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3 votes
1 answer
161 views

Behavior of invariants under reduction mod p

Let $R$ be a finitely generated $\mathbb{Z}$-algebra with an [edit: linear algebraic] action of $G(\mathbb{Z})$ where $G$ is a split simply-connected semisimple group. Then for any prime $p$ we have a ...
0 votes
0 answers
141 views

Stabilizers in the action of $\mathrm{GL}(n, \mathbb Z)$ on $\mathbb Z^n$

How can we calculate effectively the subgroups of $G: = \mathrm{GL}(n, \mathbb Z)$ which fix pointwise a given submodule $S$ of $\mathbb Z^n$ in the action of $G$ on $\mathbb Z^n$ by left ...
5 votes
1 answer
605 views

Unipotent completion of free group

Whilst I am reading articles on unipotent completion to understand its basic construction, I found something confusing. Let $F$ be a free group of rank 2 whose generating letters are $x$ and $y$ and ...
4 votes
1 answer
1k views

Maximal ideal of group ring

Let $R$ be a finite commutative ring with identity and $G$ an finite abelian group. Is there any more conditions (on $R$ or on $G $) under which we can characterize maximal ideals of group ring $RG $, ...
3 votes
2 answers
2k views

Is there an almost-direct product decomposition for disconnected reductive algebraic groups?

$\textbf{Some definitions:}$ Let $G$ be an algebraic group (for me that is the complex points of an affine algebraic group). We say $G$ is reductive if its unipotent radical (maximal connected normal ...
2 votes
2 answers
756 views

Abstract Commensurator Group of $\mathbb{Z}^n$ $Comm(\mathbb{Z}^n)\cong GL(n,\mathbb{Q})$?

Hello! In a paper I read that $\mathrm{Comm}(\mathbb{Z}^n)\cong \mathrm{GL}(n,\mathbb{Q})$. Why is that true? How can I find an isomorphism of this groups? I know that $\mathrm{Aut}(\mathbb{Z}^n)\...