1
vote
0answers
31 views

Online Finite Field Calculator

I need to find an online Finite Field calculator with the following operations: Inverse SqrRoot Mult Div I have found one a couple of days ago but lost the url, and cannot find it now. Any ...
4
votes
2answers
65 views

Existence of nontrivial unit in $\mathbb{Q}[G]$, where $G$ is finite.

Suppose $G$ is a finite group of order $|G|>1$, and $\mathbb{Q}[G]$ is the group ring. I'm curious about an example of a nontrivial invertible element, i.e., one that is not of the form $ag$, ...
6
votes
1answer
78 views

Minimal counterexamples of the isomorphism problem for integral group rings

The isomorphism problem for integral group rings asks if two finite groups $G,H$ are isomorphic when their integral group rings $\mathbb{Z}[G]$, $\mathbb{Z}[H]$ are isomorphic. Quite a lot has been ...
1
vote
1answer
27 views

Let $F$ be a finite field and $\tau$ an element of $F$. Prove that there exists $a,b\in F$ such that $\tau=a^2+b^2$.

Let $F$ be a finite field and $\tau$ an element of $F$. Prove that there exists $a,b\in F$ such that $\tau=a^2+b^2$. It suffices to prove for the case $F=\mathbb{Z}_p$. How to prove?
2
votes
1answer
93 views

If the ring of integers corresponds to finite cyclic groups, what infinite rings do the other finite groups correspond to?

Consider $\mathbb{Z}_n$, where $n$ is any positive integer. Then there is a subgroup of $\mathbb{Z}_n$ for each divisor $d$ of $n$, so that the number of subgroups equals the number of divisors of ...
0
votes
2answers
77 views

Additive group of a finite ring of square free order is cyclic

$R$ is a finite ring of square free order $n>1$. How to show by the basis theorem for finite Abelian groups that additive group of $R$ is cyclic group of order $n$?
2
votes
2answers
28 views

Ring $\mathbb{Z}/2mnr \mathbb{Z}$ unit, identity, orders

Let $p$ be a prime number which doesn't divide $2mnr$. So $p$ is a unit in the ring $\mathbb{Z}/2mnr \mathbb{Z}$ and $q=p^k$ for a certain $k \in \mathbb{Z}$ Could you explain to me why then: 1) ...
1
vote
0answers
61 views

$S_k$ action on $A/I$

Let $S_2$ be a finite group of order $2$ and let $S_2$ act on $k[x,y]$ by interchanging $x$ and $y$, where $k=\overline{k}$. Then since $$ R = \left( \dfrac{k[x,y]}{(x+y)} \right)^{S_2} = ...
2
votes
0answers
35 views

$k[V]^G = \widetilde{A}$ where $\widetilde{A}$ is the normalization of $A$

Let $V$ be a finite dimensional vector space over $k =\overline{k}$ and let $G$ be a subgroup of $GL(V)$ so that $k[V]^G$ is finitely generated. Let $A$ be a subring of $k[V]^G$ that is finitely ...
11
votes
1answer
500 views

Subrings of finite index and units

If $S$ is a subring of $R$ of finite index as an abelian group, does it follow that the subgroup of units $S^\star$ has finite index in $R^\star$ as a multiplicative group?