How do I find the associated, minimal and embedded prime ideals?
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Given the ideal $I=(x^3,xy,xz^2)subseteq k[x,y,z]$ how do I find the associated, minimal and embedded prime ideals?
I got the minimal primary decomposition to be $I=(x,y)cap (x^3,z^2)$ so that the associated primes are $(x,y)$ and $(x,z)$, by the first uniqueness theorem.
I'm not sure how I go about finding the rest. I know that the minimal prime ideals have to be the minimal of the associated prime ideals while the rest are embedded, but which is the minimal?
abstract-algebra commutative-algebra
asked Oct 15 at 18:39
Isomorphic Twin
30735
add a comment |
up vote
-1
down vote
favorite
Given the ideal $I=(x^3,xy,xz^2)subseteq k[x,y,z]$ how do I find the associated, minimal and embedded prime ideals?
I got the minimal primary decomposition to be $I=(x,y)cap (x^3,z^2)$ so that the associated primes are $(x,y)$ and $(x,z)$, by the first uniqueness theorem.
I'm not sure how I go about finding the rest. I know that the minimal prime ideals have to be the minimal of the associated prime ideals while the rest are embedded, but which is the minimal?
abstract-algebra commutative-algebra
asked Oct 15 at 18:39
Isomorphic Twin
30735
1
Something must be wrong in your decomposition, or the given ideal has other generators: $yz^2$ belongs to the decomposition, but is not in $I$.
– user26857
Oct 15 at 21:59
@user26857 Does $(x)cap (x^3,y,z^2)$ look correct? Making $(x)$ and $(x,y,z)$ the associated primes where $(x)$ is the minimal prime and $(x,y,z)$ the embedded prime?
– Isomorphic Twin
Oct 29 at 18:07
1
Yes, this sounds right.
– user26857
Oct 29 at 22:29
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Given the ideal $I=(x^3,xy,xz^2)subseteq k[x,y,z]$ how do I find the associated, minimal and embedded prime ideals?
I got the minimal primary decomposition to be $I=(x,y)cap (x^3,z^2)$ so that the associated primes are $(x,y)$ and $(x,z)$, by the first uniqueness theorem.
I'm not sure how I go about finding the rest. I know that the minimal prime ideals have to be the minimal of the associated prime ideals while the rest are embedded, but which is the minimal?
abstract-algebra commutative-algebra
asked Oct 15 at 18:39
Isomorphic Twin
30735
Given the ideal $I=(x^3,xy,xz^2)subseteq k[x,y,z]$ how do I find the associated, minimal and embedded prime ideals?
I got the minimal primary decomposition to be $I=(x,y)cap (x^3,z^2)$ so that the associated primes are $(x,y)$ and $(x,z)$, by the first uniqueness theorem.
I'm not sure how I go about finding the rest. I know that the minimal prime ideals have to be the minimal of the associated prime ideals while the rest are embedded, but which is the minimal?
abstract-algebra commutative-algebra
abstract-algebra commutative-algebra
asked Oct 15 at 18:39
Isomorphic Twin
30735
asked Oct 15 at 18:39
Isomorphic Twin
30735
asked Oct 15 at 18:39
Isomorphic Twin
30735
asked Oct 15 at 18:39
Isomorphic Twin
30735
asked Oct 15 at 18:39
Isomorphic Twin
30735
30735
1
Something must be wrong in your decomposition, or the given ideal has other generators: $yz^2$ belongs to the decomposition, but is not in $I$.
– user26857
Oct 15 at 21:59
@user26857 Does $(x)cap (x^3,y,z^2)$ look correct? Making $(x)$ and $(x,y,z)$ the associated primes where $(x)$ is the minimal prime and $(x,y,z)$ the embedded prime?
– Isomorphic Twin
Oct 29 at 18:07
1
Yes, this sounds right.
– user26857
Oct 29 at 22:29
add a comment |
1
Something must be wrong in your decomposition, or the given ideal has other generators: $yz^2$ belongs to the decomposition, but is not in $I$.
