Skew Laurent polynomial ring ($A[t,t^{-1};\sigma ]$)

Adjoining an inverse of $t$ to the skew polynomial ring of an automorphism allows finite sums ${\sum }_{j\in \mathbb{Z}}{a}_j{t}^j$, with $ta=\sigma (a)t$. Every element can be moved into the polynomial subring by multiplying by a sufficiently large power of $t$. Consequently, for a right ideal $I$ of this ring, $I$ is generated by its intersection with the polynomial subring: if $f{t}^N$ lies there, then $f=(f{t}^N)t^{-N}$. This proves transfer of right Noetherianity from the polynomial subring, and multiplication on the left proves the left-handed version.

Let $A$ have no zero divisors and $\sigma$ be an automorphism. For nonzero Laurent polynomials $f=\sum a_i{x}^i$ and $g=\sum b_j{x}^j$ with greatest exponents $m,n$, the coefficient of $x^{m+n}$ in $fg$ is $a_m{\sigma }^m(b_n)$. No other exponent pair contributes to that highest degree. Both factors are nonzero, so their product is nonzero, proving the noncommutative domain property in every characteristic.

For the Integer Heisenberg group, choose $x=I+E_{12}$, $y=I+E_{23}$ and central $z=I+E_{13}$. Every element has a unique form $y^c{z}^b{x}^a$, and $xy=zyx$. The group algebra is therefore the displayed skew Laurent polynomial ring, with $\sigma (z)=z$. The base is a commutative Laurent polynomial ring and an integral domain; the skew Hilbert basis theorem gives both-sided Noetherianity and the leading coefficient of a product proves that there are no zero divisors.