5
$\begingroup$

In A. Bondal, M. van den Bergh's paper, Generators and representability of functors in commutative and noncommutative geometry , the "reduction principle" of quasi-compact, quasi-separated schemes is shown. It is stated as follows.

Assume $X = U_1 \cup U_2 $ with $U_1, U_2$ open and put $U_{12} = U_1 \cap U_2$.

Let $P$ be a property satisfied by some schemes such that

(1) $P$ is true for affine schemes.

(2) If $P$ holds for $U_1, U_2, U_{12}$ as above, then it holds for $X$.

Then $P$ holds for all quasi-compact quasi-separated schemes.

I understood this statement. However, the next Remark3.3.2 says

It is easy to see that the class of quasi-compact quasi-separated schemes is the biggest class of schemes to which the reduction principle is applicable (for all properties $P$).

I'm not sure why this characterization of quasi-compact, quasi-separated schemes holds. Is there an obvious property $P$ which satisfies above properties and holds only for quasi-compact, quasi-separated schemes?

$\endgroup$
3
  • $\begingroup$ Presumably induction has something to do with it, hence quasi-compactness, and I gather quasi-separatedness is related to intersections of open affines being affine. But an expert should really step in and answer this. $\endgroup$
    – David Roberts
    Feb 10, 2014 at 9:24
  • 4
    $\begingroup$ “Is there an obvious property $P$ which satisfies above properties and holds only for quasi-compact, quasi-separated schemes?” — How about the property qcqs? (I.e., being quasi-compact and quasi-seperated.) $\endgroup$
    – jmc
    Feb 10, 2014 at 9:39
  • 1
    $\begingroup$ As far as I can tell, the same characterization holds if you replace "schemes" with "morphisms". $\endgroup$
    – S. Carnahan
    Feb 10, 2014 at 11:11

1 Answer 1

9
$\begingroup$

Yes, such a property $P$ exists, namely "being qcqs" (i.e., quasi-compact and quasi-separated). It is trivial that qcqs answers your question, if it indeed satisfies your conditions on $P$.

Let's check the conditions:

  1. $P$ is true for affine schemes. http://stacks.math.columbia.edu/tag/01S7

  2. If $P$ holds for $U_{1}$, $U_{2}$, $U_{12}$ as above, then it holds for $X$. The qc part is (easy) topology. For qs one can use http://stacks.math.columbia.edu/tag/01KO, and see that #3 is satisfied for $X$. Indeed, take $S = \mathrm{Spec}(\mathbb{Z})$, and cover it with the trivial covering $\mathrm{Spec}(\mathbb{Z})$. The rest of the lemma says that we have to cover $X$ with affine opens $V_{j}$: well, we may (and do) choose the $V_{j}$ to all be in $U_{1}$ or $U_{2}$. All we need to check is that $V_{j} \cap V_{j'}$ is covered by a finite number of affine open subsets of $X$. Well, take any such cover $W_{k}$. Suppose $V_{j} \subset U_{i}$, and $V_{j'} \subset U_{i'}$. We have $$ V_{j} \cap V_{j'} = (V_{j} \cap U_{i}) \cap (U_{i'} \cap V_{j'}) = V_{j} \cap (U_{i} \cap U_{i'}) \cap V_{j'}. $$ On the right hand side everything is qcqs (either because affine, or by assumption) [edit:] and all intersections take place in qcqs schemes (namely $U_{i}$ and $U_{i'}$) [/edit]. Hence so is the intersection, and therefore $V_{j} \cap V_{j'}$ is qcqs (in particular qc). This allows us to take a finite subcover of $W_{k}$.


Edit: The last paragraph was not very well-phrased. The point is that a priori $V_{j} \cap V_{j'}$ need not be qcqs (even though both $V_{j}$ and $V_{j'}$ are qcqs). After all, the intersection takes place in $X$, and we do not know that $X$ is qcqs. (Indeed, it is what we are trying to prove.) However, we know that $V_{j} \subset U_{i}$, and $V_{j'} \subset U_{i'}$, and this makes it possible to rewrite the intersection into intersections taking place in qcqs schemes. And of those, we know that the intersection is qcqs: item (6) of http://stacks.math.columbia.edu/tag/01KU.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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