# Is the group ring of a finite cyclic group a “PID”?

Let $G=\langle\sigma\rangle$ be a group wirtten multiplicatively. Assume that $|G|=n$. The group ring of $G$ (denoted by $\mathbb{Z}[G]$) is defined as $$\mathbb{Z}[G]=\{\sum_{i=0}^{n-1}a_ig^i\text{ }|\text{ }a_i\in\mathbb{Z}\}$$

The sum of two elements of $\mathbb{Z}[G]$ is defined as follows: $$\sum_{i=0}^{n-1}a_ig^i+\sum_{i=0}^{n-1}b_ig^i=\sum_{i=0}^{n-1}(a_i+b_i)g^i$$

The product of two elements of $\mathbb{Z}[G]$ is defined as follows: $$(\sum_{i=0}^{n-1}a_ig^i)\cdot (\sum_{j=0}^{n-1}b_jg^j)=\sum_{i,j}(a_ib_j)g^{i+j}$$

Notice that $\mathbb{Z}[G]$ is not an integral domain: $$(1-\sigma)\cdot(1+\sigma+\dots+\sigma^{n-1})=0$$ Notice also that $\mathbb{Z}[G]$ is commutative.

My question:

Are all the ideals of $\mathbb{Z}[G]$ generated by a single element?

• If you factor $\mathbb Z [\mathbb Z/p]$ by cyclotomic polynomial in $\sigma$, you get $\mathbb Z [\zeta_p]$, which can be not a PID, but image of PI ring is PI. – xsnl Apr 7 '17 at 20:38
• This particular case is pretty easy because you get pretty simple extension presentation with familiar factor. In general integral group rings are horribly complicated objects, and AFAIK even for nilpotent p-groups your question is still open (in a sense that upper bound of ideal ranks are unknown). – xsnl Apr 7 '17 at 20:55

Suppose $n=2$.
Then $Z[G]$ is isomorphic to $Z[x]/(x^2-1)$.
But $Z[x]/(x^2-1)$ has some non-principal ideals$\,-\,$for example the ideal $(2,x-1)$.