|X_0)$ is a variety in the sense of Serre's FAC.

Conversely let $V$ be a variety over $k$ in Serre's sense. By this question, there exists a reduced separated scheme $X$ of finite type over $k$ such that $V$ is identified with $(X_0, \mathcal{O}_X|X_0)$. Such $X$ is essentially unique(i.e. they are all $k$-isomorphic).

Let $V$ be a projective variety. Let $W$ be a variety in Serre's sense. Let $p\colon V\times W \rightarrow W$ be the projection map. We will prove that $p$ is a closed map. There exists a projective scheme $X$ over $k$ such that $V$ is isomoprphic to $(X_0, \mathcal{O}_X|X_0)$. Similiarly there exists a scheme $Y$ over $k$ such that $W$ is isomoprphic to $(Y_0, \mathcal{O}_Y|Y_0)$. Let $\pi\colon X\times_k Y \rightarrow Y$ be the projection. Since $X$ is proper over $k$(Hartshorne, Theorem 4.9, Chapter II, or the theorem below), $\pi$ is a closed morphism. Since $(X\times_k Y)_0 = X_0 \times Y_0$ as a set, $p$ is a closed map by this question.

Let $f\colon V \rightarrow W$ be a morphism. Let $\Gamma = \{(x, f(x)) \in V \times W|\ x \in V\}$. Let $\Delta = \{(y, y) \in W \times W|\ y \in W\}$. Consider the map $f \times id_W\colon V\times W \rightarrow W \times W$. Then $\Gamma = (f \times id_W)^{-1}(\Delta)$. Since $\Delta$ is closed in $W\times W$, $\Gamma$ is closed in $V\times W$. Hence $f(V) = p(\Gamma)$ is closed in $W$. Since any closed subset of $V$ is a projective variety, $f$ is a closed map as desired.

Theorem The projective scheme $\mathbf{P}_{\mathbf{Z}}^n$ is proper over $\mathbf{Z}$.

Proof(EGA II, theorem 5.5.3): We need to show that the projection $\mathbf{P}_{\mathbf{Z}}^n \times Y \rightarrow Y$ is a closed morphism for any scheme $Y$ over $\mathbf{Z}$. Since the problem is local on $Y$, we may assume that $Y$ is affine. Let $Y = Spec(A)$. Then $\mathbf{P}_{\mathbf{Z}}^n \times Y = \mathbf{P}_A^n = Proj(A[X_0,\dots,X_n))$. Let $Z$ be a closed subset of $Proj(A[X_0,\dots,X_n))$. Then $Z$ is isomorphic to $Proj(A[X_0,\dots,X_n]/I)$, where $I$ is a homogeneous ideal. Let $S = A[X_0,\dots,X_n]/I$. Let $f\colon Proj(S) \rightarrow Y$ be the canonical homomorphism. It suffices to prove that $f(Proj(S))$ is closed in $Y$.

Let $y \in Y$. $f^{-1}(y) = Proj(S)\times_Y Spec (k(y)) = Proj(S\otimes_A k(y))$. Hence $f^{-1}(y) = \emptyset$ if and only if $S_n\otimes_A k(y) = 0$ for all sufficiently large $n$.Since $S_n$ is an $A$-module of finite type, $S_n\otimes_A \mathcal{O}_y$ is an $\mathcal{O}_y$-module of finite type. Hence $S_n\otimes_A k(y) = 0$ is equivalent to $S_n\otimes_A \mathcal{O}_y = 0$ by Nakayama's lemma. This is equivalent to that $\mathfrak{p}_y$ does not contain $\mathfrak{a}_n$, where $\mathfrak{p}_y$ is the prime ideal corresponding to $y$ and $\mathfrak{a}_n$ is the annihilator of $A$-module $S_n$. Since $S_n = S_{n}S_1$, $\mathfrak{a}_n \subset \mathfrak{a}_{n+1}$. Let $\mathfrak{a} = \bigcup_n \mathfrak{a}_n$. Then $f(Proj(S)) = V(\mathfrak{a})$. QED

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