ag.algebraic geometry - 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.