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On an auxiliary lemma for the structure theorem on Dedekind domains
Clash Royale CLAN TAG#URR8PPP
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I'm searching for a proof of the following statement
Let $M$ be a $P$-torsion module over a Dedekind domain $R$, and let $x_1 in M$ such that $operatornameAnn_R(x_1) = Ann_R(M)$. If there exist $y_2 dots y_s in M/Rx_1$ such that
$$
fracMRx_1 simeq Ry_2 oplus cdots oplus Ry_s
$$
then there exist $x_2, dots x_s in M$ such that $operatornameAnn_R(x_i) = operatornameAnn_R(y_i)$, $pi(x_i) = y_i$ with $pi: M to M/Rx_1$ the projection, and
$$
M simeq Rx_1 oplus cdots oplus Rx_s
$$
I'm somewhat familiar with the lemma for principal domains, in which for a given $y in M/Rx_1$ we first show that it can be chosen $z in M$ with $operatornameAnn_R(z) = operatornameAnn_R(y)$ and $pi(z) = y$, from which the ulterior condition follows.
However, the aforementioned construction uses strongly that if $operatornameAnn_R(M) = p^mR$, then $p^m$ annihilates $M/Rx_1$ and so $p^my = ax_1$ for some $a in R$.
My knowledge on Dedekind domains is very rudimentary so I would really appreciate if you could either provide a proof of this statement or a reference in which I could find one. Maybe the same be done for these type of rings? That is, if $operatornameAnn_R(M) = mathcalP^m$, can we find $p in mathcalP$ with $p^my = ax_1$? I think I could take it from there following the proof for PIDs.
modules proof-explanation dedekind-domain
edited Jul 29 at 11:14
Fabio Lucchini
5,56911025
asked Jul 28 at 0:06
Guido A.
3,642624
add a comment |Â
up vote
2
down vote
favorite
I'm searching for a proof of the following statement
Let $M$ be a $P$-torsion module over a Dedekind domain $R$, and let $x_1 in M$ such that $operatornameAnn_R(x_1) = Ann_R(M)$. If there exist $y_2 dots y_s in M/Rx_1$ such that
$$
fracMRx_1 simeq Ry_2 oplus cdots oplus Ry_s
$$
then there exist $x_2, dots x_s in M$ such that $operatornameAnn_R(x_i) = operatornameAnn_R(y_i)$, $pi(x_i) = y_i$ with $pi: M to M/Rx_1$ the projection, and
$$
M simeq Rx_1 oplus cdots oplus Rx_s
$$
I'm somewhat familiar with the lemma for principal domains, in which for a given $y in M/Rx_1$ we first show that it can be chosen $z in M$ with $operatornameAnn_R(z) = operatornameAnn_R(y)$ and $pi(z) = y$, from which the ulterior condition follows.
However, the aforementioned construction uses strongly that if $operatornameAnn_R(M) = p^mR$, then $p^m$ annihilates $M/Rx_1$ and so $p^my = ax_1$ for some $a in R$.
My knowledge on Dedekind domains is very rudimentary so I would really appreciate if you could either provide a proof of this statement or a reference in which I could find one. Maybe the same be done for these type of rings? That is, if $operatornameAnn_R(M) = mathcalP^m$, can we find $p in mathcalP$ with $p^my = ax_1$? I think I could take it from there following the proof for PIDs.
modules proof-explanation dedekind-domain
edited Jul 29 at 11:14
Fabio Lucchini
5,56911025
asked Jul 28 at 0:06
Guido A.
3,642624
Try to localize at $P $.
– Fabio Lucchini
Jul 28 at 6:53
@FabioLucchini would you mind expandind on that a bit? Unfortunately I haven't learned about localization yet.
– Guido A.
