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This example wanted to show if $k$ is a field, $X$ the affine line with a double point as in Ex 2.3.6, then X is not separated.

It argued that $X$(product over $k$)$X$ is affine plane with double axes and four origins. It is not closed because all four origins are in the closure of the diagonal.So the diagonal is not a closed immersion.

But I don't know how to display the product scheme explicitly. Although it looks natural, usually the underlying space of the product scheme may not be the product of the corresponding topological spaces.

I tried to construct it from the method in arbitrary product schemes but got messy. Can anyone help me with that?

Another question is if $ k$ is assumed to be algebraically closed here?

1 Answers1

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Because I'm too lazy to explain Hartshorne's sketchy proof, here is an alternative proof (which regard as the most canonical one):

In general, a scheme $X$ is separated if and only if for all open affine schemes $U,V \subseteq X$ the canonical morphism $\Delta : U \cap V \to U \times V$ is a closed immersion (in particular $U \cap V$ is affine). Here you only need the direction $\Rightarrow$, which follows by taking the base change of $\Delta : X \to X \times X$ with respect to $U \times V \to X \times X$.

If $X$ is the gluing of two affine lines $U,V$ along the complement of the origin, then $U \cap V $ is the complement of the origin, and $U \times V$ is the product of two affine lines. Clearly the image is not closed.

All this works for every base field.

Martin Brandenburg
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  • Could it be that the image of $U \cap V $ in $U\times V$, which is the plane, is the hyperbola, which is closed? – Jordi Apr 09 '18 at 22:18