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Existence of direct summand isomorphic to $mathbbQ_p[P]$
Clash Royale CLAN TAG#URR8PPP
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Let $P$ be a cyclic group of order $p^n$. Let $M$ be a $mathbbZ_p[P]$-module and $M_p :=mathbbQ_potimes M$ be the associated $mathbbQ_p[P]$-module. When will $M_p$ have a direct summand isomorphic to $mathbbQ_p[P]$?
We know $mathbbQ[P]simeq bigoplus_dmid p^nmathbbQ(zeta_d)$.
If $M_psimeq bigoplus_dmid p^nM_p,d$ where $M_p,d$ is a vector space over $mathbbQ(zeta_d)otimes mathbbQ_p$ and $m_d$ is its dimension, then $M_psimeq bigoplus_d (mathbbQ(zeta_d)otimes mathbbQ_p)^m_d$ as $mathbbQ_p[P]$-modules. When does $M_p$ have a free rank one direct summand?
abstract-algebra group-theory number-theory modules representation-theory
edited Jul 23 at 20:35
Fabio Lucchini
5,57911025
asked Jul 23 at 18:45
debanjana
375111
add a comment |Â
up vote
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down vote
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Let $P$ be a cyclic group of order $p^n$. Let $M$ be a $mathbbZ_p[P]$-module and $M_p :=mathbbQ_potimes M$ be the associated $mathbbQ_p[P]$-module. When will $M_p$ have a direct summand isomorphic to $mathbbQ_p[P]$?
We know $mathbbQ[P]simeq bigoplus_dmid p^nmathbbQ(zeta_d)$.
If $M_psimeq bigoplus_dmid p^nM_p,d$ where $M_p,d$ is a vector space over $mathbbQ(zeta_d)otimes mathbbQ_p$ and $m_d$ is its dimension, then $M_psimeq bigoplus_d (mathbbQ(zeta_d)otimes mathbbQ_p)^m_d$ as $mathbbQ_p[P]$-modules. When does $M_p$ have a free rank one direct summand?
abstract-algebra group-theory number-theory modules representation-theory
edited Jul 23 at 20:35
Fabio Lucchini
5,57911025
asked Jul 23 at 18:45
debanjana
375111
1
Perhaps I do not understand the question. Is the answer not just: when $m_dgeq 1$ for all $dmid p^n$? The ring $mathbfQ_p[P]$ is the direct product of fields $mathbfQ_p(zeta_d)$, $dmid p^n$.
– Keenan Kidwell
Jul 24 at 18:42
Sure, but are there some (not very strict) conditions on say the $dim_mathbbQ_p M_p$ so that we know this will happen?
– debanjana
Jul 24 at 20:04
Imposing some condition on the dimension of $M_p$ as a $mathbfQ_p$-vector space will not be enough to ensure that $M_p$ has a submodule (equivalently, direct summand) isomorphic to $mathbfQ_p[P]$. Take $M=mathbfZ_p^n$ for any $ngeq 1$ with trivial $P$-action. Then $dim_mathbfQ_p(M_p)=dim_mathbfQ_p(mathbfQ_p^n)=n$. Can you elaborate on what you might be looking for, or if there is something you specifically want to do with $M_p$?
– Keenan Kidwell
Jul 24 at 20:14
$K$ be a number field and $F/K$ be a cyclic $p$-power Galois extension. $E$ be an elliptic curve defined over $K$. $S(E/F)$ be the Selmer group and $M$ be its Pontryagin dual. Then Mazur-Rubin (2007) show that under some natural conditions, if $dim M_p$ is odd then $m_dgeq 1$ for all $d$. If $R(E/F)$ is a subgroup of $S(E/F)$ (in particular, the fine Selmer group) and we define $M^prime$ to be its Pontryagin dual etc. then can one draw conclusions as to when $M^prime_p$ will have a direct summand isomorphic to $mathbbQ_p[P]$. Maybe using the semi-simplicity of $mathbbQ_p[P]$?
