Zero Divisors in the Group Ring Z[C2]

Detailed Solution

Write $x=e-g$ and $y=e+g$. Multiply using distributivity and the relations $e\cdot e=e$, $e\cdot g=g\cdot e=g$, and $g\cdot g=e$ (since $g^2=e$): $$ xy=(e-g)(e+g)=e\cdot e+e\cdot g-g\cdot e-g\cdot g. $$ Substituting the relations, $$ xy=e+g-g-e=0. $$ So $xy=0$ even though neither $x$ nor $y$ is the zero element of $\mathbb Z[C_2]$ (their coefficient vectors are $(1,-1)$ and $(1,1)$, both nonzero). This shows that $\mathbb Z[C_2]$ has genuine zero divisors, despite the fact that the coefficient ring $\mathbb Z$ itself is an integral domain. This is a standard illustration of why group rings of groups with torsion (elements of finite order greater than $1$, like $g$ here) typically fail to be integral domains: the relation $g^2=e$ forces $(e-g)(e+g)=e^2-g^2=e-e=0$, mirroring a difference-of-squares factorization of zero. Thus the correct option is $\boxed{\mathrm{A}}$.