[FOM] Grothendieck foundations: Zariski and coherent cohomology
Colin McLarty
colin.mclarty at case.edu
Tue Nov 29 09:07:10 EST 2011
I can confirm one conjecture from my previous post, but the proof is
more involved than I expected so it leaves me a bit less sure of the
other conjecture.
Harvey's suggestion of using ZF[0] (with suitable choice principle) is
working really well. In the present case I began with a quick and
dirty idea for n=1 or 2, and when it shook out it was n=0. So this is
at the strength of 2nd order arithmetic.
Specifically, the case is derived functor cohomology for all sheaves
of modules on any Noetherian scheme. This includes coherent
cohomology of Noetherian schemes, which is the tool of Hartshorne's
book _Algebraic Geometry_ and the central tool of all cohomological
number theory. It would be nice to also get etale cohomology on this
foundation, and I still suspect that can be done. But I do not know.
On the other hand, this foundation does not prove existence of some
important examples: real complex and p-adic numbers. It proves the
theorems hold for them if they exist. That is another issue. This
foundation does prove existence of all the "arithmetic schemes"
(schemes of finite type over the integers).
I sketch the key issues in the proof, as I now have it. It brings
coherent cohomology to the ambit of reverse math, where I have not
followed.
1) It currently takes global choice (over ZF[0]) to prove every
module over any Noetherian ring has an injective embedding. We use a
kind of Zorn's lemma argument for submodules of a given module,
without supposing there is a set of all those submodules. I tried
hard to get by with less choice, but I have no evidence now that it
*cannot* be done with less.
2) The natural context for this foundation is the Noetherian case.
That is the most important case in practice, especially for coherent
cohomology. The point for us is that ZF[0] even without choice proves
every set has a set of all its finite subsets, so every ring has a set
of all its finitely generated ideals -- which is all ideals, for a
Noetherian ring. An obvious inner model of countable sets shows that
this foundation does not prove all countable rings have sets of all
their ideals.
3) The chief result now is that every sheaf of modules on a Noetherian
scheme has an injective embedding, not only quasi-coherent modules.
This goes by proving the structure sheaf of rings on any Noetherian
module has a set of all sheaf ideals, not only quasi-coherent ideals.
And that works by proving that every sheaf of ideals on the sheaf of
rings R is determined by finite data relative to R, so there is a set
of all those data. The simplest proof I can find uses Krull's
Principle Ideal Theorem, roughly saying any single equation on an
algebraic space defines a subspace of codimension 1 or 0. It serves
to sharply limit how far a sheaf of ideals can deviate from being
quasi-coherent, in the Noetherian case.
colin
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