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David Madore science forum beginner
Joined: 23 Feb 2005
Posts: 24

Posted: Thu Jul 20, 2006 12:04 pm Post subject:
descent for projectivity



I am looking for a reference for the following fact (assuming it is
true, or else the closest which can be found...):
If K/k is an arbitrary field extension, X an algebraic variety over k
such that X_K (extension of scalars from k to K) is projective, then
X is projective.
I believe I can prove it, though I haven't checked all details and
could possibly be wrong. A sketch of proof goes like this:
First, we can assume that K over k is of finite type (essentially by
EGA IV.. So we can reduce to two cases: a purely transcendental
extension k(t)/k and a finite (algebraic) extension. In the former
case, again presumably by results in EGA IV.8 we can assume X is
projective over a nonempty open subset of Spec k[t], in which we
can find a closed point. So all reduces to the case of a finite
extension. By SGA1 exp VIII corollary 4.8 ("proper" descends by
fpqc base change), X is proper because X_K is proper (and Spec K >
Spec k is fpqc). Now consider p: X_K > X the projection morphism,
which is finite (hence proper) and (faithfully) flat: take D an
ample divisor on X_K, and let D' = p_*(D) its pushforward. Apply
the intersection formula (or whatever it's called) to compare
Y.p_*(D) with p^*(Y).D (if Y is a cycle in X) and the
NakaiMoishezon criterion to prove that D' is ample.
This is all rather tedious (writing down all the details was too much
of a pain, I eventually gave up), and I'm surprised I can't find a
convenient reference for this statement which looks like a basic fact
every algebraic geometer would need eventually. It is strange that
SGA1 mentions fpqc descent for properness but not for projectivity.

David A. Madore
(david.madore@ens.fr,
http://www.dma.ens.fr/~madore/ ) 

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Jannick Asmus science forum Guru
Joined: 25 Mar 2005
Posts: 312

Posted: Thu Jul 20, 2006 3:15 pm Post subject:
Re: descent for projectivity



On 20.07.2006 14:04, David Madore wrote:
Quote:  I am looking for a reference for the following fact (assuming it is
true, or else the closest which can be found...):
If K/k is an arbitrary field extension, X an algebraic variety over k
such that X_K (extension of scalars from k to K) is projective, then
X is projective.
I believe I can prove it, though I haven't checked all details and
could possibly be wrong. A sketch of proof goes like this:
First, we can assume that K over k is of finite type (essentially by
EGA IV.. So we can reduce to two cases: a purely transcendental
extension k(t)/k and a finite (algebraic) extension. In the former
case, again presumably by results in EGA IV.8 we can assume X is
projective over a nonempty open subset of Spec k[t], in which we
can find a closed point. So all reduces to the case of a finite
extension. By SGA1 exp VIII corollary 4.8 ("proper" descends by
fpqc base change), X is proper because X_K is proper (and Spec K 
Spec k is fpqc). Now consider p: X_K > X the projection morphism,
which is finite (hence proper) and (faithfully) flat: take D an
ample divisor on X_K, and let D' = p_*(D) its pushforward. Apply
the intersection formula (or whatever it's called) to compare
Y.p_*(D) with p^*(Y).D (if Y is a cycle in X) and the
NakaiMoishezon criterion to prove that D' is ample.
This is all rather tedious (writing down all the details was too much
of a pain, I eventually gave up), and I'm surprised I can't find a
convenient reference for this statement which looks like a basic fact
every algebraic geometer would need eventually. It is strange that
SGA1 mentions fpqc descent for properness but not for projectivity.

David,
have a look at EGA II.6.6.5 giving you the result for finite extensions K/k.
Could you give the exact reference giving you the descent for strictly
transcendental extensions? Thanks.
Best wishes,
J. 

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David Madore science forum beginner
Joined: 23 Feb 2005
Posts: 24

Posted: Thu Jul 20, 2006 5:45 pm Post subject:
Re: descent for projectivity



Jannick Asmus in litteris <e9o6ll$7c9$1@dizzy.math.ohiostate.edu>
scripsit:
Quote:  have a look at EGA II.6.6.5 giving you the result for finite extensions K/k.

Great! Thanks. I was fooled by the fact that the section on
projectivity is II.5, not II.6 (someday someone should write a decent
index for EGA).
Interestingly, there is a forward reference to EGA V (which was never
completed, although there are rough notes for it circulating on the
Web) for the case when K/k is arbitrary.
Quote:  Could you give the exact reference giving you the descent for strictly
transcendental extensions? Thanks.

Hmmm... I really didn't check that part too carefully (my concern was
mostly whether the algebraic inseparable case works), but the idea I
had in mind was to use theorem IV.8.10.5(xiii) upon writing Spec(k(t))
(which is the generic point of A^1_k) as the projective limit of the
non empty open sets U in A^1_k: this would say that X is projective
over k(t) if and only if it is projective over some actual open set U,
and then we can specialize to some closed point in U. (Assuming this
works, thanks are due to Jo\"el Riou, who taught me how useful this
theorem is for this sort of things. If it doesn't work, though, I'm
the one to blame.)

David A. Madore
( david.madore@ens.fr,
http://www.madore.org/~david/ ) 

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Jannick Asmus science forum Guru
Joined: 25 Mar 2005
Posts: 312

Posted: Fri Jul 21, 2006 11:27 am Post subject:
Re: descent for projectivity



On 20.07.2006 19:45, David Madore wrote:
Quote:  Jannick Asmus in litteris <e9o6ll$7c9$1@dizzy.math.ohiostate.edu
scripsit:
have a look at EGA II.6.6.5 giving you the result for finite extensions K/k.
Great! Thanks. I was fooled by the fact that the section on
projectivity is II.5, not II.6 (someday someone should write a decent
index for EGA).
Interestingly, there is a forward reference to EGA V (which was never
completed, although there are rough notes for it circulating on the
Web) for the case when K/k is arbitrary.

You might check EGA IV.9.1.5 to find the general result you are looking
for. I think this is a version which was supposed to appear in EGA V.
Best wishes,
J. 

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