Jul 28 at 16:23
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm searching for a proof of the following statement
Let $M$ be a $P$-torsion module over a Dedekind domain $R$, and let $x_1 in M$ such that $operatornameAnn_R(x_1) = Ann_R(M)$. If there exist $y_2 dots y_s in M/Rx_1$ such that
$$
fracMRx_1 simeq Ry_2 oplus cdots oplus Ry_s
$$
then there exist $x_2, dots x_s in M$ such that $operatornameAnn_R(x_i) = operatornameAnn_R(y_i)$, $pi(x_i) = y_i$ with $pi: M to M/Rx_1$ the projection, and
$$
M simeq Rx_1 oplus cdots oplus Rx_s
$$
I'm somewhat familiar with the lemma for principal domains, in which for a given $y in M/Rx_1$ we first show that it can be chosen $z in M$ with $operatornameAnn_R(z) = operatornameAnn_R(y)$ and $pi(z) = y$, from which the ulterior condition follows.
However, the aforementioned construction uses strongly that if $operatornameAnn_R(M) = p^mR$, then $p^m$ annihilates $M/Rx_1$ and so $p^my = ax_1$ for some $a in R$.
My knowledge on Dedekind domains is very rudimentary so I would really appreciate if you could either provide a proof of this statement or a reference in which I could find one. Maybe the same be done for these type of rings? That is, if $operatornameAnn_R(M) = mathcalP^m$, can we find $p in mathcalP$ with $p^my = ax_1$? I think I could take it from there following the proof for PIDs.
modules proof-explanation dedekind-domain
edited Jul 29 at 11:14
Fabio Lucchini
5,56911025
asked Jul 28 at 0:06
Guido A.
3,642624
I'm searching for a proof of the following statement
Let $M$ be a $P$-torsion module over a Dedekind domain $R$, and let $x_1 in M$ such that $operatornameAnn_R(x_1) = Ann_R(M)$. If there exist $y_2 dots y_s in M/Rx_1$ such that
$$
fracMRx_1 simeq Ry_2 oplus cdots oplus Ry_s
$$
then there exist $x_2, dots x_s in M$ such that $operatornameAnn_R(x_i) = operatornameAnn_R(y_i)$, $pi(x_i) = y_i$ with $pi: M to M/Rx_1$ the projection, and
$$
M simeq Rx_1 oplus cdots oplus Rx_s
$$
I'm somewhat familiar with the lemma for principal domains, in which for a given $y in M/Rx_1$ we first show that it can be chosen $z in M$ with $operatornameAnn_R(z) = operatornameAnn_R(y)$ and $pi(z) = y$, from which the ulterior condition follows.
However, the aforementioned construction uses strongly that if $operatornameAnn_R(M) = p^mR$, then $p^m$ annihilates $M/Rx_1$ and so $p^my = ax_1$ for some $a in R$.
My knowledge on Dedekind domains is very rudimentary so I would really appreciate if you could either provide a proof of this statement or a reference in which I could find one. Maybe the same be done for these type of rings? That is, if $operatornameAnn_R(M) = mathcalP^m$, can we find $p in mathcalP$ with $p^my = ax_1$? I think I could take it from there following the proof for PIDs.
modules proof-explanation dedekind-domain
edited Jul 29 at 11:14
Fabio Lucchini
5,56911025
asked Jul 28 at 0:06
Guido A.
3,642624
edited Jul 29 at 11:14
Fabio Lucchini
5,56911025
edited Jul 29 at 11:14
Fabio Lucchini
5,56911025
edited Jul 29 at 11:14
Fabio Lucchini
5,56911025
5,56911025
asked Jul 28 at 0:06
Guido A.
3,642624
asked Jul 28 at 0:06
Guido A.
3,642624
asked Jul 28 at 0:06
Guido A.
3,642624
3,642624
Try to localize at $P $.
– Fabio Lucchini
Jul 28 at 6:53
@FabioLucchini would you mind expandind on that a bit? Unfortunately I haven't learned about localization yet.
– Guido A.
Jul 28 at 16:23
add a comment |Â
Try to localize at $P $.
