Show that $K [\Bbb Z]$ is an integral domain.

Question

Let $K$ be a field. Then show that the group ring $K [\Bbb Z]$ is an integral domain.

How do I show that $K[\Bbb Z]$ contains no divisor of zero? Any help will be highly appreciated.

Thanks in advance.

Answer 1

Notice that $K[\mathbb{Z}]$ is isomorphic (as $K$-algebra) to the Laurent polynomial ring $K[X,X^{-1}]$ (identify $n\in\mathbb{Z}$ with $X^n$), which is an integral domain, being the localization of $K[X]$, an integral domain, in the multiplicative set consisting of the non-negative powers of X.

This is actually an elementary instance of the complicated Kaplansky's conjecture, stated more in general when $\mathbb{Z}$ is replaced by a torsion-free group $G$. You can look at https://mathoverflow.net/q/79559/158845 to have an idea about the state of art and the techniques implied.

EDIT: to avoid the use of localization, one can just observe, as precisely pointed out by @reuns in a comment, that $K[X,X^{-1}]$ is a subring of $K(X)$, the field of rational functions in $X$ over $K$. $K(X)$ is just the field of fractions of the integral domain $K[X]$ (have you heard about "field of fractions"? I can go more in detail about this easy construction).