# Relation between varieties in the sense of Serre's FAC and algebraic schemes

This is a generalization of Hartshorne, Proposition 2.6 and Proposition 4.10, Chapter II.

We fix an algebraically closed field $k$.

Let $X$ be a topological space. We denote by $\mathcal{F}_X$ the sheaf of $k$-valued functions on $X$. We regard $\mathcal{F}_X$ as a sheaf of $k$-algebras in the obvious way.

Let $X$ be a Zariski closed subset of $k^n$ for some integer $n \ge 0$. Let $I(X) = \{f \in k[x_1,\dots,x_n]| f(p) = 0$ for every $p \in X\}$. Let $A = k[x_1,\dots,x_n]/I(X)$. Let $U$ be an open subset of $X$. Let $f\colon U \rightarrow k$ be a function. We say $f$ is regular at a point $p$ of $U$ if there exist an open neighborhood $V$ of $p$ contained in $U$ and $g, h \in A$ such that $h$ does not vanish at every point of $V$ and $f(x) = g(x)/h(x)$ for every $x \in V$. We say $f$ is regular on $U$ if $f$ is regular at every point of $U$. Let $\Gamma(U)$ be the set of regular functions on $U$. Clearly $U \rightarrow \Gamma(U)$ defines a subsheaf $\mathcal{O}_X$ of $\mathcal{F}_X$. We call the pair $(X, \mathcal{O}_X)$ an affine variety. By abuse of notation, we usually say $X$ is an affine variety. Note that we don't assume $X$ is irreducible.

Let $X$ be a topological space. Let $\mathcal{O}_X$ be a $k$-algebra subsheaf of $\mathcal{F}_X$. Suppose $(X, \mathcal{O}_X)$ satisfies the following conditions.

(1) $X$ is covered by a finite number of open subsets $U_i$.

(2) Each $(U_i, \mathcal{O}_X|U_i)$ is isomorphic to an affine variety.

Then $X$ is called a prevariety.

Let $X, Y$ be prevarieties. $X\times Y$ becomes a prevariety in the obvious way. Suppose the diagonal subset $\Delta_X = \{(x, x)|\ x \in X\}$ is closed in $X\times X$. Then $X$ is called a variety.

Let $X, Y$ be prevarieties. Let $f\colon X \rightarrow Y$ be a continuous map. Suppose $\psi\circ f$ is regular on $f^{-1}(U)$ for every open subset $U$ of $Y$ and every regular function $\psi$ on $U$. Then $f$ is called a morphism. Thus prevarieties(resp. varieties) form a category.

Let $X$ be a topological space. Let $t(X)$ be the set of irreducible closed subsets of $X$. If $Y$ is a closed subset of $X$, $t(Y) \subset t(X)$. We can define a topology on $t(X)$ by taking closed sets as the subsets of the form $t(Y)$, where $Y$ is a closed subset of $X$. Let $f\colon X \rightarrow Y$ be a continuous map. We define $t(f)\colon t(X) \rightarrow t(Y)$ by $t(f)(Z) = cl(f(Z))$, where $cl$ means closure. Thus $t$ is a functor on topological spaces. We define a map $\alpha\colon X \rightarrow t(X)$ by $\alpha(x) = cl(\{x\})$. It is easy to see that $U \rightarrow \alpha^{-1}(U)$ is a bijection from the set of open subsets of $t(X)$ to the set of open subsets of $X$.

Are the following assertions true?

(1) For every variety $V$ in the above sense, $(t(V), \alpha_*(\mathcal{O}_V)$ is a reduced separated scheme of finite type over $k$.

(2) There is a fully faithful and essentially epimorpfic(*) functor $t\colon Var(k) \rightarrow Sch(k)$, where $Var(k)$ is the category of varieties over $k$ and $Sch(k)$ is the category of reduced separated schemes of finite type over $k$.

(*) A functor $F\colon \mathcal{C} \rightarrow \mathcal{D}$ is essentially epimorpfic if, for every $Y\in Ob(\mathcal{D})$, there exists $X \in Ob(\mathcal{C})$ such that $F(X)$ is isomorphic to $Y$.

-
The answer is in Wedhorn and Görtz book – user40276 Mar 26 '15 at 0:17