– Fabio Lucchini
Jul 28 at 6:53
@FabioLucchini would you mind expandind on that a bit? Unfortunately I haven't learned about localization yet.
– Guido A.
Jul 28 at 16:23
Try to localize at $P $.
– Fabio Lucchini
Jul 28 at 6:53
Try to localize at $P $.
– Fabio Lucchini
Jul 28 at 6:53
@FabioLucchini would you mind expandind on that a bit? Unfortunately I haven't learned about localization yet.
– Guido A.
Jul 28 at 16:23
@FabioLucchini would you mind expandind on that a bit? Unfortunately I haven't learned about localization yet.
– Guido A.
Jul 28 at 16:23
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$newcommandideal[1]mathfrak#1DeclareMathOperatorAnnAnn$Let $R$ be a Dedekind domain, $ideal p$ be a prime ideal in $R$ and $M$ be a $ideal p$-torsion $R$-module and $x_1in M$ such that $Ann_R(x_1)=Ann_R(M)$.
Then $Ann_R(M)supseteqideal p^n$ for some $ninBbb N$, hence $M$ can regarded as a $R/ideal p^n$-module.
Let $R_ideal p$ denote the localization at $ideal p$ of $R$.
Then $R_ideal p$ is a PID, $ideal m=R_ideal pideal p$ is a maximal ideal and $R_ideal p/ideal m^ncong R/ideal p^n$.
Consequently, $M$ is a $ideal m$-torsion $R_ideal p$-module and hence the proof valid for PIDs applyies.
answered Jul 29 at 12:05
Fabio Lucchini
5,56911025
1
Hi Fabio, thanks a lot for taking the time to answer, I really appreciate it. This is a lot to digest for me, since I'm not very familiar with these concepts. I'll accept the answer as soon as I get the time to give this a careful read, whether I fully comprehend it or not.
– Guido A.
Jul 29 at 18:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$newcommandideal[1]mathfrak#1DeclareMathOperatorAnnAnn$Let $R$ be a Dedekind domain, $ideal p$ be a prime ideal in $R$ and $M$ be a $ideal p$-torsion $R$-module and $x_1in M$ such that $Ann_R(x_1)=Ann_R(M)$.
Then $Ann_R(M)supseteqideal p^n$ for some $ninBbb N$, hence $M$ can regarded as a $R/ideal p^n$-module.
Let $R_ideal p$ denote the localization at $ideal p$ of $R$.
Then $R_ideal p$ is a PID, $ideal m=R_ideal pideal p$ is a maximal ideal and $R_ideal p/ideal m^ncong R/ideal p^n$.
Consequently, $M$ is a $ideal m$-torsion $R_ideal p$-module and hence the proof valid for PIDs applyies.
answered Jul 29 at 12:05
Fabio Lucchini
5,56911025
1
Hi Fabio, thanks a lot for taking the time to answer, I really appreciate it. This is a lot to digest for me, since I'm not very familiar with these concepts. I'll accept the answer as soon as I get the time to give this a careful read, whether I fully comprehend it or not.
– Guido A.
Jul 29 at 18:39
add a comment |Â
up vote
1
down vote
accepted
$newcommandideal[1]mathfrak#1DeclareMathOperatorAnnAnn$Let $R$ be a Dedekind domain, $ideal p$ be a prime ideal in $R$ and $M$ be a $ideal p$-torsion $R$-module and $x_1in M$ such that $Ann_R(x_1)=Ann_R(M)$.
Then $Ann_R(M)supseteqideal p^n$ for some $ninBbb N$, hence $M$ can regarded as a $R/ideal p^n$-module.
Let $R_ideal p$ denote the localization at $ideal p$ of $R$.
Then $R_ideal p$ is a PID, $ideal m=R_ideal pideal p$ is a maximal ideal and $R_ideal p/ideal m^ncong R/ideal p^n$.
