Consider F―p, the algebraic closure of Fp. I want to see that: for every proper...

maduregimc

maduregimc

Answered

2022-01-14

Consider Fp, the algebraic closure of Fp. I want to see that: for every proper subfield KFp,FpK is not a finite extension.
It is known that, and can be somewhat easily shown that Fp=n1Fpn.
Now, if any of the proper subfields have the form Fpn, it is easy enough to see that FpFpn(α1,,αm) for some αi by going high up enough, i.e, to some big enough m such that αiFpmFp
The problem is characterizing the proper subfields. Is every subfield of Fp going to have this form? Can we have an infinite intermediate subfield?

Answer & Explanation

Annie Levasseur

Annie Levasseur

Expert

2022-01-15Added 30 answers

Step 1
As Slade explained (+1) this follows either from the known structure of the automorphism group of Fp or from a theorem of Artin & Schreier stating that any finite extension KK, with the bigger field algebraically closed, is of the form K=K(1).
Step 2
An elementary argument can also be given. Assume that FpK is a finite extension. Then that extension is Galois. It is obviously normal, and it is also separable because every element of Fp belongs to a finite field, and hence is a zero of a separable polynomial over the ' field (and hence also over K). By basic Galois theory this implies that Fp has an automorphism of a finite order.
Daniel Cormack

Daniel Cormack

Expert

2022-01-16Added 34 answers

The Galois group here is isomorphic to G=Z^, the profinite integers. Since G is torsion-free, there are no finite index subfields.
This also follow from Artin-Schreier, though I don't know offhand of an extremely clean and short proof in the case of finite fields.

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