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I would like to understand at least one of the several existing approaches to algebraic geometry over $\mathbb{F}_1$ (the field with one element). Is there an example of an "interesting" theorem that can be formulated purely in the language of ordinary schemes, but which can be proved using algebraic geometry over $\mathbb{F}_1$?

Of course, the interpretation of the word "interesting" is entirely up to your own taste. An example in which the theorem cannot be proved using "classical" methods would be most desirable, but examples where (one of) the theories of schemes over $\mathbb{F}_1$ gives an alternative proof of an already known result would also be very much appreciated.

[On a related note, perhaps there should be a tag "naive-question" for situations like this one.]

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4 Answers 4

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I'm confident that the answer to the original question is no. There are hardly any theorems at all in the subject, much less ones with external applications! In other words, if no further progress is ever made in any of the directions people have pursued, everything will likely be forgotten (which would not make it so unusual an area). What attracts people to these things is not a track record of existing applications but the possibility of exciting future ones. So investing time in the subject is something of a gamble---it might pay off if you're good at divining the future (or if you have insider information), or you might end up wasting a lot of time.

I don't mean to be too pessimistic. I for one have high hopes for certain directions (!), but I think it's best to see clearly what you'd be getting into.

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    $\begingroup$ Geez, Jim, and you tell us that now!? ;-P $\endgroup$
    – javier
    Commented May 4, 2010 at 13:03
  • $\begingroup$ Well, which directions do you have high hopes for? $\endgroup$
    – Kevin H. Lin
    Commented May 4, 2010 at 14:07
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    $\begingroup$ Jim is pursuing an approach via $\Lambda$-rings, Witt rings, and related techniques, which you can learn more about by looking at his arxiv publications, for example. It is less formal than some other aprpoaches, which is one of its merits. $\endgroup$
    – Emerton
    Commented May 4, 2010 at 15:38
  • $\begingroup$ Thanks Jim! This is exactly the type of answer I was looking for. I was $hoping$ for something more optimistic, but I won't blame reality for being what it is. $\endgroup$
    – senti_today
    Commented May 4, 2010 at 16:26
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Connes et al. motivate their study of geometry over $\mathbb F_1$ in part by an idea that it may help to prove Riemann's Hypothesis. Some people, such as Mochizuki I think, hope that this study may help to prove the $abc$ conjecture.

These are all potential, not yet real, applications. But hey, if one of these works...

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    $\begingroup$ A small elaboration: as far as the abc conjecture goes, the hope as I understand it is that a good enough theory of F_1 will be equipped with a good enough theory of differentiation with F_1 as a base field that one can somehow imitate the proof of the Mason-Stothers theorem. $\endgroup$
    – Qiaochu Yuan
    Commented May 4, 2010 at 16:59
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To elaborate on the possible connection with the Riemann hypothesis:

One of the parts of the Weil conjectures (which were proved by Deligne, and which also follow from Grothendieck's "standard conjectures") states:

If $X$ is a smooth projective variety of dimension $n$ over $\mathbb{F}_q$, then its zeta function $Z_X(t)$ has the form $$\frac{P_1(t) P_3(t) \cdots P_{2n-1}(t)}{P_0(t)P_2(t) \cdots P_{2n}(t)},$$ where each $P_i(t)$ has the form $\prod_j (1-\alpha_{ij}t)$, and such that each $\alpha_{ij}$ is an algebraic integer such that $|\alpha_{ij}| = q^{i/2}$.

The zeta function $Z_X(t)$ of a variety $X$ is defined to be $\exp(\sum N_r t^r/r)$, where $N_r$ is the number of $\mathbb{F}_{q^r}$-points of $X$.

Write $\zeta_X(s) = Z_X(q^{-s})$. Then the above statement implies that the zeros and poles of $\zeta_X(s)$ lie on the lines given by $\mathfrak{Re}(s) = i/2$ with $i=1, \dots, 2n$. The function $\zeta_X(s)$ is also equal to $$\prod_{x \in X} \frac{1}{1-|k(x)|^{-s}}$$ where $|k(x)|$ is the order of the residue field of $x$ (which is always finite).

Now if $X = \operatorname{Spec} \mathbb{Z}$ were somehow a smooth projective variety of dimension 1 over $\operatorname{Spec} \mathbb{F}_1$, then $\zeta_X(s)$ would be the Riemann zeta function, and...

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  • $\begingroup$ Nice explanation! Some of the main problems that one faces at this point are, ignoring for a moment the wild jungle of different approaches, that no matter what your definition of $\mathbb{F}_1$ variety is, $\text{Spec}\mathbb{Z}$ cannot be even of finite type, and moreover nobody has yet come with a nice idea of what on earth a "smooth" $\mathbb{F}_1$ variety is. $\endgroup$
    – javier
    Commented May 4, 2010 at 13:56
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    $\begingroup$ @javier Could you possibly elaborate on why we know that $\mathrm{Spec}\mathbb{Z}$ cannot be of finite type, no matter what our definition of $\mathbb{F}_1$ is? Sorry if this is an overly simple question. $\endgroup$
    – Tim
    Commented Mar 1, 2016 at 1:50
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I'm not aware of such a theorem, but you should take a look into absolute zeta-functions, which hope to prove a lot about zeta-functions — perhaps up to Riemann's Hypothesis.

Since I don't know much, I better refer you to another MO question:

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