Mixing
version of Nakayama's lemma for group rings
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 '18 at 19:29
E. Blioch
5015
|
show 1 more comment
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 '18 at 19:29
E. Blioch
5015
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 '18 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 '18 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 '18 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 '18 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 '18 at 20:39
|
show 1 more comment
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 '18 at 19:29
E. Blioch
5015
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 '18 at 19:29
E. Blioch
5015
asked Nov 20 '18 at 19:29
E. Blioch
5015
asked Nov 20 '18 at 19:29
E. Blioch
5015
asked Nov 20 '18 at 19:29
E. Blioch
5015
asked Nov 20 '18 at 19:29
E. Blioch
5015
5015
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 '18 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 '18 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 '18 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 '18 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 '18 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 '18 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 '18 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 '18 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 '18 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 '18 at 20:39
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 '18 at 20:39
|
show 1 more comment
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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 '18 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 '18 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 '18 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 '18 at 20:38
For future discussions, let's agree to simplify our life and substitute $Y$ for $X-c$.
– darij grinberg
Nov 20 '18 at 20:39