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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.

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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 '10 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 '10 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 '10 at 11:33

2 Answers 2

up vote 1 down vote accepted

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

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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?

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That's exactly a concrete description of my comment/answer even in the general case (no connectedness or reducedness hypotheses): those idempotents correspond to a finite decomposition of $R$ into factor rings, and your unit in the $j$th factor ring of $R[X,1/X]$ is the product of the $j$th component of $r$ times $X^{i_j}$ times something which is 1 mod nilpotents. I think my proof in terms of fibering over Spec($R$) is elegant enough. :) –  Boyarsky Jul 8 '10 at 15:23
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Um, thank you very much for the explanation! I have to admit I did not understand how to interpret your answer before. –  Seb Jul 8 '10 at 15:36
    
Thanks for the reference to that article, it may be useful for me! (btw, I think you should accept your own answer!) –  Jose Brox Jul 8 '10 at 22:48

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