For $G=\mathbb{Z}^2\rtimes \mathbb{Z}$, $Spec(\mathbb{Z}G)$=? 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}$
 A: 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 $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$.
