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Is there a ring with $\mathbb{Z}$ as its group of units?

More generally, does anyone know of a sufficient condition for a group to be the group of units for some ring?

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$k[X,1/X]$ where $k$ is the 2-element field. – Noam D. Elkies Sep 12 '11 at 3:53
A necessary condition is of course that $-1 = 1$ – Fernando Muro Sep 12 '11 at 5:53
The group $G$ is always contained in the group of units of the group ring $R[G]$, when $R$ is a commutative ring with unit. I don't know precise conditions for when they are equal, but here's a reference: – Mark Grant Sep 12 '11 at 6:58
I was thinking also to look at the group ring $\mathbb{F}_2(G)$; but the units of this ring are strictly more than just the elements of $G$, for example, if $G$ is a finite $2$-group of order greater than $2$. See the second paragraph of – Jesse Elliott Sep 12 '11 at 8:30
Since the last part of the question has been asked again in a slightly different way, I thought I would add a comment that an example of a group which is not the group of units of any ring is the cyclic group of order 5. This is a nice exercise. – Tobias Kildetoft Oct 19 '11 at 8:23

1 Answer 1

up vote 16 down vote accepted

The example provided by Noam answers the first question. The second question is very old and, indeed, too general. See e.g. the notes to Chapter XVIII (page 324) of the book "László Fuchs: Pure and applied mathematics, Volume 2; Volume 36". In particular, rings with cyclic groups of units have been studied by RW Gilmer [Finite rings having a cyclic multiplicative group of units, Amer. J. Math 85 (1963), 447-452], by K. E. Eldridge, I. Fischer [D.C.C. rings with a cyclic group of units, Duke Math. J. 34 (1967), 243-248] and by KR Pearson and JE Schneider [J. Algebra 16 (1970) 243-251].

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