All Questions
6 questions
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)\...