[FOM] Grothendieck foundations progress, and a posted error

Colin McLarty colin.mclarty at case.edu
Mon Sep 12 11:26:41 EDT 2011

I have sent the math arXiv an updated version of the paper, adding
this result about cohomology with just one powerset of the naturals:

ZFC[1], which is ZFC without the power set axiom but positing that the
naturals have a power set, proves every countable module on a
countable ring has cohomology groups of all finite orders.  Added as
Section 3.6.2.  The powerset serves to form function sets between
countable (not finitely generated) Abelian groups--notably the
underlying Abelian groups of polynomial rings.

The result fits nicely in this context.  But it will get its full
value when extended to cohomology of schemes, and I am still working
to see the best low-order arithmetic approach to arithmetic schemes
for cohomology.

As Harvey has pointed out, ZFC[n] differs from (n+2)-order arithmetic
in using less coding to express common theorems.  For me, anyway, that
is a huge advantage, I can use ZFC[n], but I would like to also know
the proof that this is interpretable in (n+2)-order arithmetic.

The error is in my last post to FOM attempting a proof (and I have to
say Jeremy Avigad and Albert Visser had already sent me better advice,
but I am stil finding my way around in this).   I succeeded at
interpreting sets of rank omega and less in second order arithmetic,
but ZFC[0] also gives any finite rank above that.   Some tree
interpretation is needed and I'm sure it is routine when you get clear
about it.

Ali Enayat pointed me to relevant discussion in Kreisel's Survey of
Proof Theory.  I have read that madly rich paper before but never with
this issue in mind and it is not a paper I will soon exhaust so I am
sure it will help.

I am still in the market for a plain published proof of mutual
interpretability of ZF[0] in second order arithmetic if one exists.
Even if I work out the details for myself first it would be good to
have a citation.

best, Colin

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