What are some examples of morphisms of schemes which are not quasi separated?

$\begingroup$ A morphism is separated if its diagonal is a closed immersion. A morphism is quasiseparated if its diagonal is quasicompact. In particular, a quasiseparated scheme over a field has the property that the intersection of two affine open subsets is quasicompact. $\endgroup$ – Anton Geraschenko Sep 29 '09 at 19:54

1$\begingroup$ Is there a finite dimensional example? $\endgroup$ – David ZureickBrown♦ Sep 30 '09 at 12:26
Suppose $U\hookrightarrow X$ is a nonquasicompact open immersion. Then you can glue two copies of $X$ together along $U$ (effectively doubling up the complement of $U$) to get a nonquasiseparated scheme $Y$.
By the assumption that $U\hookrightarrow X$ is not quasicompact, there is some open affine $W$ of $X$ such that the intersection of $W$ and $U$ is not quasicompact. So there are two copies of $W$ sitting inside of $Y$ (one for each copy of $X$). The intersection of these two is exactly $U\cap W$, which is not quasicompact. So we found two open affines in $Y$ whose intersection is not quasicompact, which shows that $Y$ is not quasiseparated.
Now we just have to find some nonquasicompact open immersions. The complement of the origin in $\mathbb A^\infty$ is one (so $\mathbb A^\infty$ with a doubled origin is a nonquasiseparated scheme).
Edit: Here's another one that I don't completely understand, but gives a finitedimensional example (zerodimensional in fact). Consider $X=Spec(\overline{\mathbb Q}\otimes_{\mathbb Q}\overline{\mathbb Q})$. Topologically, $X=Gal(\overline{\mathbb Q}/\mathbb Q)$, with the profinite topology (perhaps somebody could explain how to see this in a comment). In particular, any point is closed, but the complement $U$ is not quasicompact, so we get another example of a nonquasicompact open immersion, so $X$ with a doubled point is nonquasiseparated.

$\begingroup$ I was just in the middle of trying the example you just edited in! Profinite topology... very nice :) I'd also like to see an explanation of this... $\endgroup$ – Andrew Critch Nov 6 '09 at 18:13

$\begingroup$ A quick explanation to the Galois example: Spec \bar{Q} > Spec Q is a protorsor, in fact a proGalois one, with Galois group Gal(\bar{Q}/Q). If we take the fiber product of the torsor against itself, we get the product of Spec \bar{Q} and the Galois group. This is settheoretically (not schemetheoretically) a disjoint union of copies of Spec \bar{Q}, indexed by the Galois group. $\endgroup$ – shenghao Oct 10 '10 at 15:13

$\begingroup$ See also mathoverflow.net/questions/30030/galoisgroupasasheaf/… for a proof. $\endgroup$ – Martin Brandenburg Oct 28 '10 at 10:13
Here is some intuitive propaganda for Anton's answer...
We know that a qsep (quasiseparated) scheme (over $\mathbb{Z}$) is precisely one where the intersection U∩V of any two open affines, U=Spec(A) and V=Spec(B), is quasicompact. Looking at compliments gives a different perspective: that their differences U\V and V\U are cut out by finitely many elements in A,B respectively, meaning that these differences are "easy to see". I'd say this justifies the following credo:
 A quasiseparated scheme is one where any two open affines are "easy to distinguish".
 A nonqsep scheme is one containing some "subtle distinction" between open affines.
The two copies of $\mathbb{A}^\infty$ in Anton's answer differ only by the origin, which is "hard to see" in that it cannot be cut out by finitely many ring elements, and I'd say using infinitely many variables to cut out one point is about the most natural way to achieve this. Thus, I like to characterize nonqsep schemes as containing "(infinitely) subtle distinctions" such as this one.
Further tinkering yields a similar way to think about a qsep morphism $f:X\to Y$. I'd say the corresponding credo is that:
 A quasiseparated morphism is one which preserves the existence of "subtle distinctions".
 A nonqsep morphism is one which destroys some "subtle distinctions".
This helps intuitivize theorems like:
(1) " Any map from a qsep scheme is qsep ", because it has no subtle distinction that can be destroyed.
(2) " If $Y$ is qsep, then $f:X\to Y$ is qsep iff $X$ is qsep ", since $f$ destroys subtle distinctions iff $X$ has them.
(3) " If $g\circ f$ is qsep, then $f$ is qsep ", since if $f$ destroyed some subtle distinction, then $g$ could not recover it.
Here is a coarse and a fine justification for this credo in each direction...
Coarse version: By 1971 EGA I 6.1.11, for any cover of Y by qsep opens $V_{i}$, $f$ is qsep iff each preimage $f^{1}(V_i)$ is qsep. Thus, $f$ is nonqsep iff there is some qsep open $V\subseteq Y$ such that $f^{1}(V)$ is nonqsep, meaning it contains some subtle distinction which is lost after application of by $f$.
Fine version: Suppose $f$ is qsep. By 1971 EGA I 6.1.9, fibre products and compositions of qsep morphisms are qsep, and any universal injection is qsep (for example any immersion). Now suppose
$S\hookrightarrow X$
$T\hookrightarrow Y$
are any universal injections such that $f_S$ factors through $T$, for example if $T$ is the schemetheoretic image of $S$. Then $T$ qsep $\Rightarrow$ $S$ qsep, hence $S$ nonqsep $\Rightarrow$ $T$ nonqsep, meaning $f$ preserves the existence of subtle distinctions in passing from any such $S$ to $T$.