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Is there a way to extract a primary decomposition of $(0)subset R/I$ given a decomposition of $Isubset R$?

0

$begingroup$

Suppose we have a (minimal) primary decomposition of an ideal $I$ in a polynomial ring $R$ over a field; for simplicity assume $I=I_1cap I_2$ where $I_j$ is $P_j$-primary. Is there a way to extract from this a primary decomposition of $(0)$ in $R/I$? If so, say $(0)=J_1capdotscap J_k$ where $J_i$ is $Q_i$-primary.

Can I just take the equality $I=I_1cap I_2$ and replace all generators by their residues mod $I$? If so, how to justify this? How are $Q_i$ and $P_j$ related?

abstract-algebra algebraic-geometry commutative-algebra ideals primary-decomposition

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asked Apr 2 at 2:41

user437309user437309

787414

$endgroup$

add a comment | 

0

$begingroup$

Suppose we have a (minimal) primary decomposition of an ideal $I$ in a polynomial ring $R$ over a field; for simplicity assume $I=I_1cap I_2$ where $I_j$ is $P_j$-primary. Is there a way to extract from this a primary decomposition of $(0)$ in $R/I$? If so, say $(0)=J_1capdotscap J_k$ where $J_i$ is $Q_i$-primary.

Can I just take the equality $I=I_1cap I_2$ and replace all generators by their residues mod $I$? If so, how to justify this? How are $Q_i$ and $P_j$ related?

abstract-algebra algebraic-geometry commutative-algebra ideals primary-decomposition

share|cite|improve this question

asked Apr 2 at 2:41

user437309user437309

787414

$endgroup$

add a comment | 

$begingroup$

Suppose we have a (minimal) primary decomposition of an ideal $I$ in a polynomial ring $R$ over a field; for simplicity assume $I=I_1cap I_2$ where $I_j$ is $P_j$-primary. Is there a way to extract from this a primary decomposition of $(0)$ in $R/I$? If so, say $(0)=J_1capdotscap J_k$ where $J_i$ is $Q_i$-primary.

Can I just take the equality $I=I_1cap I_2$ and replace all generators by their residues mod $I$? If so, how to justify this? How are $Q_i$ and $P_j$ related?

abstract-algebra algebraic-geometry commutative-algebra ideals primary-decomposition

share|cite|improve this question

asked Apr 2 at 2:41

user437309user437309

787414

$endgroup$

share|cite|improve this question

asked Apr 2 at 2:41

user437309user437309

787414

share|cite|improve this question

asked Apr 2 at 2:41

user437309user437309

787414

asked Apr 2 at 2:41

user437309user437309

787414

asked Apr 2 at 2:41

user437309user437309

787414

1 Answer
1

active

oldest

votes

1

$begingroup$

Here are three results that together solve your problem:

1. For $Isubset R$ an ideal, there's a bijective correspondence between ideals of $R$ containing $I$ and ideals in $R/I$, given by sending $Jsupset I mapsto J/I$. (Correspondence theorem)




2. If $Isubset J$ are ideals in $R$, then $(R/I)/(I/J)cong R/J$, and the maps are the obvious maps. (Third isomorphism theorem)




3. An ideal $Isubset R$ is primary iff the only zero divisors of $R/I$ are nilpotent. (Definition of primary ideal)

So the images of the ideals $I_1$ and $I_2$ in $R/I$ are still primary, since $R/I_i cong (R/I)/(I_i/I)$ and thus the condition on zero divisors being nilpotent is still satisfied. Further, any primary decomposition of $(0)subset R/I$ gives rise bijectively to a primary decomposition of $Isubset R$ and vice-versa via 1. In short, your final line about "[replacing] all generators by their residue mod $I$" is correct.

share|cite|improve this answer

answered Apr 2 at 4:36

KReiserKReiser

10.2k21435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So in particular, my terminology, $P_k=Q_k$ for $k=1,2$ by the third isomorphism theorem? (And $J_k=I_k/I$)
  $endgroup$
  – user437309
  Apr 2 at 23:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming I've read your post right (it's a little tough in places), $P_k$ cannot equal $Q_k$ since they're ideals of different rings.
  $endgroup$
  – KReiser
  Apr 3 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But in both cases we are dealing with primary decompositions of $R$-modules (in the first case the module is $Isubset R$, in the second case it's $0subset R/I$). Shouldn't the components of either decomposition be $P$-primary ideals for some prime ideal $Psubset R$? I thought such $P$s live in one and the same ring $R$.
  $endgroup$
  – user437309
  Apr 3 at 0:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In my experience, it is typical that when one writes about primary decompositions of ideals in rings, one means the primary decomposition of that ideal in that ring. If you mean something else, you should say that specifically.
  $endgroup$
  – KReiser
  Apr 3 at 2:25
  

  
  
  
  

add a comment | 

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1

$begingroup$

Here are three results that together solve your problem:

1. For $Isubset R$ an ideal, there's a bijective correspondence between ideals of $R$ containing $I$ and ideals in $R/I$, given by sending $Jsupset I mapsto J/I$. (Correspondence theorem)




2. If $Isubset J$ are ideals in $R$, then $(R/I)/(I/J)cong R/J$, and the maps are the obvious maps. (Third isomorphism theorem)




