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Let $X$ be an $S$-scheme with structural morphism given by $f : X \to S$. The image of the diagonal morphism $\Delta : X \to X \times_S X$ is contained in the subset $Z := \{ z \in X \times_S X : p(z) = q(z) \} \subset X \times_S X$ where $p, q$ are the projection maps.

Is $Z$ closed in general? Is it furthermore the closure of $\Delta(X)$?

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up vote 8 down vote accepted

In general, $Z$ will not be closed. As an example, consider $X=\mathbb{A}^1_k$, $S=\text{Spec}(k)$, where $k$ is a field. Then we have $X\times_S X=\mathbb{A}^2_k$ is the affine plane over $k$. Let $C$ be an irreducible plane curve which is not a vertical or horizontal line, or the diagonal. The curve $C$ has a generic point $x_C\in X\times_S X$, which is not closed. One can check that $p(x_C)=q(x_C)$ is the generic point of $X$, so $x_C \in Z$. However, the closure of $\{x_C\}$ contains all of the (closed) points lying on the curve $C$, so by hypothesis, contains a point not on the diagonal. This point will not be in $Z$, so $Z$ is not closed.

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Dear Pink Elephants , I hope you will pardon my curiosity but how come you are capable of writing such a perfect answer and have had no activity in algebraic geometry in the three months that you have been a member here? – Georges Elencwajg Jul 24 '12 at 23:37
Is it obvious that $p(x_c)=q(x_c)?$ and is the generic point of $X?$ – Ehsan M. Kermani Jul 25 '12 at 0:35
@GeorgesElencwajg I like to think about a lot of different kinds of math – Julian Rosen Jul 25 '12 at 2:06
@ehsanmo Because $p$ is continuous, $p(\overline{\{x_C\}})\subset \overline{\{p(x_C)\}}$. This means that $p(x_C)$ must be a point of $X$ which contains $p(C)$ in its closure. Because $C$ is irreducible and not contained in a vertical fiber, we must have that $p(C)$ is dense in $X$. It follows that $f(x_C)$ is the generic point of $X$. Likewise for $q$. – Julian Rosen Jul 25 '12 at 2:08

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