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!
Why is the affine line with a doubled point not a separated scheme?
9
$\begingroup$
algebraic-geometry
schemes
sheaf-theory
-
0What are origins, and why they should be 4? – 2012-10-31
2 Answers
18
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.