12
$\begingroup$

As far as I know there is many "suggestions" of what should be a "field with one element" $\mathbf{F}_{1}$.

My question is the following:

How we should think or what should be the "absolute Galois group" of the "hypothetical" $\mathbf{F}_{1}$ ?

$\endgroup$
  • 1
    $\begingroup$ See answers to this question: mathoverflow.net/q/6508/2024. $\endgroup$ – Dror Speiser Dec 28 '16 at 20:09
  • 1
    $\begingroup$ Class field theory suggests ${\mathbb R}$ (with addition). $\endgroup$ – Felipe Voloch Dec 28 '16 at 22:29
  • $\begingroup$ @FelipeVoloch Could you develop your suggestion please (comment or answer) $\endgroup$ – Muhammed Ali Dec 29 '16 at 9:03
10
$\begingroup$

The Galois group of the maximal abelian extension of $\mathbb Q$ (or any number field) is given (class field theory) as the quotient of the idele class group by the connected component of the identity which is isomorphic to $\mathbb R$. If there is an ${\mathbb F}_1$ then its extensions provide "constant field extensions" of $\mathbb Q$. So, to be compatible with class field theory and keep the analogy with function fields, $\mathbb R$ must be at least the abelianization of the absolute Galois group of ${\mathbb F}_1$. But finite fields are abelian, so this suggests the answer. A slightly different perspective is that Weil constructed canonically and functorially an extension of the absolute Galois group of any number field by ${\mathbb R}$ (the Weil group) and that should be the extension one would get by allowing constant field extensions again.

$\endgroup$
  • $\begingroup$ $\mathbb F_1$, whatever it is or isn't, is weird. $\endgroup$ – Wojowu Dec 29 '16 at 12:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.