10
$\begingroup$

For any prime p, one has the Frobenius homomorphism Fp defined on rings of characteristic p.

Is there any kind of object, say U, with a "universal Frobenius map" F such that for any prime p and any ring R of characteristic p we can view the Frobenius Fp over R as "the" base change of F from U to R?

I have the following picture in mind: In some sense it should be possible to view the category of Z-algebras as a sheaf of categories over Spec Z such that the fibre over Spec Fp is just the category of F_p-algebras. A natural transformation f of the identity functor on the category of Z-algebras should restrict to a natural transformation fp of the identity functor on the category of Fp-algebras. In this naive picture one cannot expect the existence of an f such that fp is the Frobenius on Fp-algebras for all primes p. But is there way to make this picture work?

Another possible way to answer my question could be the following: Is there a classifying topos of, say, algebras with a Frobenius action? By this I mean the following: Is there a topos E with a fixed ring object R and an algebra A over it and an R-linear endomorphism f of A such that for any other topos E' with similar data R', A' there is a unique morphism of topoi E' -> E that pulls back R, A to R', A' and such that f is pulled back to the Frobenius fp of A' in case R' is of prime characteristic.

(Feel free to modify my two pictures to make them work.)

Improve this question
$\endgroup$
4
  • $\begingroup$ I may have initially misinterpreted this question. Marc, are you asking if there is some $f_U:U\to U$ such that for any ring $R$ of characteristic $p$, there is a unique morphism $u_R:Spec(R)\to U$ such that the pullback of $f_U$ along $u_R$ is the Frobenius ($p$th power) map $f_R:R\to R$? $\endgroup$
    – Andrew Critch
    Commented Nov 26, 2009 at 9:26
  • $\begingroup$ Something like this, yes. I have the following picture in mind: In some sense it should be possible to view the category of Z-modules as a sheaf of categories over Spec Z such that the fibre over Spec F_p is just the category of F_p-modules. A natural transformation f of the identity functor on the category of Z-modules should restrict to a natural transformation f_p of the identity functor on the category of F_p-modules. In this naive picture one cannot expect the existence of an f such that f_p is the Frobenius on F_p-modules for all primes p. But is there way to make this picture work? $\endgroup$
    – Marc Nieper-Wißkirchen
    Commented Nov 26, 2009 at 12:19
  • $\begingroup$ I suggest editing your question to include one or more precise formulations, and also a discussion like the comment you just wrote; MO users like to see partial progress in question statements, because it both motivates and provides a starting place for others. I'd like to see this question get more upvotes so it attracts attention and gets an answer. $\endgroup$
    – Andrew Critch
    Commented Nov 26, 2009 at 21:53
  • $\begingroup$ Improved my question. $\endgroup$
    – Marc Nieper-Wißkirchen
    Commented Nov 28, 2009 at 21:17

2 Answers 2

Reset to default
3
$\begingroup$

I don't really get the categorical picture of what you are asking, but it feels something very similar to the relation between finite fields of characteristic $p$ and the ring (of characteristic 0) of $p$-typical Witt vectors.

You might want to have a look at Borger and Wieland work on pleythistic algebras. The "lifting of all Frobeniuses at the same time" gives you an structure of $\Lambda$-ring, so the paper The basic geometry of Witt vectors by Borger might also be useful.

Improve this answer
$\endgroup$
1
$\begingroup$

It seems that you need 'a field with one element' here. There is a very readable paper by Manin on the subject (probably, this is "Lectures on zeta functions and motives (according to Deninger and Kurokawa)", Astérisque 228 (4): 121–163). Probably the survey http://arxiv.org/abs/0909.0069 suggested at What is the field with one element? could also help.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .