Mixing
When is $S^G [G] cong S$ as a $S^G[G]$-module?
$begingroup$
Let $S$ be a ring and take a finite group $G$ acting on $S$. Put $R = S^G$, the fixed subring. When is $S$ a free rank-$1$ $R[G]$-module? That is, when do we have $S cong R[G]$ as $R[G]$-modules?
I am sure this is known, and I am aware of the normal basis theorem for the case of fields, but this is more general. I am interested in a complete characterization of when this proeprty holds, or any partial results that do not merely apply to fields.
For one example, I conjecture that this holds for integrally closed domains. For an extension of number fields $mathcal{O}_L / mathcal{O}_K$ with $L/K$ Galois, I would suspect this to follow from the normal basis theorem.
Edit: my guess here was shown to be false in the comments.
Does anyone know any cool results on the matter more general than the normal basis theorem?
Thanks for any help.
galois-theory invariant-theory
asked Jan 26 at 18:20
Dean YoungDean Young
1,739721
$endgroup$
|
show 3 more comments
$begingroup$
Let $S$ be a ring and take a finite group $G$ acting on $S$. Put $R = S^G$, the fixed subring. When is $S$ a free rank-$1$ $R[G]$-module? That is, when do we have $S cong R[G]$ as $R[G]$-modules?
I am sure this is known, and I am aware of the normal basis theorem for the case of fields, but this is more general. I am interested in a complete characterization of when this proeprty holds, or any partial results that do not merely apply to fields.
For one example, I conjecture that this holds for integrally closed domains. For an extension of number fields $mathcal{O}_L / mathcal{O}_K$ with $L/K$ Galois, I would suspect this to follow from the normal basis theorem.
Edit: my guess here was shown to be false in the comments.
Does anyone know any cool results on the matter more general than the normal basis theorem?
Thanks for any help.
galois-theory invariant-theory
asked Jan 26 at 18:20
Dean YoungDean Young
1,739721
$endgroup$
1
$begingroup$
amazon.co.uk/…
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:22
$begingroup$
Presumably $G$ acts by ring automorphisms :-)
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:20
$begingroup$
Equally trivially, you probably want $G$ to act faithfully, i.e. only $1_G$ acts via $id_S$.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:26
1
$begingroup$
Can you realize $Bbb{Z}[i]simeq_{C_2}Bbb{Z}[C_2]$ with the non-trivial element of $C_2$ acting by compex conjugation? If ${a+bi,a-bi}$ is a normal basis consisting of algebraic integers (so $abneq0$), then their $Bbb{Z}$-span is of an even index $2ab$ in $Bbb{Z}[i]$. In other words, it cannot be all of $Bbb{Z}[i]$?
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:46
1
$begingroup$
I wanted to emphasize that the action is by ring automorphisms as opposed to group automorphisms. For example, we have $C_2$ acting on $Bbb{Z}$, where the generator acts by negating everything. The set of fixed points then consists of zero alone, which is no good. Sorry about the level of pedantry :-) Anyway, I still don't see a way of writing $Bbb{Z}[i]$ as $Bbb{Z}[C_2]$ if the generator is to act via complex conjugation.
$endgroup$
– Jyrki Lahtonen
Jan 30 at 8:39
|
show 3 more comments
$begingroup$
Let $S$ be a ring and take a finite group $G$ acting on $S$. Put $R = S^G$, the fixed subring. When is $S$ a free rank-$1$ $R[G]$-module? That is, when do we have $S cong R[G]$ as $R[G]$-modules?
I am sure this is known, and I am aware of the normal basis theorem for the case of fields, but this is more general. I am interested in a complete characterization of when this proeprty holds, or any partial results that do not merely apply to fields.
For one example, I conjecture that this holds for integrally closed domains. For an extension of number fields $mathcal{O}_L / mathcal{O}_K$ with $L/K$ Galois, I would suspect this to follow from the normal basis theorem.
Edit: my guess here was shown to be false in the comments.
