# Separated and Finite Type Scheme over an Algebraically Closed Field

Let $$(X,\mathcal{O}_X)$$ be a separated scheme and of finite type over an algebraically closed field $$k$$. The fact that $$X$$ is separated means that the image of $$X$$ under the diagonal morphism $$\Delta: X \rightarrow X \times_{Spec(k)} X$$ is closed.

1. Do i have this correct?
2. Why is this so important, i.e. $$\Delta(X)$$ being closed in $$X \times_{Spec(k)} X$$ (i am looking for just some high level intuition)?

Next, i am trying to understand what "finite type over $$k$$" really means. Applying Hartshorne's definition, i seem to get something like that: consider the underlying continuous map of topological spaces $$f:X \rightarrow Spec(k)$$. Then $$f^{-1}(Spec(k))$$ can be covered by a finite number of open affine sets $$Spec(A_i)$$, where each $$A_i$$ is a finitely generated $$k$$-algebra. But since $$Spec(k)$$ is a one element set, we must have that $$f^{-1}(Spec(k))=X$$ and so finite type over $$k$$ is equivalent to saying that $$X$$ can be covered by a finite number of open affine sets $$Spec(A_i)$$ with $$A_i$$ being a finitely generated $$k$$-algebra. Am i understanding the situation correctly?

## 3 Answers

As you note, finite type means that $X$ is the union of finitely opens each of which is the Spec of a finite type $k$-algebra. This is an abstraction of a basic finiteness property of quasi-projective varieties.

Separatedness is the analogue, for the Zariski topology, of being Hausdorff, and like Hausdorfness, it often plays a basic role in arguments. As one example, if $f:X \to Y$ is a morphism of $k$-schemes and $Y$ is separated, then the graph $\Gamma_f \hookrightarrow X \times Y$ will be a closed subscheme.

If $X$ is a scheme of finite type over an algebraically closed field $k$ then one can show that for any affine open set Spec$A$ we have $A$ is a finitely generated $k$ algebra. Also all the restriction maps are $k$ linear.

2) Why is this so important, i.e. $Δ(X)$ being closed in $X×_{Spec(k)}X$?

One thing missing from a previous answer that being separated is the analogue of being Hausdorff, is that the categorical condition above is exactly the same categorical description of what a Hausdorff space is.

To be precise, in the category of topological spaces, with a terminal object being the single point $*$ just like $Spec(k)$, it is an exercise in point set topology that a space $X$ is Hausdorff iff $Δ(X)=\{(x,x):x\in X\}$ is closed as a subspace in $X×_{*}X=X\times X$ under the product topology.