What is the algebraic closure of the field with one element? 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...
 A: 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).
A: 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.
