# Group of units of a field

1. Nov 21, 2008

Given a field F, I know that if F is finite, then its group of units F* is cyclic. I'm trying to prove the converse: if F* is cyclic, then F is finite.

I have no idea where to start; I've tried a few things and they didn't get me anywhere. I know that if F is infinite and F* is cyclic, then F* is isomorphic to Z, but I can't figure out how that might form a contradiction.

2. Nov 21, 2008

Well, I have something that works provided char F ≠ 2.

Suppose that F is infinite and that F* is cyclic; then F* is isomorphic to Z. Let P be the prime subfield of F (i.e. the smallest subfield of F), so that P is isomorphic to Zp if char F = p is prime, and to Q if char F = 0. Then P* is a nontrivial subgroup of F* if char F ≠ 2, so is itself infinite and cyclic. But if char F = p > 2, then P* is isomorphic to Zp* which is finite, and if char F = 0, then P* is isomorphic to Q* which is not cyclic; in either case we get a contradiction.

3. Nov 21, 2008

### gel

and if char F = 2, then as F is infinite and can't contain the field with 4 elements, it must contain an element x transcendental over the prime subfield P, but P(x)* is not cyclic.

In fact, if F* is finitely generated then F is finite. Why restrict to cyclic?