Does anyone know any cool results on the matter more general than the normal basis theorem?
Thanks for any help.
galois-theory invariant-theory
asked Jan 26 at 18:20
Dean YoungDean Young
1,739721
$endgroup$
Let $S$ be a ring and take a finite group $G$ acting on $S$. Put $R = S^G$, the fixed subring. When is $S$ a free rank-$1$ $R[G]$-module? That is, when do we have $S cong R[G]$ as $R[G]$-modules?
I am sure this is known, and I am aware of the normal basis theorem for the case of fields, but this is more general. I am interested in a complete characterization of when this proeprty holds, or any partial results that do not merely apply to fields.
For one example, I conjecture that this holds for integrally closed domains. For an extension of number fields $mathcal{O}_L / mathcal{O}_K$ with $L/K$ Galois, I would suspect this to follow from the normal basis theorem.
Edit: my guess here was shown to be false in the comments.
Does anyone know any cool results on the matter more general than the normal basis theorem?
Thanks for any help.
galois-theory invariant-theory
galois-theory invariant-theory
asked Jan 26 at 18:20
Dean YoungDean Young
1,739721
asked Jan 26 at 18:20
Dean YoungDean Young
1,739721
edited Jan 30 at 14:18
Dean Young
asked Jan 26 at 18:20
Dean YoungDean Young
1,739721
asked Jan 26 at 18:20
Dean YoungDean Young
1,739721
asked Jan 26 at 18:20
Dean YoungDean Young
1,739721
1,739721
1
$begingroup$
amazon.co.uk/…
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:22
$begingroup$
Presumably $G$ acts by ring automorphisms :-)
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:20
$begingroup$
Equally trivially, you probably want $G$ to act faithfully, i.e. only $1_G$ acts via $id_S$.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:26
1
$begingroup$
Can you realize $Bbb{Z}[i]simeq_{C_2}Bbb{Z}[C_2]$ with the non-trivial element of $C_2$ acting by compex conjugation? If ${a+bi,a-bi}$ is a normal basis consisting of algebraic integers (so $abneq0$), then their $Bbb{Z}$-span is of an even index $2ab$ in $Bbb{Z}[i]$. In other words, it cannot be all of $Bbb{Z}[i]$?
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:46
1
$begingroup$
I wanted to emphasize that the action is by ring automorphisms as opposed to group automorphisms. For example, we have $C_2$ acting on $Bbb{Z}$, where the generator acts by negating everything. The set of fixed points then consists of zero alone, which is no good. Sorry about the level of pedantry :-) Anyway, I still don't see a way of writing $Bbb{Z}[i]$ as $Bbb{Z}[C_2]$ if the generator is to act via complex conjugation.
$endgroup$
– Jyrki Lahtonen
Jan 30 at 8:39
|
show 3 more comments
1
$begingroup$
amazon.co.uk/…
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:22
$begingroup$
Presumably $G$ acts by ring automorphisms :-)
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:20
$begingroup$
Equally trivially, you probably want $G$ to act faithfully, i.e. only $1_G$ acts via $id_S$.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:26
1
$begingroup$
Can you realize $Bbb{Z}[i]simeq_{C_2}Bbb{Z}[C_2]$ with the non-trivial element of $C_2$ acting by compex conjugation? If ${a+bi,a-bi}$ is a normal basis consisting of algebraic integers (so $abneq0$), then their $Bbb{Z}$-span is of an even index $2ab$ in $Bbb{Z}[i]$. In other words, it cannot be all of $Bbb{Z}[i]$?
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:46
1
$begingroup$
I wanted to emphasize that the action is by ring automorphisms as opposed to group automorphisms. For example, we have $C_2$ acting on $Bbb{Z}$, where the generator acts by negating everything. The set of fixed points then consists of zero alone, which is no good. Sorry about the level of pedantry :-) Anyway, I still don't see a way of writing $Bbb{Z}[i]$ as $Bbb{Z}[C_2]$ if the generator is to act via complex conjugation.
