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Associated primes of the square of a monomial ideal

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0

$begingroup$

Consider the ideal $J=(xy,yz,zx)$ in $R=mathbb C[x,y,z]$. How to show that $(x,y,z) in mathrmAss_R (R/J^2)$ ?

algebraic-geometry polynomials commutative-algebra monomial-ideals primary-decomposition

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edited Mar 21 at 19:28

user26857

39.5k124283

asked Mar 21 at 13:26

unouno

1314

$endgroup$

add a comment | 

0

$begingroup$

Consider the ideal $J=(xy,yz,zx)$ in $R=mathbb C[x,y,z]$. How to show that $(x,y,z) in mathrmAss_R (R/J^2)$ ?

algebraic-geometry polynomials commutative-algebra monomial-ideals primary-decomposition

share|cite|improve this question

edited Mar 21 at 19:28

user26857

39.5k124283

asked Mar 21 at 13:26

unouno

1314

$endgroup$

add a comment | 

$begingroup$

Consider the ideal $J=(xy,yz,zx)$ in $R=mathbb C[x,y,z]$. How to show that $(x,y,z) in mathrmAss_R (R/J^2)$ ?

algebraic-geometry polynomials commutative-algebra monomial-ideals primary-decomposition

share|cite|improve this question

edited Mar 21 at 19:28

user26857

39.5k124283

asked Mar 21 at 13:26

unouno

1314

$endgroup$

share|cite|improve this question

edited Mar 21 at 19:28

user26857

39.5k124283

asked Mar 21 at 13:26

unouno

1314

share|cite|improve this question

edited Mar 21 at 19:28

user26857

39.5k124283

asked Mar 21 at 13:26

unouno

1314

edited Mar 21 at 19:28

user26857

39.5k124283

edited Mar 21 at 19:28

user26857

39.5k124283

asked Mar 21 at 13:26

unouno

1314

asked Mar 21 at 13:26

unouno

1314

1 Answer
1

active

oldest

votes

1

$begingroup$

A down-to-earth method is: find an element whose annihilator is $(x, y, z)$.

An element that seems to work here is $xyz+J^2$. It is nonzero in $R/J^2$, since $J^2$ consists only of polynomials whose terms of degree $leq 3$ are $0$.

We have
beginalign*
x cdot xyz=x^2yz=(xy)(xz) in J^2, \
y cdot xyz=xy^2z=(xy)(yz) in J^2, \
z cdot xyz=xyz^2=(xz)(yz) in J^2,
endalign*
showing that $(x, y, z) subseteq mathrmAnn (xyz + J^2)$. On the other hand, $(x, y, z)$ is a maximal ideal and $1 notin mathrmAnn (xyz + J^2)$, so we conclude equality.

share|cite|improve this answer

edited Mar 21 at 13:48

answered Mar 21 at 13:39

Pavel ČoupekPavel Čoupek

4,57611126

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... is there a similar way to show $(x,y,z) in Ass_R (R/J^n), forall n ge 3$ ? I know this is true by some high tech method, but I would really like to see an elementary proof ...
  $endgroup$
  – uno
  Mar 21 at 21:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @uno: Sure, just take e.g. $(xy)^n-2cdot(xyz)=x^n-1y^n-1z$. This is still not $0$ in $R/J^n$ by the same degree argument, but $(x, y, z)$ kills it by the very same computation as above (simply because $(xy)^n-2 in J^n-2$), so you can again conclude that the annihilator is precisely $(x, y, z)$.
  $endgroup$
  – Pavel Čoupek
  Mar 21 at 21:58
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

A down-to-earth method is: find an element whose annihilator is $(x, y, z)$.

An element that seems to work here is $xyz+J^2$. It is nonzero in $R/J^2$, since $J^2$ consists only of polynomials whose terms of degree $leq 3$ are $0$.

