Tagged Questions
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
60 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
33 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
435 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?
