MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Inspired by a recent question on the multiplicative group of fields. Necessary conditions include that there are at most $n$ solutions to $x^n = 1$ in such a group and that any finite subgroup is cyclic. Is this sufficient? (Edit: Well, no, it's not, since the only such groups which are finite are the cyclic groups of order one less than a prime power. Hmm.)

share|cite|improve this question
Regarding your edit: Yes, the finite case is known, so your question is about the infinite case, right? – Jonas Meyer Jan 26 '10 at 6:37
Perhaps not what you're looking for (and I do not really know their contents) but the following look interesting: and – Jonas Meyer Jan 26 '10 at 7:04
In the case of number fields or function fields for instance, units would mean the units in the ring of integers. – Anweshi Jan 26 '10 at 21:02
I don't think I've ever heard anyone use "units" in that way. – Qiaochu Yuan Jan 27 '10 at 3:27
A (closed) duplicate with further answers can be found here:… – YCor Apr 30 '15 at 7:23
up vote 17 down vote accepted

The following paper claims an answer to this question:

Dicker, R. M. A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field. Proc. London Math. Soc. (3) 18 1968 114--124.

You can find it here:

One might hope for a more aesthetically appealing characterization. I don't know if such a thing has ever been given.

share|cite|improve this answer
Thanks! Maybe I should have just asked for nice necessary conditions... – Qiaochu Yuan Jan 26 '10 at 14:46
@qiaochu: I wonder why you are satisfied with this answer, because I already mentioned this paper in the other question which you refer to. – Martin Brandenburg Jan 27 '10 at 22:29
Oops; I don't seem to have read that question too closely. Sorry! – Qiaochu Yuan Feb 1 '10 at 4:07
Dear Pete: it seems that the link is broken. Do you still have the article? I would love to see it. – René Apr 28 '15 at 22:00

Another characterization is theorem 2.1 in this paper on the field with one element:

If H is a commutative group, let H+ be H together with a new element 0. To give a field structure on H+ is equivalent to giving a bijection s:H+ --> H+ that commutes with all of its conjugates-by-H.

Maybe this is similar to Dicker's characterization? Dicker mentions the operation x --> 1-x, while the s in Connes-Consani is meant to be x --> x + 1.

share|cite|improve this answer

Your Answer


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.