We have
beginalign*
x cdot xyz=x^2yz=(xy)(xz) in J^2, \
y cdot xyz=xy^2z=(xy)(yz) in J^2, \
z cdot xyz=xyz^2=(xz)(yz) in J^2,
endalign*
showing that $(x, y, z) subseteq mathrmAnn (xyz + J^2)$. On the other hand, $(x, y, z)$ is a maximal ideal and $1 notin mathrmAnn (xyz + J^2)$, so we conclude equality.

share|cite|improve this answer

edited Mar 21 at 13:48

answered Mar 21 at 13:39

Pavel ČoupekPavel Čoupek

4,57611126

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... is there a similar way to show $(x,y,z) in Ass_R (R/J^n), forall n ge 3$ ? I know this is true by some high tech method, but I would really like to see an elementary proof ...
  $endgroup$
  – uno
  Mar 21 at 21:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @uno: Sure, just take e.g. $(xy)^n-2cdot(xyz)=x^n-1y^n-1z$. This is still not $0$ in $R/J^n$ by the same degree argument, but $(x, y, z)$ kills it by the very same computation as above (simply because $(xy)^n-2 in J^n-2$), so you can again conclude that the annihilator is precisely $(x, y, z)$.
  $endgroup$
  – Pavel Čoupek
  Mar 21 at 21:58
  
  

  
  
  
  

add a comment | 

1

$begingroup$

A down-to-earth method is: find an element whose annihilator is $(x, y, z)$.

An element that seems to work here is $xyz+J^2$. It is nonzero in $R/J^2$, since $J^2$ consists only of polynomials whose terms of degree $leq 3$ are $0$.

We have
beginalign*
x cdot xyz=x^2yz=(xy)(xz) in J^2, \
y cdot xyz=xy^2z=(xy)(yz) in J^2, \
z cdot xyz=xyz^2=(xz)(yz) in J^2,
endalign*
showing that $(x, y, z) subseteq mathrmAnn (xyz + J^2)$. On the other hand, $(x, y, z)$ is a maximal ideal and $1 notin mathrmAnn (xyz + J^2)$, so we conclude equality.

share|cite|improve this answer

edited Mar 21 at 13:48

answered Mar 21 at 13:39

Pavel ČoupekPavel Čoupek

4,57611126

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... is there a similar way to show $(x,y,z) in Ass_R (R/J^n), forall n ge 3$ ? I know this is true by some high tech method, but I would really like to see an elementary proof ...
  $endgroup$
  – uno
  Mar 21 at 21:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @uno: Sure, just take e.g. $(xy)^n-2cdot(xyz)=x^n-1y^n-1z$. This is still not $0$ in $R/J^n$ by the same degree argument, but $(x, y, z)$ kills it by the very same computation as above (simply because $(xy)^n-2 in J^n-2$), so you can again conclude that the annihilator is precisely $(x, y, z)$.
  $endgroup$
  – Pavel Čoupek
  Mar 21 at 21:58
  
  

  
  
  
  

add a comment | 

$begingroup$

A down-to-earth method is: find an element whose annihilator is $(x, y, z)$.

An element that seems to work here is $xyz+J^2$. It is nonzero in $R/J^2$, since $J^2$ consists only of polynomials whose terms of degree $leq 3$ are $0$.

We have
beginalign*
x cdot xyz=x^2yz=(xy)(xz) in J^2, \
y cdot xyz=xy^2z=(xy)(yz) in J^2, \
z cdot xyz=xyz^2=(xz)(yz) in J^2,
endalign*
showing that $(x, y, z) subseteq mathrmAnn (xyz + J^2)$. On the other hand, $(x, y, z)$ is a maximal ideal and $1 notin mathrmAnn (xyz + J^2)$, so we conclude equality.

share|cite|improve this answer

edited Mar 21 at 13:48

answered Mar 21 at 13:39

Pavel ČoupekPavel Čoupek

4,57611126

$endgroup$

share|cite|improve this answer

edited Mar 21 at 13:48

answered Mar 21 at 13:39

Pavel ČoupekPavel Čoupek

4,57611126

answered Mar 21 at 13:39

Pavel ČoupekPavel Čoupek

4,57611126

answered Mar 21 at 13:39

Pavel ČoupekPavel Čoupek

4,57611126