Proving the Converse: Finite Field with Cyclic Group of Units

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The discussion centers on proving that if the group of units F* of a field F is cyclic, then the field F must be finite. The user explores the implications of F being infinite and F* being cyclic, leading to contradictions based on the characteristics of the field. Specifically, they analyze cases where the characteristic of F is either a prime number greater than 2 or zero, demonstrating that these scenarios lead to non-cyclic subgroups, thus confirming that F must be finite if F* is cyclic. The conclusion emphasizes that if F* is finitely generated, then F is indeed finite.

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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.
 
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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.
 
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?
 

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