Is the scheme Spec of $\mathbb C$-algebra always separated?

Question

Is the scheme $\operatorname{Spec}A$ separated for every $\mathbb C$-algebra $A$? It is true for finitely generated $A$ and it seems to me that this is also true for algebras which are not, but would like to make sure. (If no, then is there some elementary reformulation of this condition in terms of properties of $A$?)

Answer 1

Yes, this is true. Separation of a scheme $X$ over $\Bbb C$ is equivalent to the diagonal map $\Delta: X\to X\times_{\Bbb C} X$ being a closed immersion. Since in the affine case this map corresponds to $A\otimes_{\Bbb C} A\to A$ by $a\otimes a'\mapsto aa'$, the fact that this morphism is surjective on rings implies that the diagonal is a closed immersion. In general, any affine morphism of schemes is separated - there is no need for finiteness hypotheses.