Wikipedia says: "The Laurent polynomial ring $R[X, X^{−1}]$ is isomorphic to the group ring of the group $\mathbb{Z}$ of integers over $R$". Can anyone offer a proof? I also don't fully understand what a group ring is, even after reading the wiki article. Further, I don't know what the multiplicative structure of the Laurent polynomials would be, as it seems that their addition is the traditional multiplication in this isomorphism.
$\endgroup$2 Answers
$\begingroup$A group ring of a ring $R$ over a group $G$ could be loosely described this way: "It's a ring that you additively generate with $R$ multiples of $G$, and then you use $G$'s multiplication to define a multiplication."
I encourage you to seek out basic exercises on group rings to develop this picture in your mind.
In this situation, we are looking for a copy of $\Bbb Z$ inside of $R[x]$. There is one obvious candidate, the powers of $x$: $\{x^i\mid i\in \Bbb Z\}$. It's easy to see that this set is group homomorphic to $\Bbb Z$ under the map $f:n\mapsto x^n$. After that, we can spread the map to every other element of the group ring by defining $f(\sum r_n n)=\sum r_nx^n$. You can then verify that this is a ring homomorphism of $R[\Bbb Z]$ with $R[x,x^{-1}]$.
This "extending" of the map is very important. This is usually stated for commutative $R$ where the statement becomes a lot simpler. Let $R$ be a commutative ring, $G$ be a group, and let $A$ be an $R$ algebra. It just says: If $f:G\to A$ is a group homomorphism of $G$ into an $R$ algebra $A$, then $f$ extends uniquely to an algebra homomorphism from $R[G]$ to $A$.
I don't recall ever seeing the statement written for noncommutative rings, but I believe this is the translation. Let $R$ be a ring, $G$ be a group, and let $A$ be another ring which is an $(R,R)$ bimodule. Suppose further that $f$ is a group homomorphism of $G$ into $A$ such that $rf(g)=f(g)r$ for every $g\in G$, $r\in R$. Then $f$ extends uniquely to a ring homomorphism from $R[G]\to A$ that is also $R$ linear on both sides.
Looking back out the map I suggested, you can see that in $R[x]$, we do assume that $rx^i=x^ir$ for all $r\in R$, $i\in \Bbb Z$, so it is satisfying the condition that $rf(g)=f(g)r$ for all $r\in R$, $g\in \Bbb Z$.
$\endgroup$ 1 $\begingroup$In a group ring each element of your group becomes a basis element. Usually one writes each element as a formal exponential to emphasize you're working in the group ring instead of the group itself. So given an integer $n$, one writes $e^n$ for the corresponding basis element. Then the group operation is transported over to the group ring, so $m+n$ becomes $e^me^n=e^{m+n}$. Notice that this is exactly how one multiplies powers of $x$ -- that is -- $x^mx^n=x^{m+n}$.
So send $e^m$ to $x^m$ for all integers $m$ and extend linearly and you've got an isomorphism between the group ring of the integers and Laurent polynomials.
As for the operations in the ring of Laurent polynomials, they are essentially the same as that of regular polynomials. You've just got negative powers to deal with as well as nonnegative powers.
Edit: Extending linearly? Once you've decided to send $e^n$ to $x^n$, you know where everything else needs to go.
An arbitrary element of $R[\mathbb{Z}]$ looks like $r_1e^{n_1}+r_2e^{n_2}+\cdots r_\ell e^{n_\ell}$ for some choice of $r_1,\dots,r_\ell \in R$ and $n_1,\dots,n_\ell \in \mathbb{Z}$. If we extend our map $e^n \mapsto x^n$ linearly, we send $r_1e^{n_1}+r_2e^{n_2}+\cdots r_\ell e^{n_\ell} \mapsto r_1x^{n_1}+r_2x^{n_2}+\cdots r_\ell x^{n_\ell}$ (linearly combinations in the "old basis" go to linear combinations in the "new basis").
$\endgroup$ 1
