version of Nakayama's lemma for group rings
up vote
1
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 7 hours ago
E. Blioch
454
add a comment |
up vote
1
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 7 hours ago
E. Blioch
454
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
7 hours ago
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
7 hours ago
@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
6 hours ago
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
6 hours ago
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
6 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
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 7 hours ago
E. Blioch
454
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 7 hours ago
E. Blioch
454
asked 7 hours ago
E. Blioch
454
asked 7 hours ago
E. Blioch
454
asked 7 hours ago
E. Blioch
454
asked 7 hours ago
E. Blioch
454
454
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
7 hours ago
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
7 hours ago
@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
6 hours ago
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
6 hours ago
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
6 hours ago
add a 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
7 hours ago
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
7 hours ago
@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
6 hours ago
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
6 hours ago
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
6 hours ago
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
7 hours ago
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
7 hours ago
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
7 hours ago
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
7 hours ago
@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
6 hours ago
@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
6 hours ago
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
6 hours ago
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
6 hours ago
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
6 hours ago
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
6 hours ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
7 hours ago
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
7 hours ago
@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
6 hours ago
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
6 hours ago
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
6 hours ago