As the title suggests, I'm interested whether $\mathbb{Z}[\mathbb{Z}/(p)]$ a PID or not. Assume $p$ is prime.
My feeling is that it is a PID, since $\mathbb{Z}/(p)$ is cyclic an morally if an ideal is generated by elements of $\mathbb{Z}/(p)$, it's enough to consider the element whose order is the GCD of the orders. But I'm unable to find a (clever) way to turn the general case into something like the easy case above.
$\mathbb Z[C_p]$ is not even a domain.
Take for instance $p=2$. Then there is an element $u\in C_2$ such that $u^2=1$ but $u\ne \pm1$.
More generally, if $G$ has an element $u$ of finite order, then $\mathbb Z[G]$ is not a domain because $(u-1)(u^{n-1}+\cdots+u+1)=u^n-1=0$. In particular, $\mathbb Z[G]$ is never a domain if $G$ is finite.
The problem of when a group ring has non-trivial zero divisors is an open problem; see wikipedia.