Laurent Polynomials

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Let $R$ be a commutative ring with identity. Is there any characterization for invertible elements of $R[x,x^{-1}]$ ?

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3 Answers

I have read about this somewhere. I think it was as follows: $\sum_{i=-n}^n a_ix^i \in R[x,x^{-1}]$ is invertible iff $\sum a_i^2$ is invertible in $R$ and for all $i \not = j$, $a_ia_j$ is nilpotent.

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And there is a very nice proof. Hints: (1) The Laurent polynomial $\sum_i a_i x^i$ is invertible if and only if the product $\left(\sum_i a_i x^i\right)\cdot \left(\sum_i a_i x^{-i}\right)$ is invertible. (2) The inverse of an invertible Laurent polynomial which lies in the subring $R\left[x+x^{-1}\right]$ must itself lie in this subring $R\left[x+x^{-1}\right]$. (3) The subring $R\left[x+x^{-1}\right]$ is isomorphic to the polynomial ring $R\left[y\right]$. (4) A polynomial over a commutative ring is invertible if and only if its constant term is invertible and all its other terms nilpotent. – darij grinberg Jan 3 at 16:26
I just wanted to post the same constructive proof (which I learned here: math.stackexchange.com/questions/147661/…) – Martin Brandenburg Jan 3 at 16:42
Actually I'm not sure about it anymore, because in the "$\sum_i a_i x^i$ invertible $\Longrightarrow$ ..." direction, it only shows that $\sum_{i-j=k} a_ia_j$ is nilpotent for every $k\neq 0$, but not that the individual $a_ia_j$ are nilpotent for $i\neq j$. But something tells me this can be fixed. – darij grinberg Jan 3 at 16:43
Ok, it can be fixed. We need to prove that $a_ia_j$ is nilpotent for every $i$ and $j$ with $i-j=k$, for every $k\neq 0$. We go by descending induction over $k$, knowing that this is tautological for $k$ large enough (since Laurent polynomials have only finitely many coefficients). Now the final hint: (5) If finitely many elements of a commutative ring are given such that the pairwise products of these elements are nilpotent ("pairwise" means "pairs of two different ones" here) and the sum of these elements is nilpotent, then each of these elements is nilpotent. – darij grinberg Jan 3 at 16:54
But $1\cdot 1$ isn't nilpotent either. – darij grinberg Jan 3 at 19:57

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Thinking geometrically in terms of the map ${\rm{Spec}}(R[x,1/x]) \rightarrow {\rm{Spec}}(R)$ and noting that being a unit amounts to being nonzero in the residue field at every prime, an element $f = \sum a_i x^i \in R[x,1/x]$ is a unit if and only if it has unit restriction to every fiber, which is to say that for every prime ideal $P$ of $R$ (with residue field $k(P)$) the image $f(P) := \sum a_i(P) x^i$ in $k(P)[x,1/x]$ is a unit. But since $k(P)$ is a field, this latter condition is exactly that $f(P)$ is a $k(P)^{\times}$-multiple of a power of $x$.

That is, there is exactly one $i$ (depending perhaps on $P$) such that $a_i(P) \ne 0$, which can be equivalently expressed as the condition that $a_i(P)a_j(P) = 0$ in $k(P)$ for all $i \ne j$ and $\sum a_i(P) \ne 0$ in $k(P)$. Varying over all $P$, this necessary and sufficient condition says exactly that (1) $a_i a_j$ is nilpotent in $R$ when $i \ne j$ and (2) $\sum a_i \in R^{\times}$.

In the presence of (1), squaring the sum in (2) (which has no effect on whether or not it is a unit) and noting that adding a nilpotent element has no effect on being a unit shows that (2) can be replaced with (2') $\sum a_i^2 \in R^{\times}$ (thereby recovering the formulation in shatich's answer).

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Invertible elements of Laurent algebras, and more generally of algebras of torsionfree, cancellable commutative monoids, are characterised in Theorem 11.3 and Corollary 11.4 of Gilmer's Commutative Semigroup Rings (Chicago Lectures in Mathematics, 1984). The proofs given there are quite accessible.

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