As it is shown in "Goertz,Whedorn - Algebraic geometry I" there is an equivalence of categories between the category of prevarieties over $k$ (field algebraically closed) and the category of integral schemes of finite type over $k$. Now the fibered product of schemes, coincides with the classical product of prevarieties, so I ask if the following proposition is true:
A prevariety $X$ over $k$ is a variety in the classical setting (i.e. the diagonal is closed in $X\times X$) if and only if $X$ is separated as scheme over $k$.
Moreover I ask if there is an equivalence of categories between:
- varieties over $k$ as separated prevarieties in the "classical setting" (closedness of the diagonal in $X\times X$)
- separated,integral schemes of finite type over $k$
Obviously, it suffices to prove the following (applied to the diagonal morphism).
Let $k$ be an algebraically closed field and let $X,Y$ be two integral schemes of finite type over $k$. Let $f : X \to Y$ be a morphism over $k$. Then $f$ is a closed embedding if and only if $f(k) : X(k) \to Y(k)$ is a closed embedding, where we endow $X(k)$ (and also $Y(k)$) with the structure of a classical variety.
The usual arguments show that we may assume that $X$ and $Y$ are affine. If $f$ is a closed immersion, then $f$ induces $X \cong V(I)$ for some ideal $I \subseteq k[Y]$. It is prime, since $X$ is a variety. But then also $X(k) \cong V(I)(k) \stackrel{!}{=} V(k)(I)$, where $V(k)$ denotes the "classical" zero set in $Y(k)$. Hence, $f(k)$ is a closed embedding. Conversely, assume that $f(k) : X(k) \to Y(k)$ is a closed embedding. Then there is an ideal $I \subseteq k[Y]$ such that $X(k) \cong V(k)(I) = V(I)(k)$. Since $X$ is a variety, $I$ is prime, hence $V(I)$ is an integral scheme. Since $X \mapsto X(k)$ is an equivalence of categories, the isomorphism lifts to a unique isomorphism of schemes $X \cong V(I)$. It commutes with the maps to $Y$, since this is the case after applying the equivalence. Hence, $f$ is a closed embedding.