All Questions
11 questions
2
votes
0
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144
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Zero divisors in the extra-special group algebra $\mathbb{R}[2^{1+6}_+]$
Can you characterize the unit-group of the real group-algebra of the extraspecial plus-type 2-group of order 128? (That is $\mathbb{R}[2_+^{1+6}]$ using Conway's notation.)
(Please choose any irrep ...
0
votes
0
answers
58
views
An action on multiplicatively antisymmetric matrix
A matrix $ Q=(q_{ij})$ is called multiplicatively antisymmetric over a field $ F $ if $ q_{ii}=1 $ and $ q_{ij}={q_{ji}}^{-1} $.Let $ \mathcal{Q} $ be the set of all $ n \times n $ multiplicatively ...
3
votes
0
answers
317
views
How to go from primitive idempotents in $\text{End}_A(M)$ to primitive idempotents in $A$?
Let $K$ be a large enough finite field, let $A$ be a finite-dimensional $K$-algebra.
Moreover, let $M$ be a finitely generated $A$-module and let $M = M_1\oplus ... \oplus M_n$ be a decomposition of $...
7
votes
1
answer
281
views
Question concerning the coefficients of block idempotents
Let $G$ be a finite group. Let $p$ be a prime number such that $p \mid |G|$.
Let Irr$(G)$ denote the set of ordinary irreducible characters of $G$.
For $\chi\in$ Irr$(G)$ define $e_{\chi} := \frac{\...
5
votes
1
answer
132
views
Integral monoid rings and Ore conditions
Consider a cancellative monoid $S$ satisfying the left Ore condition, so it embeds in a group $G=S^{-1}S$. Consider also the integral monoid rings $\mathbb Z[S]$ and $\mathbb Z[G]$.
I have two, ...
2
votes
1
answer
310
views
Flatness of submodules of free modules
Are submodules of free $\mathbb{Z}[G]$-modules flat? if not what conditions on $G$ makes it true? $G$ is an infinite group.
If $\mathbb{Z}[G]$ is a Prüfer domain then this is true. Can a group ring $...
5
votes
0
answers
187
views
Any f.p. faithful simple module over a primitive group ring?
Recall that a ring $R$ is primitive if it has a faithful simple left module. Let $G$ be a countable discrete group and $R=\mathbb{k}G$, where $\mathbb{k}$ denotes some field or $\mathbb{Z}$.
There ...
3
votes
0
answers
65
views
Intersections of generating sets of subalgebras
Let $A$ be a finitely generated, finitely presented, Noetherian, unital algebra over the complex numbers, which has no zero divisors. We do not assume that $A$ is commutative however.
Moreover, let $...
0
votes
0
answers
308
views
Basic question about power series and complete group algebras
This is a pretty basic question, but I suspect it might be too exotic for math.stackexchange.
Let $\mathbb{Z}_p$ be the $p$-adic integers. For free pro-$p$ group $F_r$ of rank $r$, we can consider ...
3
votes
1
answer
220
views
Intersection of Maximal Left Ideals with Finite Dimensional Quotient
Let $\Gamma$ be a finitely generated group and let $A=\mathbb{C}[\Gamma]$ be the corresponding group algebra over $\mathbb{C}$. Let $X$ be the set of all maximal left ideals of $A$ and let $X_0=\{I \...
64
votes
4
answers
8k
views
What is the current status of the Kaplansky zero-divisor conjecture for group rings?
Let $K$ be a field and $G$ a group. The so called zero-divisor conjecture for group rings asserts that the group ring $K[G]$ is a domain if and only if $G$ is a torsion-free group.
A couple of good ...