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...



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


