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How to show, that the affine line with a split point is not a separated scheme? Hartshorne writes something about this point in product, but it is not product in topological spaces category! Give the most strict proof!

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Should "separable" be "separated?" –  Matt Oct 30 '12 at 23:20
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You can compute the fibre products locally to see that there must be four origins in $X\times X.$ To see that the diagonal is not closed, consider the intersection of the diagonal with the canonical open charts of $X\times X.$ –  Andrew Oct 30 '12 at 23:22
    
What are origins, and why they should be 4? –  user46336 Oct 31 '12 at 4:07
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up vote 8 down vote accepted

Let $X$ be the affine line with the origin doubled. More precisely, if we let $Z = \mathbb A^1$ and $U = \mathbb A^1 \setminus \{0\},$ then $X$ is the union of two copies of $Z$ in which the two copies of $U$ are identified in the obvious way. There are two obvious maps $Z \to X$ (corresponding to the two copies of $Z$ of which $X$ is the union), and they are distinct, but they coincide when restricted to $U$.

These two maps induce a map $Z \to X \times X$, and the above discussion shows that preimage of the diagonal is exactly equal to $U$. Since $U$ is not closed in $Z$, we conclude that the diagonal is not closed in $X\times X$. Thus $X$ is not separated.

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