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

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A morphism is separated if its diagonal is a closed immersion. A morphism is quasi-separated if its diagonal is quasi-compact. In particular, a quasi-separated scheme over a field has the property that the intersection of two affine open subsets is quasi-compact. –  Anton Geraschenko Sep 29 '09 at 19:54
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Is there a finite dimensional example? –  David Zureick-Brown Sep 30 '09 at 12:26

2 Answers 2

up vote 11 down vote accepted

Suppose $U\hookrightarrow X$ is a non-quasi-compact open immersion. Then you can glue two copies of $X$ together along $U$ (effectively doubling up the complement of $U$) to get a non-quasi-separated scheme $Y$.

By the assumption that $U\hookrightarrow X$ is not quasi-compact, there is some open affine $W$ of $X$ such that the intersection of $W$ and $U$ is not quasi-compact. 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 quasi-compact. So we found two open affines in $Y$ whose intersection is not quasi-compact, which shows that $Y$ is not quasi-separated.

Now we just have to find some non-quasi-compact open immersions. The complement of the origin in $\mathbb A^\infty$ is one (so $\mathbb A^\infty$ with a doubled origin is a non-quasi-separated scheme).

Edit: Here's another one that I don't completely understand, but gives a finite-dimensional example (zero-dimensional 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 quasi-compact, so we get another example of a non-quasi-compact open immersion, so $X$ with a doubled point is non-quasi-separated.

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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... –  Andrew Critch Nov 6 '09 at 18:13
    
+1: nice answer. –  Paul Balmer May 9 '10 at 0:53
    
A quick explanation to the Galois example: Spec \bar{Q} --> Spec Q is a pro-torsor, in fact a pro-Galois 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 set-theoretically (not scheme-theoretically) a disjoint union of copies of Spec \bar{Q}, indexed by the Galois group. –  shenghao Oct 10 '10 at 15:13
    

Here is some intuitive propaganda for Anton's answer...

We know that a qsep (quasi-separated) 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 quasi-compact. 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 quasi-separated scheme is one where any two open affines are "easy to distinguish".
  • A non-qsep 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 non-qsep 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 quasi-separated morphism is one which preserves the existence of "subtle distinctions".
  • A non-qsep 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 non-qsep iff there is some qsep open $V\subseteq Y$ such that $f^{-1}(V)$ is non-qsep, 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 scheme-theoretic image of $S$. Then $T$ qsep $\Rightarrow$ $S$ qsep, hence $S$ non-qsep $\Rightarrow$ $T$ non-qsep, meaning $f$ preserves the existence of subtle distinctions in passing from any such $S$ to $T$.

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