version of Nakayama's lemma for group rings
up vote
2
down vote
favorite
I have the following rather complicated setting and I would like to know if something that resembles Nakayama's lemma can be proved in this setting, but I can make no progress with it:
Let $G$ be a finite group and let $K$ be a field of characteristic $0$. Assume that $M$ is a module over the polynomial ring $K[X]$, that $M$ is finitely generated as a $K$-module and also that there exists an action of the group $G$ on $M$, so that $M$ becomes a module the group ring $(K[X])[G]$.
I know in addition that the annihilator of $M$ in $K[X]$ is of the form $(X-c)^d$ for some $c in K$, $d>0$ and that the set of elements in $M$ annihilated by $(X-c)$ is a cyclic $K[G]$-module (also the action of $X$ and $G$ commute). Does it follow that $M$ is cyclic as a $(K[X])[G]$-module?
commutative-algebra modules representation-theory
asked Nov 20 at 19:29
E. Blioch
505
|
show 1 more comment
up vote
2
down vote
favorite
I have the following rather complicated setting and I would like to know if something that resembles Nakayama's lemma can be proved in this setting, but I can make no progress with it:
Let $G$ be a finite group and let $K$ be a field of characteristic $0$. Assume that $M$ is a module over the polynomial ring $K[X]$, that $M$ is finitely generated as a $K$-module and also that there exists an action of the group $G$ on $M$, so that $M$ becomes a module the group ring $(K[X])[G]$.
I know in addition that the annihilator of $M$ in $K[X]$ is of the form $(X-c)^d$ for some $c in K$, $d>0$ and that the set of elements in $M$ annihilated by $(X-c)$ is a cyclic $K[G]$-module (also the action of $X$ and $G$ commute). Does it follow that $M$ is cyclic as a $(K[X])[G]$-module?
commutative-algebra modules representation-theory
asked Nov 20 at 19:29
E. Blioch
505
So $M$ is not just a $Kleft[Xright]$-module, but actually a $Kleft[Xright] / left(X-cright)^d$-module. Note that $Kleft[Xright] / left(X-cright)^d$ is a local ring, with its maximal ideal generated by $X-c$. This may help in applying regular Nakayama.
– darij grinberg
Nov 20 at 20:08
Is the set of elements in $M$ annihilated by $left(X-cright)$ isomorphic (as a $Kleft[Xright]left[Gright]$-module) to $M / left(X-cright)M$ ? Or, at least, is there a surjective $Kleft[Xright]left[Gright]$-module homomorphism frm the former to the latter? I suspect so; this would reduce your question to the standard Nakayama lemma.
– darij grinberg
Nov 20 at 20:11
@darijgrinberg I can see the obvious isomorphism $M/N cong (X-c)M$ (where $N$ is the set of elements annihilated by $X-c$), but I don't see how to construct the 'swapped' version $M/(X-c)M cong N$ you suggest
– E. Blioch
Nov 20 at 20:16
OK, I see why $M / left(X-cright) M cong N$ as $Kleft[Xright]$-modules: Both sides are $Kleft[Xright] / left(X-cright)$-modules, which is the same as $K$-vector spaces, so it suffices to prove that their dimensions are equal; but this follows from looking at the Jordan normal form of a nilpotent endomorphism (the endomorphism being the action of $X-c$ on $M$). But this isomorphism is not canonical, so it is not clear at all that it will play well with the action of $G$.
– darij grinberg
Nov 20 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 at 20:39
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following rather complicated setting and I would like to know if something that resembles Nakayama's lemma can be proved in this setting, but I can make no progress with it:
Let $G$ be a finite group and let $K$ be a field of characteristic $0$. Assume that $M$ is a module over the polynomial ring $K[X]$, that $M$ is finitely generated as a $K$-module and also that there exists an action of the group $G$ on $M$, so that $M$ becomes a module the group ring $(K[X])[G]$.
I know in addition that the annihilator of $M$ in $K[X]$ is of the form $(X-c)^d$ for some $c in K$, $d>0$ and that the set of elements in $M$ annihilated by $(X-c)$ is a cyclic $K[G]$-module (also the action of $X$ and $G$ commute). Does it follow that $M$ is cyclic as a $(K[X])[G]$-module?
commutative-algebra modules representation-theory
asked Nov 20 at 19:29
E. Blioch
505
I have the following rather complicated setting and I would like to know if something that resembles Nakayama's lemma can be proved in this setting, but I can make no progress with it:
Let $G$ be a finite group and let $K$ be a field of characteristic $0$. Assume that $M$ is a module over the polynomial ring $K[X]$, that $M$ is finitely generated as a $K$-module and also that there exists an action of the group $G$ on $M$, so that $M$ becomes a module the group ring $(K[X])[G]$.
