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0

$begingroup$

Let $R$ be a Noetherian ring and $M$ a f.g. $R$-module. Let $M'$ be a proper submodule of $M$ and let $M'=cap_i=1^n M_i$ be a primary decomposition with $M_i$ a $P_i$-primary submodule.

Part (c) of Theorem 3.10 in Eisenbud (1995 edition) on p.95 says

If the intersection is minimal, in the sense that there is no such intersection with fewer terms, then each associated prime of $M/M'$ is equal to $P_i$ for exactly one index $i$. In this case, if $P_i$ is minimal over $operatornameAnn(M/M')$, then $M_i$ is the $P_i$-primary component of $M'$.

According to the list of errata, one should replace "of $M'$." with "of $0$ in $M'$."

According to the terminology here (taken from Eisenbud), the last claim (with corrections) says that $M_i=ker(M'to M_P_i')$.

However, if we look at the proof, what is really proved there is that $M_i/M'=ker(M/M_ito (M/M_i)_P_i)$ (in fact, he only proves $M_i=ker (Mto M_P_i)$ since he assumes $M'=0$, but I don't want to assume that and I want to consider the statement in the full generality). Thus the statement "$M_i$ is the $P_i$-primary component of $0$ in $M'$" claimed in Theorem 3.10(c) is false, right? And even the statement "$M_i$ is the $P_i$-primary component of $0$ in $M$" is false? The correct statement should be "$M_i/M'=ker(M/M_ito (M/M_i)_P_i)$", I believe? Can it be expressed in words in terms of $P_i$-primary submodules?

abstract-algebra commutative-algebra modules ideals primary-decomposition

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edited 2 days ago

Bernard

122k741116

asked 2 days ago

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$endgroup$

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0

$begingroup$

Let $R$ be a Noetherian ring and $M$ a f.g. $R$-module. Let $M'$ be a proper submodule of $M$ and let $M'=cap_i=1^n M_i$ be a primary decomposition with $M_i$ a $P_i$-primary submodule.

Part (c) of Theorem 3.10 in Eisenbud (1995 edition) on p.95 says

If the intersection is minimal, in the sense that there is no such intersection with fewer terms, then each associated prime of $M/M'$ is equal to $P_i$ for exactly one index $i$. In this case, if $P_i$ is minimal over $operatornameAnn(M/M')$, then $M_i$ is the $P_i$-primary component of $M'$.

According to the list of errata, one should replace "of $M'$." with "of $0$ in $M'$."

According to the terminology here (taken from Eisenbud), the last claim (with corrections) says that $M_i=ker(M'to M_P_i')$.

However, if we look at the proof, what is really proved there is that $M_i/M'=ker(M/M_ito (M/M_i)_P_i)$ (in fact, he only proves $M_i=ker (Mto M_P_i)$ since he assumes $M'=0$, but I don't want to assume that and I want to consider the statement in the full generality). Thus the statement "$M_i$ is the $P_i$-primary component of $0$ in $M'$" claimed in Theorem 3.10(c) is false, right? And even the statement "$M_i$ is the $P_i$-primary component of $0$ in $M$" is false? The correct statement should be "$M_i/M'=ker(M/M_ito (M/M_i)_P_i)$", I believe? Can it be expressed in words in terms of $P_i$-primary submodules?

abstract-algebra commutative-algebra modules ideals primary-decomposition

share|cite|improve this question

edited 2 days ago

Bernard

122k741116

asked 2 days ago

user437309user437309

734313

$endgroup$

add a comment | 

$begingroup$

Let $R$ be a Noetherian ring and $M$ a f.g. $R$-module. Let $M'$ be a proper submodule of $M$ and let $M'=cap_i=1^n M_i$ be a primary decomposition with $M_i$ a $P_i$-primary submodule.

Part (c) of Theorem 3.10 in Eisenbud (1995 edition) on p.95 says

If the intersection is minimal, in the sense that there is no such intersection with fewer terms, then each associated prime of $M/M'$ is equal to $P_i$ for exactly one index $i$. In this case, if $P_i$ is minimal over $operatornameAnn(M/M')$, then $M_i$ is the $P_i$-primary component of $M'$.

According to the list of errata, one should replace "of $M'$." with "of $0$ in $M'$."

According to the terminology here (taken from Eisenbud), the last claim (with corrections) says that $M_i=ker(M'to M_P_i')$.

However, if we look at the proof, what is really proved there is that $M_i/M'=ker(M/M_ito (M/M_i)_P_i)$ (in fact, he only proves $M_i=ker (Mto M_P_i)$ since he assumes $M'=0$, but I don't want to assume that and I want to consider the statement in the full generality). Thus the statement "$M_i$ is the $P_i$-primary component of $0$ in $M'$" claimed in Theorem 3.10(c) is false, right? And even the statement "$M_i$ is the $P_i$-primary component of $0$ in $M$" is false? The correct statement should be "$M_i/M'=ker(M/M_ito (M/M_i)_P_i)$", I believe? Can it be expressed in words in terms of $P_i$-primary submodules?

abstract-algebra commutative-algebra modules ideals primary-decomposition

share|cite|improve this question

edited 2 days ago

Bernard

122k741116

asked 2 days ago

user437309user437309

734313

$endgroup$

share|cite|improve this question

edited 2 days ago

Bernard

122k741116

asked 2 days ago

user437309user437309

734313

share|cite|improve this question

edited 2 days ago

Bernard

122k741116

asked 2 days ago

user437309user437309

734313

edited 2 days ago

Bernard

122k741116

edited 2 days ago

Bernard

122k741116

asked 2 days ago

user437309user437309

734313

asked 2 days ago

user437309user437309

734313