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What are the units in $R[X,X^{-1}]$, where $R$ is a commutative ring with $1$? I know that the question for polynomial rings is a standard textbook exercise. However, I couldn't find a reference for Laurent polynomials, since most people only seem to consider coefficients in an integral domain.

edited Jul 25 2010 at 5:33

Victor Protsak
7,751●1●30●44

asked Jul 8 2010 at 9:08

Seb
143●5

How does the standard proof for $R[X]$ go? We may be able to extend that proof, since $R[X,X^{-1}]$ is a localization of $R[X]$. – Jose Brox Jul 8 2010 at 10:44

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The units of $R[X]$ are $R^*+rad(R)(X)$ One direction is clear, and for the other one use the result for integral domains and that $rad(R)$ is the intersection of the prime ideals. Finding the units of $R[X,X^{-1}]$ is not so easy because we have to determine all divisors of $X^n$ in $R[X]$ ... I work on it. – Martin Brandenburg Jul 8 2010 at 10:50

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Look at Spec($R[x,1/x]) \rightarrow$ Spec($R$): fibers are Spec($k[x,1/x]$) with $k$ a field. So $\sum a_i x^i$ is a unit iff for all $p \in $ Spec($R$) some $a_i(p)$ is a unit & others vanish; i.e., some $a_i$ is a unit on Spec($R$) near $p$ and so others vanish near $p$. Thus, if $U_i$ is the open where $a_i$ is a unit then these are a separation of Spec($R$) and $a_j$ is nilpotent on $U_i$ for $j \ne i$. Answer: $R=\prod R_i$ with $a_i$ having unit $i$th component and nilpotent $j$th for $j \ne i$. So if Spec($R$) is connected and reduced then $R[x,1/x]=R^{\times} x^{\mathbb{Z}}$. – Boyarsky Jul 8 2010 at 11:33

Bill Dubuque · Accepted Answer · 2010-07-27 17:40:32Z UTC

You can find a more general result in the paper [1], which determines the units and nilpotents in arbitrary group rings $\rm R[G]$ where $\rm G$ is a unique-product group - which includes ordered groups. As the author remarks, his note was prompted by an earlier paper [2] which explicitly treats the Laurent case.

1 Erhard Neher. Invertible and Nilpotent Elements in the Group Algebra of a Unique Product Group
Acta Appl Math (2009) 108: 135-139
http://dx.doi.org/10.1007/s10440-008-9370-8
http://homepage.uibk.ac.at/~c70202/jordan/archive/note/note.pdf

2 Ottmar Loos. Remarks on Holger P. Petersson's "Idempotent 2-by-2 matrices" http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf

Seb · Answer 2 · 2010-07-08 15:04:48Z UTC

Thanks for your comments! A colleague just pointed me to an article by Karpilovsky ('On finite generation of unit groups of commutative group rings'). Translated into our setup, this gives the following result:

Take $r \in R^\times$, $a_i \in R$ nilpotent ($i \in \mathbb{Z}$), $k\ge 0$,$e_1,\dots,e_k \in R$ orthogonal idempotents that sum up to $1$, and fix $i_1,\dots,i_k \in \mathbb{Z}$. Then the element $$ r (1+\sum_i a_i X^i )(e_1X^{i_1} + \dots + e_kX^{i_k} ) $$ is a unit in $R[X,X^{-1}]$ and all units arise in this way.

In particular, all units are of the form $rX^j$ with $r \in R^\times$iff $R$ is reduced and connected, as Boyarsky pointed out.

I was hoping for a nice condition on the coefficients for the polynomial (as in the case $R[X]$). Maybe someone still sees how to simplify this statement or elegantly prove it in this setup?

What are the units in the ring of Laurent polynomials?

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