I know in addition that the annihilator of $M$ in $K[X]$ is of the form $(X-c)^d$ for some $c in K$, $d>0$ and that the set of elements in $M$ annihilated by $(X-c)$ is a cyclic $K[G]$-module (also the action of $X$ and $G$ commute). Does it follow that $M$ is cyclic as a $(K[X])[G]$-module?
commutative-algebra modules representation-theory
commutative-algebra modules representation-theory
asked Nov 20 at 19:29
E. Blioch
505
asked Nov 20 at 19:29
E. Blioch
505
asked Nov 20 at 19:29
E. Blioch
505
asked Nov 20 at 19:29
E. Blioch
505
asked Nov 20 at 19:29
E. Blioch
505
505
So $M$ is not just a $Kleft[Xright]$-module, but actually a $Kleft[Xright] / left(X-cright)^d$-module. Note that $Kleft[Xright] / left(X-cright)^d$ is a local ring, with its maximal ideal generated by $X-c$. This may help in applying regular Nakayama.
– darij grinberg
Nov 20 at 20:08
Is the set of elements in $M$ annihilated by $left(X-cright)$ isomorphic (as a $Kleft[Xright]left[Gright]$-module) to $M / left(X-cright)M$ ? Or, at least, is there a surjective $Kleft[Xright]left[Gright]$-module homomorphism frm the former to the latter? I suspect so; this would reduce your question to the standard Nakayama lemma.
– darij grinberg
Nov 20 at 20:11
@darijgrinberg I can see the obvious isomorphism $M/N cong (X-c)M$ (where $N$ is the set of elements annihilated by $X-c$), but I don't see how to construct the 'swapped' version $M/(X-c)M cong N$ you suggest
– E. Blioch
Nov 20 at 20:16
OK, I see why $M / left(X-cright) M cong N$ as $Kleft[Xright]$-modules: Both sides are $Kleft[Xright] / left(X-cright)$-modules, which is the same as $K$-vector spaces, so it suffices to prove that their dimensions are equal; but this follows from looking at the Jordan normal form of a nilpotent endomorphism (the endomorphism being the action of $X-c$ on $M$). But this isomorphism is not canonical, so it is not clear at all that it will play well with the action of $G$.
– darij grinberg
Nov 20 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 at 20:39
|
show 1 more comment
So $M$ is not just a $Kleft[Xright]$-module, but actually a $Kleft[Xright] / left(X-cright)^d$-module. Note that $Kleft[Xright] / left(X-cright)^d$ is a local ring, with its maximal ideal generated by $X-c$. This may help in applying regular Nakayama.
– darij grinberg
Nov 20 at 20:08
Is the set of elements in $M$ annihilated by $left(X-cright)$ isomorphic (as a $Kleft[Xright]left[Gright]$-module) to $M / left(X-cright)M$ ? Or, at least, is there a surjective $Kleft[Xright]left[Gright]$-module homomorphism frm the former to the latter? I suspect so; this would reduce your question to the standard Nakayama lemma.
– darij grinberg
Nov 20 at 20:11
@darijgrinberg I can see the obvious isomorphism $M/N cong (X-c)M$ (where $N$ is the set of elements annihilated by $X-c$), but I don't see how to construct the 'swapped' version $M/(X-c)M cong N$ you suggest
– E. Blioch
Nov 20 at 20:16
OK, I see why $M / left(X-cright) M cong N$ as $Kleft[Xright]$-modules: Both sides are $Kleft[Xright] / left(X-cright)$-modules, which is the same as $K$-vector spaces, so it suffices to prove that their dimensions are equal; but this follows from looking at the Jordan normal form of a nilpotent endomorphism (the endomorphism being the action of $X-c$ on $M$). But this isomorphism is not canonical, so it is not clear at all that it will play well with the action of $G$.
– darij grinberg
Nov 20 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 at 20:39
So $M$ is not just a $Kleft[Xright]$-module, but actually a $Kleft[Xright] / left(X-cright)^d$-module. Note that $Kleft[Xright] / left(X-cright)^d$ is a local ring, with its maximal ideal generated by $X-c$. This may help in applying regular Nakayama.
