# Is $\mathbb{Z}[\mathbb{Z}/(p)]$ a PID?

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.

• If you mod out the ideal generated by $1-\overline{1}$ then you get a ring isomorphic to $\Bbb{Z}(\zeta_p)$, $\zeta_p=e^{2\pi i/p}$ (the isomorphism maps $\overline{k}$ to $\zeta_p^k$). This is the ring of integers $\Bbb{Q}(\zeta_p)$. That is probably known to have class number $h=1$ for many a choice of $p$, but it's complicated. May even be the case that the answer to that question is not known fully? – Jyrki Lahtonen Mar 10 '16 at 17:42
• Äsh. I meant to mod out the ideal generated by the sum $s=1+u+u^2+\cdots+u^{p-1}\in \Bbb{Z}[C_p]$ (or, if you insist on additive notation, $s=\sum_{k=0}^{p-1}1\cdot\overline{k}.$) – Jyrki Lahtonen Mar 10 '16 at 18:09
• @JyrkiLahtonen I was about to ask you what was that notation, but I found that relation on some old note of mine and therefore I got what you meant – Luigi M Mar 10 '16 at 20:25

$\mathbb Z[C_p]$ 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.