– debanjana
Jul 24 at 21:56
add a comment |Â
up vote
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up vote
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Let $P$ be a cyclic group of order $p^n$. Let $M$ be a $mathbbZ_p[P]$-module and $M_p :=mathbbQ_potimes M$ be the associated $mathbbQ_p[P]$-module. When will $M_p$ have a direct summand isomorphic to $mathbbQ_p[P]$?
We know $mathbbQ[P]simeq bigoplus_dmid p^nmathbbQ(zeta_d)$.
If $M_psimeq bigoplus_dmid p^nM_p,d$ where $M_p,d$ is a vector space over $mathbbQ(zeta_d)otimes mathbbQ_p$ and $m_d$ is its dimension, then $M_psimeq bigoplus_d (mathbbQ(zeta_d)otimes mathbbQ_p)^m_d$ as $mathbbQ_p[P]$-modules. When does $M_p$ have a free rank one direct summand?
abstract-algebra group-theory number-theory modules representation-theory
edited Jul 23 at 20:35
Fabio Lucchini
5,57911025
asked Jul 23 at 18:45
debanjana
375111
Let $P$ be a cyclic group of order $p^n$. Let $M$ be a $mathbbZ_p[P]$-module and $M_p :=mathbbQ_potimes M$ be the associated $mathbbQ_p[P]$-module. When will $M_p$ have a direct summand isomorphic to $mathbbQ_p[P]$?
We know $mathbbQ[P]simeq bigoplus_dmid p^nmathbbQ(zeta_d)$.
If $M_psimeq bigoplus_dmid p^nM_p,d$ where $M_p,d$ is a vector space over $mathbbQ(zeta_d)otimes mathbbQ_p$ and $m_d$ is its dimension, then $M_psimeq bigoplus_d (mathbbQ(zeta_d)otimes mathbbQ_p)^m_d$ as $mathbbQ_p[P]$-modules. When does $M_p$ have a free rank one direct summand?
abstract-algebra group-theory number-theory modules representation-theory
edited Jul 23 at 20:35
Fabio Lucchini
5,57911025
asked Jul 23 at 18:45
debanjana
375111
edited Jul 23 at 20:35
Fabio Lucchini
5,57911025
edited Jul 23 at 20:35
Fabio Lucchini
5,57911025
edited Jul 23 at 20:35
Fabio Lucchini
5,57911025
5,57911025
asked Jul 23 at 18:45
debanjana
375111
asked Jul 23 at 18:45
debanjana
375111
asked Jul 23 at 18:45
debanjana
375111
375111
1
Perhaps I do not understand the question. Is the answer not just: when $m_dgeq 1$ for all $dmid p^n$? The ring $mathbfQ_p[P]$ is the direct product of fields $mathbfQ_p(zeta_d)$, $dmid p^n$.
– Keenan Kidwell
Jul 24 at 18:42
Sure, but are there some (not very strict) conditions on say the $dim_mathbbQ_p M_p$ so that we know this will happen?
– debanjana
Jul 24 at 20:04
Imposing some condition on the dimension of $M_p$ as a $mathbfQ_p$-vector space will not be enough to ensure that $M_p$ has a submodule (equivalently, direct summand) isomorphic to $mathbfQ_p[P]$. Take $M=mathbfZ_p^n$ for any $ngeq 1$ with trivial $P$-action. Then $dim_mathbfQ_p(M_p)=dim_mathbfQ_p(mathbfQ_p^n)=n$. Can you elaborate on what you might be looking for, or if there is something you specifically want to do with $M_p$?
