Skew Laurent Polynomial Ring. Let $R$ be a ring and $R[x^{\pm 1}]$ the Laurent Polynomial Ring. 


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*$R[x^{\pm 1}]$ is a domain since $R$ is. How to show this?


Let $R$ be a ring and $R[x^{\pm 1}]$ the Laurent Polynomial Ring. If further $x^{\pm 1}$ act on $R$ via an automorphism of rings $\kappa$, such that $xr=\kappa(r)x$. Then we call $R[x^{\pm 1}, \kappa]$ Skew Laurent Polynomial Ring.


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*Would you agree with this definition of Skew Laurent Polynomial Ring?

*How can we extend my first question to: Since $R$ is a domain, $R[x^{\pm 1}, \kappa]$ is domain too.

 A: You appear to be overlooking a key aspect of the definition. In general (see, for example, Goodearl and Warfield "An Introduction to Noncommutative Noetherian Rings, 2nd Edition"), a Skew Laurent extension of $R$ is a ring $T=R[x^{\pm 1},\kappa]$ satisfying:
(1) $R\subseteq T$;
(2) $x$ is an invertible element of $T$;
(3) $T$ is a free left $R$-module with basis $\{1,x,x^{-1},x^2,x^{-2},\ldots\}$;
(4) $xr=\kappa(r)x$ for all $r\in R$.
In particular you need (3) to prove your domain question. Hint: Look at the coefficients!
A: The standard argument for showing that $R[x]$ is a domain if $R$ is a domain is to look at leading coefficients of polynomials (the coefficient of the highest power of $x$.)
Notice that according to polynomial multiplication, the leading coefficent of $fg$ is the product of leading coefficients of $f$ and of $g$. This means that $fg$ has degree at least $deg(f)+deg(g)$, and so there's no way the product could be zero.
Now, what about the Laurent polynomials? 
Hint: everything in $R[x,x^{-1}]$ can be written as $x^ip(x)$ where $i$ is an integer and $p(x)$ is an element of $R[x]$. (Don't forget that $x$ is a unit in this ring.)

For skew polynomial rings, you can perform exactly the same sort of argument. While the leading coefficient is no longer just the product of leading coefficients, it is still nonzero, and so degrees of polynomials are still defined. Finally, $x$ is still a unit in the skew polynomial ring.
