Mixing
Embedded primes in a finitely generated algebra over an algebraically closed field
$begingroup$
Let $F$ be an algebraically closed field. Consider the polynomial ring $F[X,Y,Z]$ and its quotient by the ideal $J=(X^aZ, XY^b)$. I want to find all the $(a,b)$, for $a,b$ positive integers, such that there are no embedded primes in the primary decomposition of zero.
Unfortunately, I don't know how to use the fact that F is alg. closed to answer the question.
I computed that, for any a, b, the primary decomposition of $J$ will be the intersection of $(X)$, $(X^a,Y^b)$, $(Z,Y^b)$ (do you agree?), and that the minimal primes are $(X)$ and $(Z,Y)$. I nowhere used the hypothesis of $F$ alg. closed. The embedded prime shouldn’t be $(X,Y)$? I'm definitely missing something here...
algebraic-geometry polynomials commutative-algebra field-theory primary-decomposition
edited Jan 26 at 20:56
user26857
39.4k124183
asked Jan 26 at 15:21
GLeGLe
455
$endgroup$
add a comment |
$begingroup$
Let $F$ be an algebraically closed field. Consider the polynomial ring $F[X,Y,Z]$ and its quotient by the ideal $J=(X^aZ, XY^b)$. I want to find all the $(a,b)$, for $a,b$ positive integers, such that there are no embedded primes in the primary decomposition of zero.
Unfortunately, I don't know how to use the fact that F is alg. closed to answer the question.
I computed that, for any a, b, the primary decomposition of $J$ will be the intersection of $(X)$, $(X^a,Y^b)$, $(Z,Y^b)$ (do you agree?), and that the minimal primes are $(X)$ and $(Z,Y)$. I nowhere used the hypothesis of $F$ alg. closed. The embedded prime shouldn’t be $(X,Y)$? I'm definitely missing something here...
algebraic-geometry polynomials commutative-algebra field-theory primary-decomposition
edited Jan 26 at 20:56
user26857
39.4k124183
asked Jan 26 at 15:21
GLeGLe
455
$endgroup$
$begingroup$
In $R = F[X,Y,Z]/(X^a,Y^b)$ the zero-divisors are $XR+YR$ and they are all nilpotent, so $(X^a,Y^b)$ is $(X^a,Y^b,(X,Y)) = (X,Y)$ primary ?
$endgroup$
– reuns
Jan 26 at 16:27
$begingroup$
So you agree with me? But I must be not getting something because otherwise why should there be the hypothesis of the alg closed field? And why should the embedded prime depend on a,b ? If my considerations are right, there is only one and it does not depend on a and b... so what am i missing?
$endgroup$
– GLe
Jan 26 at 19:06
$begingroup$
I agree with your primary decomposition. However, if $a>1$ there is an embedded prime (which one?), while for $a=1$ this doesn't happen. ($F$ algebraically closed is not necessary.)
$endgroup$
– user26857
Jan 26 at 20:47
$begingroup$
Ah yes, for a=1 there are no embedded primes because $(X)$ is minimal right? But still it is very strange that the alg closed hypothesis is not necessary....
$endgroup$
– GLe
Jan 26 at 20:52
add a comment |
$begingroup$
Let $F$ be an algebraically closed field. Consider the polynomial ring $F[X,Y,Z]$ and its quotient by the ideal $J=(X^aZ, XY^b)$. I want to find all the $(a,b)$, for $a,b$ positive integers, such that there are no embedded primes in the primary decomposition of zero.
Unfortunately, I don't know how to use the fact that F is alg. closed to answer the question.
I computed that, for any a, b, the primary decomposition of $J$ will be the intersection of $(X)$, $(X^a,Y^b)$, $(Z,Y^b)$ (do you agree?), and that the minimal primes are $(X)$ and $(Z,Y)$. I nowhere used the hypothesis of $F$ alg. closed. The embedded prime shouldn’t be $(X,Y)$? I'm definitely missing something here...
algebraic-geometry polynomials commutative-algebra field-theory primary-decomposition
edited Jan 26 at 20:56
user26857
39.4k124183
asked Jan 26 at 15:21
GLeGLe
455
$endgroup$
Let $F$ be an algebraically closed field. Consider the polynomial ring $F[X,Y,Z]$ and its quotient by the ideal $J=(X^aZ, XY^b)$. I want to find all the $(a,b)$, for $a,b$ positive integers, such that there are no embedded primes in the primary decomposition of zero.
Unfortunately, I don't know how to use the fact that F is alg. closed to answer the question.
I computed that, for any a, b, the primary decomposition of $J$ will be the intersection of $(X)$, $(X^a,Y^b)$, $(Z,Y^b)$ (do you agree?), and that the minimal primes are $(X)$ and $(Z,Y)$. I nowhere used the hypothesis of $F$ alg. closed. The embedded prime shouldn’t be $(X,Y)$? I'm definitely missing something here...
algebraic-geometry polynomials commutative-algebra field-theory primary-decomposition
algebraic-geometry polynomials commutative-algebra field-theory primary-decomposition
edited Jan 26 at 20:56
user26857
39.4k124183
asked Jan 26 at 15:21
GLeGLe
455
edited Jan 26 at 20:56
user26857
39.4k124183
asked Jan 26 at 15:21
GLeGLe
455
edited Jan 26 at 20:56
user26857
39.4k124183
edited Jan 26 at 20:56
user26857
39.4k124183
edited Jan 26 at 20:56
user26857
39.4k124183
39.4k124183
asked Jan 26 at 15:21
GLeGLe
455
asked Jan 26 at 15:21
GLeGLe
455
asked Jan 26 at 15:21
GLeGLe
455
455
$begingroup$
In $R = F[X,Y,Z]/(X^a,Y^b)$ the zero-divisors are $XR+YR$ and they are all nilpotent, so $(X^a,Y^b)$ is $(X^a,Y^b,(X,Y)) = (X,Y)$ primary ?