$endgroup$
– Jyrki Lahtonen
Jan 30 at 8:39
1
1
$begingroup$
amazon.co.uk/…
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:22
$begingroup$
amazon.co.uk/…
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:22
$begingroup$
Presumably $G$ acts by ring automorphisms :-)
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:20
$begingroup$
Presumably $G$ acts by ring automorphisms :-)
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:20
$begingroup$
Equally trivially, you probably want $G$ to act faithfully, i.e. only $1_G$ acts via $id_S$.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:26
$begingroup$
Equally trivially, you probably want $G$ to act faithfully, i.e. only $1_G$ acts via $id_S$.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:26
1
1
$begingroup$
Can you realize $Bbb{Z}[i]simeq_{C_2}Bbb{Z}[C_2]$ with the non-trivial element of $C_2$ acting by compex conjugation? If ${a+bi,a-bi}$ is a normal basis consisting of algebraic integers (so $abneq0$), then their $Bbb{Z}$-span is of an even index $2ab$ in $Bbb{Z}[i]$. In other words, it cannot be all of $Bbb{Z}[i]$?
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:46
$begingroup$
Can you realize $Bbb{Z}[i]simeq_{C_2}Bbb{Z}[C_2]$ with the non-trivial element of $C_2$ acting by compex conjugation? If ${a+bi,a-bi}$ is a normal basis consisting of algebraic integers (so $abneq0$), then their $Bbb{Z}$-span is of an even index $2ab$ in $Bbb{Z}[i]$. In other words, it cannot be all of $Bbb{Z}[i]$?
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:46
1
1
$begingroup$
I wanted to emphasize that the action is by ring automorphisms as opposed to group automorphisms. For example, we have $C_2$ acting on $Bbb{Z}$, where the generator acts by negating everything. The set of fixed points then consists of zero alone, which is no good. Sorry about the level of pedantry :-) Anyway, I still don't see a way of writing $Bbb{Z}[i]$ as $Bbb{Z}[C_2]$ if the generator is to act via complex conjugation.
$endgroup$
– Jyrki Lahtonen
Jan 30 at 8:39
$begingroup$
I wanted to emphasize that the action is by ring automorphisms as opposed to group automorphisms. For example, we have $C_2$ acting on $Bbb{Z}$, where the generator acts by negating everything. The set of fixed points then consists of zero alone, which is no good. Sorry about the level of pedantry :-) Anyway, I still don't see a way of writing $Bbb{Z}[i]$ as $Bbb{Z}[C_2]$ if the generator is to act via complex conjugation.
$endgroup$
– Jyrki Lahtonen
Jan 30 at 8:39
|
show 3 more comments
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1
$begingroup$
amazon.co.uk/…
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:22
$begingroup$
Presumably $G$ acts by ring automorphisms :-)
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:20
$begingroup$
Equally trivially, you probably want $G$ to act faithfully, i.e. only $1_G$ acts via $id_S$.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:26
1
$begingroup$
Can you realize $Bbb{Z}[i]simeq_{C_2}Bbb{Z}[C_2]$ with the non-trivial element of $C_2$ acting by compex conjugation? If ${a+bi,a-bi}$ is a normal basis consisting of algebraic integers (so $abneq0$), then their $Bbb{Z}$-span is of an even index $2ab$ in $Bbb{Z}[i]$. In other words, it cannot be all of $Bbb{Z}[i]$?
$endgroup$
– Jyrki Lahtonen
Jan 29 at 8:46
1
$begingroup$
I wanted to emphasize that the action is by ring automorphisms as opposed to group automorphisms. For example, we have $C_2$ acting on $Bbb{Z}$, where the generator acts by negating everything. The set of fixed points then consists of zero alone, which is no good. Sorry about the level of pedantry :-) Anyway, I still don't see a way of writing $Bbb{Z}[i]$ as $Bbb{Z}[C_2]$ if the generator is to act via complex conjugation.
$endgroup$
– Jyrki Lahtonen
Jan 30 at 8:39