# Rational numbers as an extension of the field with one element?

Greetings. I would love to have a field $\mathbb F$ which is a subfield of the field of rational numbers $\mathbb Q$, and such that the Galois group $Gal (\mathbb Q / \mathbb F)$ has preferably infinitely many elements.

While there is no such field $\mathbb F$, since $\mathbb Q$ has no proper subfields at all, I've recently heard of this field $\mathbb F_1$ with one element concept.

As far as I understand there is no definition which would be set in stone for this object, at least not yet. My question to those who know the subject: does any of the currently studied definitions of $\mathbb F_1$ allow for realization of $\mathbb F_1$ as a "subfield" of $\mathbb Q$ in some sense?

• Thanks for the kind words. I'd just like to emphasize (not to suggest that you meant or said otherwise) that these analogies ($\mathrm{Gal}(\bar{\mathbf{Q}}/\mathbf{F}_1)$ and so on) should be taken with a grain of salt. For instance, it is good to think of the functor $\mathbf{Q}\otimes -$ from $\Lambda$-rings to $\mathbf{Q}$-algebras as the functor $\mathbf{Q}\otimes_{\mathbf{F}_1} -$, but this functor does not have the descent property, which is what Galois theory is really used for. So while it is reasonable to call $\mathbf{Q}$ an $\mathbf{F}_1$-algebra, it's a bit abusive to speak of... – JBorger Mar 24 '10 at 1:41
• the relative Galois theory. This is in contrast to usual field theory, where for any extension $L/K$, the functor $L\otimes_K -$ from $K$-algebras to $L$-algebras always has the descent property; but it is similar to general ring theory where, for instance, the functor $\mathbf{Q}\otimes_{\mathbf{Z}}-$ does not have the descent property. The moral of the story is that, from the $\Lambda$-ring point of view, $\mathbf{F}_1$ fails to have lots of properties that usual fields (as opposed to rings) have. So it's a bit safer to think of this $\mathbf{F}_1$ as a generalized ring, and so... – JBorger Mar 24 '10 at 1:46