Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

Imo the best theory today for the field with one element is Borger's proposal to consider Lambda-rings and use their Lambda-structure as a substitute for descent from the integers (or rationals) to the field F1 with one element.

Some examples of this philosophy are contained in the nice short paper by Borger and Bart de Smit (arXiv:0801.2352) 'Galois theory and integral models of Lambda-rings'.

Lambda-rings finite etale over the rationals Q are finite discrete sets equipped with a continuous action of the monoid Gal(Qbar/Q) x N' where N' are the positive integers under multiplication. This suggest that the Galois monoid Gal(Qbar/F1) = Gal(Qbar/Q) x N'.

Likewise, Lambda-rings over Q having an integral Lambda-model correspond to finite sets with a continuous action of the monoid Zhat, that is the set of profinite integers as a topological monoid under multiplication. This suggests that the absolute Galois monoid of F1, that is Gal(F1bar/F1) = Zhat.

share|improve this answer
4  
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
2  
at times field-specific concepts in descent theory do not translate so well. In retrospect, this is not so surprising, because the importance of fields, as opposed to rings, is mostly of a psychological or historical nature (IMHO). –  JBorger Mar 24 '10 at 1:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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