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Fix some algebraically closed field k. Consider the morphims $\psi, \phi:A^1 \rightarrow A^1\sqcup{0'}$ where $\phi(k):=k^-1$ if k is different from zero and $\phi(k):=0'$. Likewise define $\psi$ as above but instead in the first case it assigns a non-zero point itself (instead of it's inverse).

The first case defines an object isomorphic to $P^1$, in the category of varieties but the second defines a pre-variety which is not a variety. Why is this?

P.S.: by variety X I mean a pre-variety who's diagonal is closed in $X\times X$.

Thank in advance guys and gals :)

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    $\begingroup$ Are you asking for a proof that the line with two origins is not affine? It Is not even separated as you point out, thus not affine. $\endgroup$
    – Derek Allums
    Jun 25, 2013 at 19:49
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    $\begingroup$ Have a look at the answer here: math.stackexchange.com/questions/225521/separated-scheme/… $\endgroup$
    – Andrew
    Jun 25, 2013 at 20:11
  • $\begingroup$ Have you tried to examine the diagonal in both cases? It's a good exercise in making sure you understand how to work with products of non-affine things. If you say where you got stuck, that helps everyone. $\endgroup$
    – TTS
    Jun 25, 2013 at 20:26
  • $\begingroup$ Cool, thanks alot guys, it was exactly as you said: an easy exersie in working with products :) $\endgroup$
    – ABIM
    Jun 28, 2013 at 0:52

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