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.

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