Consequently, $M$ is a $ideal m$-torsion $R_ideal p$-module and hence the proof valid for PIDs applyies.
answered Jul 29 at 12:05
Fabio Lucchini
5,56911025
1
Hi Fabio, thanks a lot for taking the time to answer, I really appreciate it. This is a lot to digest for me, since I'm not very familiar with these concepts. I'll accept the answer as soon as I get the time to give this a careful read, whether I fully comprehend it or not.
– Guido A.
Jul 29 at 18:39
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$newcommandideal[1]mathfrak#1DeclareMathOperatorAnnAnn$Let $R$ be a Dedekind domain, $ideal p$ be a prime ideal in $R$ and $M$ be a $ideal p$-torsion $R$-module and $x_1in M$ such that $Ann_R(x_1)=Ann_R(M)$.
Then $Ann_R(M)supseteqideal p^n$ for some $ninBbb N$, hence $M$ can regarded as a $R/ideal p^n$-module.
Let $R_ideal p$ denote the localization at $ideal p$ of $R$.
Then $R_ideal p$ is a PID, $ideal m=R_ideal pideal p$ is a maximal ideal and $R_ideal p/ideal m^ncong R/ideal p^n$.
Consequently, $M$ is a $ideal m$-torsion $R_ideal p$-module and hence the proof valid for PIDs applyies.
answered Jul 29 at 12:05
Fabio Lucchini
5,56911025
$newcommandideal[1]mathfrak#1DeclareMathOperatorAnnAnn$Let $R$ be a Dedekind domain, $ideal p$ be a prime ideal in $R$ and $M$ be a $ideal p$-torsion $R$-module and $x_1in M$ such that $Ann_R(x_1)=Ann_R(M)$.
Then $Ann_R(M)supseteqideal p^n$ for some $ninBbb N$, hence $M$ can regarded as a $R/ideal p^n$-module.
Let $R_ideal p$ denote the localization at $ideal p$ of $R$.
Then $R_ideal p$ is a PID, $ideal m=R_ideal pideal p$ is a maximal ideal and $R_ideal p/ideal m^ncong R/ideal p^n$.
Consequently, $M$ is a $ideal m$-torsion $R_ideal p$-module and hence the proof valid for PIDs applyies.
answered Jul 29 at 12:05
Fabio Lucchini
5,56911025
edited Jul 29 at 16:21
answered Jul 29 at 12:05
Fabio Lucchini
5,56911025
answered Jul 29 at 12:05
Fabio Lucchini
5,56911025
answered Jul 29 at 12:05
Fabio Lucchini
5,56911025
5,56911025
1
Hi Fabio, thanks a lot for taking the time to answer, I really appreciate it. This is a lot to digest for me, since I'm not very familiar with these concepts. I'll accept the answer as soon as I get the time to give this a careful read, whether I fully comprehend it or not.
– Guido A.
Jul 29 at 18:39
add a comment |Â
1
Hi Fabio, thanks a lot for taking the time to answer, I really appreciate it. This is a lot to digest for me, since I'm not very familiar with these concepts. I'll accept the answer as soon as I get the time to give this a careful read, whether I fully comprehend it or not.
– Guido A.
Jul 29 at 18:39
1
1
Hi Fabio, thanks a lot for taking the time to answer, I really appreciate it. This is a lot to digest for me, since I'm not very familiar with these concepts. I'll accept the answer as soon as I get the time to give this a careful read, whether I fully comprehend it or not.
– Guido A.
Jul 29 at 18:39
Hi Fabio, thanks a lot for taking the time to answer, I really appreciate it. This is a lot to digest for me, since I'm not very familiar with these concepts. I'll accept the answer as soon as I get the time to give this a careful read, whether I fully comprehend it or not.
– Guido A.
Jul 29 at 18:39
add a comment |Â
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Try to localize at $P $.
– Fabio Lucchini
Jul 28 at 6:53
@FabioLucchini would you mind expandind on that a bit? Unfortunately I haven't learned about localization yet.
– Guido A.
Jul 28 at 16:23