– Keenan Kidwell
Jul 24 at 20:14
$K$ be a number field and $F/K$ be a cyclic $p$-power Galois extension. $E$ be an elliptic curve defined over $K$. $S(E/F)$ be the Selmer group and $M$ be its Pontryagin dual. Then Mazur-Rubin (2007) show that under some natural conditions, if $dim M_p$ is odd then $m_dgeq 1$ for all $d$. If $R(E/F)$ is a subgroup of $S(E/F)$ (in particular, the fine Selmer group) and we define $M^prime$ to be its Pontryagin dual etc. then can one draw conclusions as to when $M^prime_p$ will have a direct summand isomorphic to $mathbbQ_p[P]$. Maybe using the semi-simplicity of $mathbbQ_p[P]$?
– debanjana
Jul 24 at 21:56
add a comment |Â
1
Perhaps I do not understand the question. Is the answer not just: when $m_dgeq 1$ for all $dmid p^n$? The ring $mathbfQ_p[P]$ is the direct product of fields $mathbfQ_p(zeta_d)$, $dmid p^n$.
– Keenan Kidwell
Jul 24 at 18:42
Sure, but are there some (not very strict) conditions on say the $dim_mathbbQ_p M_p$ so that we know this will happen?
– debanjana
Jul 24 at 20:04
Imposing some condition on the dimension of $M_p$ as a $mathbfQ_p$-vector space will not be enough to ensure that $M_p$ has a submodule (equivalently, direct summand) isomorphic to $mathbfQ_p[P]$. Take $M=mathbfZ_p^n$ for any $ngeq 1$ with trivial $P$-action. Then $dim_mathbfQ_p(M_p)=dim_mathbfQ_p(mathbfQ_p^n)=n$. Can you elaborate on what you might be looking for, or if there is something you specifically want to do with $M_p$?
– Keenan Kidwell
Jul 24 at 20:14
$K$ be a number field and $F/K$ be a cyclic $p$-power Galois extension. $E$ be an elliptic curve defined over $K$. $S(E/F)$ be the Selmer group and $M$ be its Pontryagin dual. Then Mazur-Rubin (2007) show that under some natural conditions, if $dim M_p$ is odd then $m_dgeq 1$ for all $d$. If $R(E/F)$ is a subgroup of $S(E/F)$ (in particular, the fine Selmer group) and we define $M^prime$ to be its Pontryagin dual etc. then can one draw conclusions as to when $M^prime_p$ will have a direct summand isomorphic to $mathbbQ_p[P]$. Maybe using the semi-simplicity of $mathbbQ_p[P]$?
– debanjana
Jul 24 at 21:56
1
1
Perhaps I do not understand the question. Is the answer not just: when $m_dgeq 1$ for all $dmid p^n$? The ring $mathbfQ_p[P]$ is the direct product of fields $mathbfQ_p(zeta_d)$, $dmid p^n$.
– Keenan Kidwell
Jul 24 at 18:42
Perhaps I do not understand the question. Is the answer not just: when $m_dgeq 1$ for all $dmid p^n$? The ring $mathbfQ_p[P]$ is the direct product of fields $mathbfQ_p(zeta_d)$, $dmid p^n$.
– Keenan Kidwell
Jul 24 at 18:42
Sure, but are there some (not very strict) conditions on say the $dim_mathbbQ_p M_p$ so that we know this will happen?
– debanjana
Jul 24 at 20:04
Sure, but are there some (not very strict) conditions on say the $dim_mathbbQ_p M_p$ so that we know this will happen?
– debanjana
Jul 24 at 20:04
Imposing some condition on the dimension of $M_p$ as a $mathbfQ_p$-vector space will not be enough to ensure that $M_p$ has a submodule (equivalently, direct summand) isomorphic to $mathbfQ_p[P]$. Take $M=mathbfZ_p^n$ for any $ngeq 1$ with trivial $P$-action. Then $dim_mathbfQ_p(M_p)=dim_mathbfQ_p(mathbfQ_p^n)=n$. Can you elaborate on what you might be looking for, or if there is something you specifically want to do with $M_p$?
