Groupring has zero divisors.

Let R be a ring, G a finite abelian group, RG the groupring of R and G, then RG has zero divisors.

Proof:
Let g be an element of G with order m. Now calculate:
(1-g)*(1+g+g^2+...+g^(m-1) =
( 1+g+g^2+...+g^(m-1) ) - ( g+g^2+...+g^m ) =
1 - g^m = 1 - 1 = 0.

(1-g) is a zero divisor.