Group ring over a cyclic group

Question

We have the group ring $R= \mathbb{Z} C_5$ over the cyclic group of order 5, generated by the element $g$.
Find an $a,b \in R$ such that $ab=0$ with $a,b \neq 0$

Since 0 is not in the cyclic group, how do you find two elements who's product is zero?

Attempt: An element $r∈R$ can be written as $r=z_0e+z_1g+z_2g^2+z_3g^3+z_4g^4$ for $z_i \in\mathbb{Z}.$
So $ab = (a_0 b_0 +a_1b_4+a_4b_1+a_2b_3+a_3b_2)e+ (z_1b_0+a_0b_1+a_2b_4+a_4b_2+2a_3b_3)g+(a_0b_2+a_2b_0+a_2b_4+a_4b_2+a_3b_4+a_4b_3)g^2+(...)g^3 +(...)g^4$
which terms would then go to zero?

Answer 1

The hint in the comments is a good one that works for cyclic groups, but there is actually one that works simply for all finite groups (or, for that matter, any group with a nontrivial finite nonidentity subgroup.)

Hint:

If $H$ is a finite subgroup of $G$ with more than one element, show that in the group ring $R[G]$

1. $e=\sum_{h\in H}h$ satisfies $e^2=e$,
2. Therefore $e(1-e)=0$
3. and neither $e$ nor $1-e$ is $0$.