Let $G$ be the group $\mathbb{Z}^2\rtimes_{\sigma} \mathbb{Z}=\langle y,z\rangle\rtimes_{\sigma}\langle x\rangle$, where $\sigma(x)=\begin{pmatrix}a, b\\c,d\end{pmatrix}\in SL_2(\mathbb{Z})$, which means that we have relations $xyx^{-1}=y^az^c, xzx^{-1}=y^bz^d$. Then we can form the group ring $R=\mathbb{Z}G$, note that since $G$ is a polycyclic-by-finite group, $R$ is a left Noetherian ring.

I am interested in what a general prime ideal $\wp$ in $R$ looks like, so I want to ask the following question:

What does Spec(R) generally look like? Especially when $\sigma(x)=\begin{pmatrix}1,0\\1,1\end{pmatrix}$ or $ ~ \sigma(x)=\begin{pmatrix}2,1\\1,1\end{pmatrix}$

  • 2
    $\begingroup$ Do people really write $Spec(R)$ for noncommutative $R$? $\endgroup$ Commented Aug 29, 2013 at 21:49
  • 2
    $\begingroup$ I am not sure, you can edit it if some better notation applies. $\endgroup$
    – Jiang
    Commented Aug 29, 2013 at 22:22
  • $\begingroup$ @F.M.: In non-commutative geometry, I'd guess they do. $\endgroup$
    – Marc Palm
    Commented Aug 30, 2013 at 8:11
  • $\begingroup$ You guess? - 5 more - $\endgroup$ Commented Aug 30, 2013 at 16:22

1 Answer 1


The natural map $\mathbb{Z} \to \mathbb{Z}G$ has central image and therefore induces a map between prime spectra $Spec(\mathbb{Z}G) \to Spec(\mathbb{Z})$. The preimage of the ideal generated by $(p)$ under this map is in a natural bijection with $Spec( kG )$ where $k = \mathbb{F}_p$ if $p$ is a prime number and $k = \mathbb{Q}$ if $p = 0$. So $Spec( \mathbb{Z}G )$ is the disjoint union of various prime spectra $Spec( kG )$, where $k$ is a field.

A prime $P$ of $kG$ is said to be faithful if $G$ embeds into the group of units of $kG/P$, or equivalently, if $G \cap (1 + P) = 1$.

If a prime of $kG$ is unfaithful, then $P^\dagger := G \cap (1 + P)$ is a non-trivial normal subgroup of $G$, and $P$ is completely determined by its image inside $k[G / P^\dagger]$, which is then a faithful prime of this smaller group ring. Thus it is enough to understand the faithful primes of $kG$.

James Roseblade published a paper in 1978 called "Prime ideals in group rings of polycyclic groups", which appeared in the Proceedings of the London Mathematical Society (I can send you a copy if you like). Although he did not completely settle the problem of classifying the faithful prime ideals in group rings $kG$ of polycyclic-by-finite groups $G$ over fields $k$, he made some serious breakthroughs in this paper.

I'd like to draw your attention to Theorem E, which says:

Let $A$ be the Zalesskii subgroup of the polycyclic group $G$. If $G$ is orbitally sound, then any faithful prime ideal of a group algebra $kG$ is controlled by $A$.

Being controlled by $A$ for a prime ideal $P$ means that $P$ is generated by its intersection with $kA$ and is therefore completely determined by an ideal in a smaller group ring. The Zalesskii subgroup $A$ is defined by the property that it contains the largest finite normal subgroup $F$ of $G$ and that $A/F$ is the centre of the Fitting (largest normal nilpotent) subgroup of $G/F$. Being orbitally sound is a technical condition that I won't go into here.

In particular, if $G$ happens to be torsion-free and nilpotent then $F$ is trivial and $A$ is just the centre of $G$, and then Roseblade's Theorem E essentially says that every faithful prime of $kG$ is controlled by the centre of $G$. This happens for example when $G = \mathbb{Z}^2 \rtimes \mathbb{Z}$ as in your question and when $\sigma$ is the matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.