# Which commutative groups are the group of units of some field?

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.)

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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: jlms.oxfordjournals.org/cgi/pdf_extract/s2-1/1/369 and iop.org/EJ/abstract/0025-5726/35/2/A05 – 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: mathoverflow.net/questions/204144/… – YCor Apr 30 '15 at 7:23

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:

http://math.uga.edu/~pete/Dicker1966.pdf

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

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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:

http://arxiv.org/pdf/0911.3537

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.

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