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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?

flag	asked Sep 12 2011 at 3:46 Tim Mercure 243●1●4

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$k[X,1/X]$ where $k$ is the 2-element field. – Noam D. Elkies Sep 12 2011 at 3:53

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A necessary condition is of course that $-1 = 1$ – Fernando Muro Sep 12 2011 at 5:53

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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: maths.ed.ac.uk/~aar/papers/higman.pdf – Mark Grant Sep 12 2011 at 6:58

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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 www.ieja.net/papers/2011/V9/13-V9-2011.pdf – Jesse Elliott Sep 12 2011 at 8:30

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@Tim: Nice question :) – Chandrasekhar Sep 13 2011 at 14:05

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Salvatore Siciliano · Accepted Answer · 2011-09-12 10:20:52Z UTC

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].

Ring with Z as its group of units?

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