– Keenan Kidwell
Jul 24 at 20:14
Imposing some condition on the dimension of $M_p$ as a $mathbfQ_p$-vector space will not be enough to ensure that $M_p$ has a submodule (equivalently, direct summand) isomorphic to $mathbfQ_p[P]$. Take $M=mathbfZ_p^n$ for any $ngeq 1$ with trivial $P$-action. Then $dim_mathbfQ_p(M_p)=dim_mathbfQ_p(mathbfQ_p^n)=n$. Can you elaborate on what you might be looking for, or if there is something you specifically want to do with $M_p$?
– Keenan Kidwell
Jul 24 at 20:14
$K$ be a number field and $F/K$ be a cyclic $p$-power Galois extension. $E$ be an elliptic curve defined over $K$. $S(E/F)$ be the Selmer group and $M$ be its Pontryagin dual. Then Mazur-Rubin (2007) show that under some natural conditions, if $dim M_p$ is odd then $m_dgeq 1$ for all $d$. If $R(E/F)$ is a subgroup of $S(E/F)$ (in particular, the fine Selmer group) and we define $M^prime$ to be its Pontryagin dual etc. then can one draw conclusions as to when $M^prime_p$ will have a direct summand isomorphic to $mathbbQ_p[P]$. Maybe using the semi-simplicity of $mathbbQ_p[P]$?
– debanjana
Jul 24 at 21:56
$K$ be a number field and $F/K$ be a cyclic $p$-power Galois extension. $E$ be an elliptic curve defined over $K$. $S(E/F)$ be the Selmer group and $M$ be its Pontryagin dual. Then Mazur-Rubin (2007) show that under some natural conditions, if $dim M_p$ is odd then $m_dgeq 1$ for all $d$. If $R(E/F)$ is a subgroup of $S(E/F)$ (in particular, the fine Selmer group) and we define $M^prime$ to be its Pontryagin dual etc. then can one draw conclusions as to when $M^prime_p$ will have a direct summand isomorphic to $mathbbQ_p[P]$. Maybe using the semi-simplicity of $mathbbQ_p[P]$?
– debanjana
Jul 24 at 21:56
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1
Perhaps I do not understand the question. Is the answer not just: when $m_dgeq 1$ for all $dmid p^n$? The ring $mathbfQ_p[P]$ is the direct product of fields $mathbfQ_p(zeta_d)$, $dmid p^n$.
– Keenan Kidwell
Jul 24 at 18:42
Sure, but are there some (not very strict) conditions on say the $dim_mathbbQ_p M_p$ so that we know this will happen?
– debanjana
Jul 24 at 20:04
Imposing some condition on the dimension of $M_p$ as a $mathbfQ_p$-vector space will not be enough to ensure that $M_p$ has a submodule (equivalently, direct summand) isomorphic to $mathbfQ_p[P]$. Take $M=mathbfZ_p^n$ for any $ngeq 1$ with trivial $P$-action. Then $dim_mathbfQ_p(M_p)=dim_mathbfQ_p(mathbfQ_p^n)=n$. Can you elaborate on what you might be looking for, or if there is something you specifically want to do with $M_p$?
– Keenan Kidwell
Jul 24 at 20:14
$K$ be a number field and $F/K$ be a cyclic $p$-power Galois extension. $E$ be an elliptic curve defined over $K$. $S(E/F)$ be the Selmer group and $M$ be its Pontryagin dual. Then Mazur-Rubin (2007) show that under some natural conditions, if $dim M_p$ is odd then $m_dgeq 1$ for all $d$. If $R(E/F)$ is a subgroup of $S(E/F)$ (in particular, the fine Selmer group) and we define $M^prime$ to be its Pontryagin dual etc. then can one draw conclusions as to when $M^prime_p$ will have a direct summand isomorphic to $mathbbQ_p[P]$. Maybe using the semi-simplicity of $mathbbQ_p[P]$?
– debanjana
Jul 24 at 21:56