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If doing geometry over $\mathbb F_p$ means also using its algebraic closure, it must be interesting to talk about the algebraic closure of $\mathbb F_1$ - the field with one element.

I saw that the finite extensions of $\mathbb F_1$ are considered as $\mu_n$, but an article by Connes et al says that it is unjustified to think of the direct limit of these. In their paper, the group ring $\mathbb Q[\mathbb Q/\mathbb Z]$ appears a lot. Maybe it's one of $\mathbb Q/\mathbb Z$, $\mathbb Q[\mathbb Q/\mathbb Z]$, $\mathbb Z[\mathbb Q/\mathbb Z]$ ?

What is the algebraic closure of the field with one element?

And then, what is $\overline{\mathbb F_1} \otimes_{\mathbb F_1}\mathbb Z$? This seems like a very interesting question...

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    $\begingroup$ Surely the correct question is: What is a "scheme over the algebraic closure of F_1"? The answer will no doubt depend on which notion of "scheme over F_1" you have in mind. $\endgroup$
    – JSE
    Commented Nov 23, 2009 at 2:29
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    $\begingroup$ You have to remember that the field with one element is not a classical "field", and notions like algebraic closure may not have much meaning without formalizing and abstracting the important properties to a category-theoretic setting. $\endgroup$
    – Harry Gindi
    Commented Nov 23, 2009 at 4:13

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There have been several questions on mathoverflow about the field with one element. Of course, such a field doesn't really exist and the discussion must fray sooner or later. So here is a different kind of answer.

Besides finite fields, which are 0-manifolds, there are only two fields which are manifolds, $\mathbb{C}$ and $\mathbb{R}$. There is a generalization of cardinality for manifolds and similar spaces, namely the geometric Euler characteristic. (This is as opposed homotopy-theoretic Euler characteristic; they are equal for compact spaces.) The geometric Euler characteristic of $\mathbb{C}$ is 1, while the geometric Euler characteristic of $\mathbb{R}$ is -1. In this sense, $\mathbb{C} = \mathbb{F}_1$ while $\mathbb{R} = \mathbb{F}_{-1}$.

It works well for some of the motivating examples of the fictitious field with one element. For instance, the Euler characteristic of the Grassmannian $\text{Gr}(k,n)$ over $\mathbb{F}_q$ is then uniformly the Gaussian binomial coefficient $\binom{n}{k}_q$.

In this interpretation, $\mathbb{F}_1$ is algebraically closed. It is also a quadratic extension of $\mathbb{F}_{-1}$; the generalized cardinality squares, as it should.

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  • $\begingroup$ Thanks! This is real food for thought. Can you think of other structures than 0-manifolds, that include, say, C and R, and have an invariant like the geometric Euler characteristic, that make F_1, in a sense, algebraically closed, or maybe not? $\endgroup$
    – Dror Speiser
    Commented Nov 25, 2009 at 2:10
  • $\begingroup$ I am not expert in the topic and I don't entirely understand your question either. What I can say is that this answer was historically part of the motivation of the Weil conjectures. Further generalizations would be tied up with cohomology theories (like etale or motivic cohomology) in the setting of algebraic geometry. $\endgroup$
    – Greg Kuperberg
    Commented Nov 25, 2009 at 2:32
  • $\begingroup$ What is meant by geometric Euler characteristic? Also, what is meant by your statement that homotopy-theoretic Euler chacteristics are "equal for compact spaces"? (Maybe I am not parsing your sentences correctly.) $\endgroup$
    – Kevin H. Lin
    Commented Dec 27, 2009 at 21:13
  • $\begingroup$ I agree that I was speaking a little loosely. I am saying that there is only one reasonable choice for the Euler characteristic of a space homeomorphic to a finite simplicial complex, the alternating sum of the number of simplices or the Betti numbers. But for a non-compact space that is sufficiently tame, you can extend Euler characteristic as a valuation, which is what I called "geometric" Euler characteristic. This differs here from the alternating sum of Betti numbers, which is "homotopy theoretic". en.wikipedia.org/wiki/Valuation_%28measure_theory%29 $\endgroup$
    – Greg Kuperberg
    Commented Dec 27, 2009 at 21:45
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The algebraic closure of F_1 is the group of all roots of unity, and, tensoring it with Z gives the integral group ring Z[mu_infty], or, if you prefer Z[Q/Z].

For a readable account (and for folklore references such as Kapranov-Smirnov) see Yu. I. Manin's "Cyclotomy and analytic geometry over F_1" (http://arxiv.org/abs/0809.1564).

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