Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Grothendieck defined retrocompact subsets of a topological space in EGA III-1. p.12. The notion of retrocompact open subsets is used in the definition of constructible subsets of a scheme which is important in algebraic geometry.

Let $X$ be a topological space. We say a subset $Z$ of $X$ is retrocompact if $Z \cap U$ is quasi-compact for every quasi-compact open subset $U$ of $X$.

If $X$ is a separated scheme, every quasi-compact open subset is retrocompact.

Suppose $X$ is a locally Noetherian scheme. Let $U$ be an open subset of $X$. Let $V$ be a quasi-compact open subset of $X$. Since $V$ is a Noetherian topological space, $U \cap V$ is quasi-compact. Hence $U$ is retrocompact.

So let us suppose $X$ is a separated scheme which is not locally Noetherian. I would like to know an example of an open subset $U$ of $X$ satisfiying the following conditions.

(1) $U$ is not quasi-compact.

(2) $U$ is retrocompact.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

The retro-compactness is a property relative to $U\to X$ and it is not very restrictive on $U$ itself. In particular, $X$ is always retro-compact in $X$.

Let $X$ be a quasi-affine (hence separated) scheme which is not quasi-compact (e.g. the complement in $\mathrm{Spec}\mathbb C[x_1,\dots, x_n, \dots]$ of the closed point defined by the maximal ideal $(x_1,\dots, x_n, \dots)$), then $X$ is retro-compact in $X$ but is not quasi-compact.

To have an example with $U\ne X$, take $U$ be the above quasi-affine scheme and take $X$ be the disjoint union of two copies of $U$.

Edit Any affine open immersion $U\to X$ is trivially retro-compact. So let $X$ be the affine punctured scheme as above, let $U=D(x_1-1)\cap X$. Then $U\to X$ is affine, but $U$ is not retro-compact. This gives an example with $X$ irreducible and $U\ne X$.

share|improve this answer
    
Thanks. I should have added a condition $U \neq X$. –  Makoto Kato Dec 12 '12 at 20:16
    
This will be harder if you ask further that $X$ be irreducible. –  user18119 Dec 12 '12 at 20:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.