– darij grinberg
Nov 20 at 20:08
So $M$ is not just a $Kleft[Xright]$-module, but actually a $Kleft[Xright] / left(X-cright)^d$-module. Note that $Kleft[Xright] / left(X-cright)^d$ is a local ring, with its maximal ideal generated by $X-c$. This may help in applying regular Nakayama.
– darij grinberg
Nov 20 at 20:08
Is the set of elements in $M$ annihilated by $left(X-cright)$ isomorphic (as a $Kleft[Xright]left[Gright]$-module) to $M / left(X-cright)M$ ? Or, at least, is there a surjective $Kleft[Xright]left[Gright]$-module homomorphism frm the former to the latter? I suspect so; this would reduce your question to the standard Nakayama lemma.
– darij grinberg
Nov 20 at 20:11
Is the set of elements in $M$ annihilated by $left(X-cright)$ isomorphic (as a $Kleft[Xright]left[Gright]$-module) to $M / left(X-cright)M$ ? Or, at least, is there a surjective $Kleft[Xright]left[Gright]$-module homomorphism frm the former to the latter? I suspect so; this would reduce your question to the standard Nakayama lemma.
– darij grinberg
Nov 20 at 20:11
@darijgrinberg I can see the obvious isomorphism $M/N cong (X-c)M$ (where $N$ is the set of elements annihilated by $X-c$), but I don't see how to construct the 'swapped' version $M/(X-c)M cong N$ you suggest
– E. Blioch
Nov 20 at 20:16
@darijgrinberg I can see the obvious isomorphism $M/N cong (X-c)M$ (where $N$ is the set of elements annihilated by $X-c$), but I don't see how to construct the 'swapped' version $M/(X-c)M cong N$ you suggest
– E. Blioch
Nov 20 at 20:16
OK, I see why $M / left(X-cright) M cong N$ as $Kleft[Xright]$-modules: Both sides are $Kleft[Xright] / left(X-cright)$-modules, which is the same as $K$-vector spaces, so it suffices to prove that their dimensions are equal; but this follows from looking at the Jordan normal form of a nilpotent endomorphism (the endomorphism being the action of $X-c$ on $M$). But this isomorphism is not canonical, so it is not clear at all that it will play well with the action of $G$.
– darij grinberg
Nov 20 at 20:38
OK, I see why $M / left(X-cright) M cong N$ as $Kleft[Xright]$-modules: Both sides are $Kleft[Xright] / left(X-cright)$-modules, which is the same as $K$-vector spaces, so it suffices to prove that their dimensions are equal; but this follows from looking at the Jordan normal form of a nilpotent endomorphism (the endomorphism being the action of $X-c$ on $M$). But this isomorphism is not canonical, so it is not clear at all that it will play well with the action of $G$.
– darij grinberg
Nov 20 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 at 20:39
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 at 20:39
|
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006780%2fversion-of-nakayamas-lemma-for-group-rings%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
So $M$ is not just a $Kleft[Xright]$-module, but actually a $Kleft[Xright] / left(X-cright)^d$-module. Note that $Kleft[Xright] / left(X-cright)^d$ is a local ring, with its maximal ideal generated by $X-c$. This may help in applying regular Nakayama.
– darij grinberg
Nov 20 at 20:08
Is the set of elements in $M$ annihilated by $left(X-cright)$ isomorphic (as a $Kleft[Xright]left[Gright]$-module) to $M / left(X-cright)M$ ? Or, at least, is there a surjective $Kleft[Xright]left[Gright]$-module homomorphism frm the former to the latter? I suspect so; this would reduce your question to the standard Nakayama lemma.
– darij grinberg
Nov 20 at 20:11
@darijgrinberg I can see the obvious isomorphism $M/N cong (X-c)M$ (where $N$ is the set of elements annihilated by $X-c$), but I don't see how to construct the 'swapped' version $M/(X-c)M cong N$ you suggest
– E. Blioch
Nov 20 at 20:16
OK, I see why $M / left(X-cright) M cong N$ as $Kleft[Xright]$-modules: Both sides are $Kleft[Xright] / left(X-cright)$-modules, which is the same as $K$-vector spaces, so it suffices to prove that their dimensions are equal; but this follows from looking at the Jordan normal form of a nilpotent endomorphism (the endomorphism being the action of $X-c$ on $M$). But this isomorphism is not canonical, so it is not clear at all that it will play well with the action of $G$.
– darij grinberg
Nov 20 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 at 20:39