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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?

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2 Answers 2

up vote 3 down vote accepted

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.

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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.

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