$endgroup$
– reuns
Jan 26 at 16:27
$begingroup$
So you agree with me? But I must be not getting something because otherwise why should there be the hypothesis of the alg closed field? And why should the embedded prime depend on a,b ? If my considerations are right, there is only one and it does not depend on a and b... so what am i missing?
$endgroup$
– GLe
Jan 26 at 19:06
$begingroup$
I agree with your primary decomposition. However, if $a>1$ there is an embedded prime (which one?), while for $a=1$ this doesn't happen. ($F$ algebraically closed is not necessary.)
$endgroup$
– user26857
Jan 26 at 20:47
$begingroup$
Ah yes, for a=1 there are no embedded primes because $(X)$ is minimal right? But still it is very strange that the alg closed hypothesis is not necessary....
$endgroup$
– GLe
Jan 26 at 20:52
add a comment |
$begingroup$
In $R = F[X,Y,Z]/(X^a,Y^b)$ the zero-divisors are $XR+YR$ and they are all nilpotent, so $(X^a,Y^b)$ is $(X^a,Y^b,(X,Y)) = (X,Y)$ primary ?
$endgroup$
– reuns
Jan 26 at 16:27
$begingroup$
So you agree with me? But I must be not getting something because otherwise why should there be the hypothesis of the alg closed field? And why should the embedded prime depend on a,b ? If my considerations are right, there is only one and it does not depend on a and b... so what am i missing?
$endgroup$
– GLe
Jan 26 at 19:06
$begingroup$
I agree with your primary decomposition. However, if $a>1$ there is an embedded prime (which one?), while for $a=1$ this doesn't happen. ($F$ algebraically closed is not necessary.)
$endgroup$
– user26857
Jan 26 at 20:47
$begingroup$
Ah yes, for a=1 there are no embedded primes because $(X)$ is minimal right? But still it is very strange that the alg closed hypothesis is not necessary....
$endgroup$
– GLe
Jan 26 at 20:52
$begingroup$
In $R = F[X,Y,Z]/(X^a,Y^b)$ the zero-divisors are $XR+YR$ and they are all nilpotent, so $(X^a,Y^b)$ is $(X^a,Y^b,(X,Y)) = (X,Y)$ primary ?
$endgroup$
– reuns
Jan 26 at 16:27
$begingroup$
In $R = F[X,Y,Z]/(X^a,Y^b)$ the zero-divisors are $XR+YR$ and they are all nilpotent, so $(X^a,Y^b)$ is $(X^a,Y^b,(X,Y)) = (X,Y)$ primary ?
$endgroup$
– reuns
Jan 26 at 16:27
$begingroup$
So you agree with me? But I must be not getting something because otherwise why should there be the hypothesis of the alg closed field? And why should the embedded prime depend on a,b ? If my considerations are right, there is only one and it does not depend on a and b... so what am i missing?
$endgroup$
– GLe
Jan 26 at 19:06
$begingroup$
So you agree with me? But I must be not getting something because otherwise why should there be the hypothesis of the alg closed field? And why should the embedded prime depend on a,b ? If my considerations are right, there is only one and it does not depend on a and b... so what am i missing?
$endgroup$
– GLe
Jan 26 at 19:06
$begingroup$
I agree with your primary decomposition. However, if $a>1$ there is an embedded prime (which one?), while for $a=1$ this doesn't happen. ($F$ algebraically closed is not necessary.)
$endgroup$
– user26857
Jan 26 at 20:47
$begingroup$
I agree with your primary decomposition. However, if $a>1$ there is an embedded prime (which one?), while for $a=1$ this doesn't happen. ($F$ algebraically closed is not necessary.)
$endgroup$
– user26857
Jan 26 at 20:47
$begingroup$
Ah yes, for a=1 there are no embedded primes because $(X)$ is minimal right? But still it is very strange that the alg closed hypothesis is not necessary....
$endgroup$
– GLe
Jan 26 at 20:52
$begingroup$
Ah yes, for a=1 there are no embedded primes because $(X)$ is minimal right? But still it is very strange that the alg closed hypothesis is not necessary....
$endgroup$
– GLe
Jan 26 at 20:52
add a comment |
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$begingroup$
In $R = F[X,Y,Z]/(X^a,Y^b)$ the zero-divisors are $XR+YR$ and they are all nilpotent, so $(X^a,Y^b)$ is $(X^a,Y^b,(X,Y)) = (X,Y)$ primary ?
$endgroup$
– reuns
Jan 26 at 16:27
$begingroup$
So you agree with me? But I must be not getting something because otherwise why should there be the hypothesis of the alg closed field? And why should the embedded prime depend on a,b ? If my considerations are right, there is only one and it does not depend on a and b... so what am i missing?
$endgroup$
– GLe
Jan 26 at 19:06
$begingroup$
I agree with your primary decomposition. However, if $a>1$ there is an embedded prime (which one?), while for $a=1$ this doesn't happen. ($F$ algebraically closed is not necessary.)
$endgroup$
– user26857
Jan 26 at 20:47
$begingroup$
Ah yes, for a=1 there are no embedded primes because $(X)$ is minimal right? But still it is very strange that the alg closed hypothesis is not necessary....
$endgroup$
– GLe
Jan 26 at 20:52