What is the algebraic closure of the field with one element? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T16:04:34Z http://mathoverflow.net/feeds/question/6508 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6508/what-is-the-algebraic-closure-of-the-field-with-one-element What is the algebraic closure of the field with one element? Dror Speiser 2009-11-23T02:03:27Z 2013-05-14T19:30:00Z <p>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.</p> <p>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 $\mathbb Q/\mathbb Z$, $\mathbb Q[\mathbb Q/\mathbb Z]$, $\mathbb Z[\mathbb Q/\mathbb Z]$ ?</p> <p>What is the algebraic closure of the field with one element?</p> <p>And then, what is $\overline{\mathbb F_1} \otimes_{\mathbb F_1}\mathbb Z$? This seems like a very interesting question...</p> http://mathoverflow.net/questions/6508/what-is-the-algebraic-closure-of-the-field-with-one-element/6745#6745 Answer by Greg Kuperberg for What is the algebraic closure of the field with one element? Greg Kuperberg 2009-11-25T00:54:44Z 2009-11-25T00:54:44Z <p>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.</p> <p>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, <code>$\mathbb{C} = \mathbb{F}_1$</code> while <code>$\mathbb{R} = \mathbb{F}_{-1}$</code>.</p> <p>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 <code>$\mathbb{F}_q$</code> is then uniformly the Gaussian binomial coefficient <code>$\binom{n}{k}_q$</code>.</p> <p>In this interpretation, <code>$\mathbb{F}_1$</code> is algebraically closed. It is also a quadratic extension of <code>$\mathbb{F}_{-1}$</code>; the generalized cardinality squares, as it should.</p> http://mathoverflow.net/questions/6508/what-is-the-algebraic-closure-of-the-field-with-one-element/9897#9897 Answer by lieven lebruyn for What is the algebraic closure of the field with one element? lieven lebruyn 2009-12-27T21:03:42Z 2009-12-27T21:03:42Z <p>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].</p> <p>For a readable account (and for folklore references such as Kapranov-Smirnov) see Yu. I. Manin's "Cyclotomy and analytic geometry over F_1" (<a href="http://arxiv.org/abs/0809.1564" rel="nofollow">http://arxiv.org/abs/0809.1564</a>).</p>