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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?

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  • $\begingroup$ 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. $\endgroup$ – xsnl Apr 7 '17 at 20:38
  • $\begingroup$ 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). $\endgroup$ – xsnl Apr 7 '17 at 20:55
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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 $(2,x-1)$.

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