1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Group of units of a field

  1. Nov 21, 2008 #1
    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. jcsd
  3. Nov 21, 2008 #2
    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.
     
  4. Nov 21, 2008 #3

    gel

    User Avatar

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Group of units of a field
  1. Groups, Rings, Fields? (Replies: 3)

Loading...