## A possible example

I suspect there is no scheme which is "two $\mathbb A^1$'s glued together along their generic points" (or "$\mathbb A^1$ with every closed point doubled"). In other words, I think that the coequalizer of the two inclusions $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ does not exist in the category of schemes. Intuitively, this coequalizer should be "too non-separated" to be a scheme.

I don't have a proof, but I thought other people might have ideas if I posted this here.

One thing I can prove is that if a coequalizer $X$ does exist, then no two closed points of $\mathbb A^1\sqcup \mathbb A^1$ map to the same point in $X$. To show this, it is enough to find functions from $\mathbb A^1\sqcup \mathbb A^1$ to other schemes which agree on the generic points but disagree on any other given pair of points. The obvious map $\mathbb A^1\sqcup \mathbb A^1\to \mathbb A^1$ separates most pairs of closed points. To see that a point on one $\mathbb A^1$ is not identified with "the same point on the other $\mathbb A^1$", consider the map from $\mathbb A^1\sqcup \mathbb A^1$ to $\mathbb A^1$ with the given point doubled.

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Here are some questions that might be helpful to answer:

> If the coequalizer above *does* exist, must the map from $\mathbb A^1\sqcup \mathbb A^1$ be surjective?

(see the related question [Can a coequalizer of schemes fail to be surjective?][1])

> Is the coequalizer of $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ in the category of **separated** schemes equal to $\mathbb A^1$? (probably)


  [1]: http://mathoverflow.net/questions/63/can-a-coequalizer-of-schemes-fail-to-be-surjective