3. An ideal $Isubset R$ is primary iff the only zero divisors of $R/I$ are nilpotent. (Definition of primary ideal)

So the images of the ideals $I_1$ and $I_2$ in $R/I$ are still primary, since $R/I_i cong (R/I)/(I_i/I)$ and thus the condition on zero divisors being nilpotent is still satisfied. Further, any primary decomposition of $(0)subset R/I$ gives rise bijectively to a primary decomposition of $Isubset R$ and vice-versa via 1. In short, your final line about "[replacing] all generators by their residue mod $I$" is correct.

share|cite|improve this answer

answered Apr 2 at 4:36

KReiserKReiser

10.2k21435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So in particular, my terminology, $P_k=Q_k$ for $k=1,2$ by the third isomorphism theorem? (And $J_k=I_k/I$)
  $endgroup$
  – user437309
  Apr 2 at 23:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming I've read your post right (it's a little tough in places), $P_k$ cannot equal $Q_k$ since they're ideals of different rings.
  $endgroup$
  – KReiser
  Apr 3 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But in both cases we are dealing with primary decompositions of $R$-modules (in the first case the module is $Isubset R$, in the second case it's $0subset R/I$). Shouldn't the components of either decomposition be $P$-primary ideals for some prime ideal $Psubset R$? I thought such $P$s live in one and the same ring $R$.
  $endgroup$
  – user437309
  Apr 3 at 0:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In my experience, it is typical that when one writes about primary decompositions of ideals in rings, one means the primary decomposition of that ideal in that ring. If you mean something else, you should say that specifically.
  $endgroup$
  – KReiser
  Apr 3 at 2:25
  

  
  
  
  

add a comment | 

1

$begingroup$

Here are three results that together solve your problem:

1. For $Isubset R$ an ideal, there's a bijective correspondence between ideals of $R$ containing $I$ and ideals in $R/I$, given by sending $Jsupset I mapsto J/I$. (Correspondence theorem)




2. If $Isubset J$ are ideals in $R$, then $(R/I)/(I/J)cong R/J$, and the maps are the obvious maps. (Third isomorphism theorem)




3. An ideal $Isubset R$ is primary iff the only zero divisors of $R/I$ are nilpotent. (Definition of primary ideal)

So the images of the ideals $I_1$ and $I_2$ in $R/I$ are still primary, since $R/I_i cong (R/I)/(I_i/I)$ and thus the condition on zero divisors being nilpotent is still satisfied. Further, any primary decomposition of $(0)subset R/I$ gives rise bijectively to a primary decomposition of $Isubset R$ and vice-versa via 1. In short, your final line about "[replacing] all generators by their residue mod $I$" is correct.

share|cite|improve this answer

answered Apr 2 at 4:36

KReiserKReiser

10.2k21435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So in particular, my terminology, $P_k=Q_k$ for $k=1,2$ by the third isomorphism theorem? (And $J_k=I_k/I$)
  $endgroup$
  – user437309
  Apr 2 at 23:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming I've read your post right (it's a little tough in places), $P_k$ cannot equal $Q_k$ since they're ideals of different rings.
  $endgroup$
  – KReiser
  Apr 3 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But in both cases we are dealing with primary decompositions of $R$-modules (in the first case the module is $Isubset R$, in the second case it's $0subset R/I$). Shouldn't the components of either decomposition be $P$-primary ideals for some prime ideal $Psubset R$? I thought such $P$s live in one and the same ring $R$.
  $endgroup$
  – user437309
  Apr 3 at 0:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In my experience, it is typical that when one writes about primary decompositions of ideals in rings, one means the primary decomposition of that ideal in that ring. If you mean something else, you should say that specifically.
  $endgroup$
  – KReiser
  Apr 3 at 2:25
  

  
  
  
  

add a comment | 

$begingroup$

Here are three results that together solve your problem:

1. For $Isubset R$ an ideal, there's a bijective correspondence between ideals of $R$ containing $I$ and ideals in $R/I$, given by sending $Jsupset I mapsto J/I$. (Correspondence theorem)




2. If $Isubset J$ are ideals in $R$, then $(R/I)/(I/J)cong R/J$, and the maps are the obvious maps. (Third isomorphism theorem)




3. An ideal $Isubset R$ is primary iff the only zero divisors of $R/I$ are nilpotent. (Definition of primary ideal)

So the images of the ideals $I_1$ and $I_2$ in $R/I$ are still primary, since $R/I_i cong (R/I)/(I_i/I)$ and thus the condition on zero divisors being nilpotent is still satisfied. Further, any primary decomposition of $(0)subset R/I$ gives rise bijectively to a primary decomposition of $Isubset R$ and vice-versa via 1. In short, your final line about "[replacing] all generators by their residue mod $I$" is correct.

share|cite|improve this answer

answered Apr 2 at 4:36

KReiserKReiser

10.2k21435

$endgroup$

share|cite|improve this answer

answered Apr 2 at 4:36

KReiserKReiser

10.2k21435

answered Apr 2 at 4:36

KReiserKReiser

10.2k21435

answered Apr 2 at 4:36

KReiserKReiser

10.2k21435