– user26857
Oct 15 at 21:59
@user26857 Does $(x)cap (x^3,y,z^2)$ look correct? Making $(x)$ and $(x,y,z)$ the associated primes where $(x)$ is the minimal prime and $(x,y,z)$ the embedded prime?
– Isomorphic Twin
Oct 29 at 18:07
1
Yes, this sounds right.
– user26857
Oct 29 at 22:29
1
1
Something must be wrong in your decomposition, or the given ideal has other generators: $yz^2$ belongs to the decomposition, but is not in $I$.
– user26857
Oct 15 at 21:59
Something must be wrong in your decomposition, or the given ideal has other generators: $yz^2$ belongs to the decomposition, but is not in $I$.
– user26857
Oct 15 at 21:59
@user26857 Does $(x)cap (x^3,y,z^2)$ look correct? Making $(x)$ and $(x,y,z)$ the associated primes where $(x)$ is the minimal prime and $(x,y,z)$ the embedded prime?
– Isomorphic Twin
Oct 29 at 18:07
@user26857 Does $(x)cap (x^3,y,z^2)$ look correct? Making $(x)$ and $(x,y,z)$ the associated primes where $(x)$ is the minimal prime and $(x,y,z)$ the embedded prime?
– Isomorphic Twin
Oct 29 at 18:07
1
1
Yes, this sounds right.
– user26857
Oct 29 at 22:29
Yes, this sounds right.
– user26857
Oct 29 at 22:29
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Just wanted to throw it out there that Macaulay2 gives a great way to find these! And there's a great online interface here. Here's the code:
You can see the language is very intuitive!
Also notice that decompose and minimalPrimes do the same thing. Hope this helps!
answered 2 days ago
Jonathan Gerhard
213
This is more a comment than an answer.
– Taroccoesbrocco
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Just wanted to throw it out there that Macaulay2 gives a great way to find these! And there's a great online interface here. Here's the code:
You can see the language is very intuitive!
Also notice that decompose and minimalPrimes do the same thing. Hope this helps!
answered 2 days ago
Jonathan Gerhard
213
This is more a comment than an answer.
– Taroccoesbrocco
2 days ago
add a comment |
up vote
0
down vote
Just wanted to throw it out there that Macaulay2 gives a great way to find these! And there's a great online interface here. Here's the code:
You can see the language is very intuitive!
Also notice that decompose and minimalPrimes do the same thing. Hope this helps!
answered 2 days ago
Jonathan Gerhard
213
This is more a comment than an answer.
– Taroccoesbrocco
2 days ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Just wanted to throw it out there that Macaulay2 gives a great way to find these! And there's a great online interface here. Here's the code:
You can see the language is very intuitive!
Also notice that decompose and minimalPrimes do the same thing. Hope this helps!
answered 2 days ago
Jonathan Gerhard
213
Just wanted to throw it out there that Macaulay2 gives a great way to find these! And there's a great online interface here. Here's the code:
You can see the language is very intuitive!
Also notice that decompose and minimalPrimes do the same thing. Hope this helps!
answered 2 days ago
Jonathan Gerhard
213
answered 2 days ago
Jonathan Gerhard
213
answered 2 days ago
Jonathan Gerhard
213
answered 2 days ago
Jonathan Gerhard
213
213
This is more a comment than an answer.
– Taroccoesbrocco
2 days ago
add a comment |
This is more a comment than an answer.
– Taroccoesbrocco
2 days ago
This is more a comment than an answer.
– Taroccoesbrocco
2 days ago
This is more a comment than an answer.
– Taroccoesbrocco
2 days ago
add a comment |
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Something must be wrong in your decomposition, or the given ideal has other generators: $yz^2$ belongs to the decomposition, but is not in $I$.
– user26857
Oct 15 at 21:59
@user26857 Does $(x)cap (x^3,y,z^2)$ look correct? Making $(x)$ and $(x,y,z)$ the associated primes where $(x)$ is the minimal prime and $(x,y,z)$ the embedded prime?
– Isomorphic Twin
Oct 29 at 18:07
1
Yes, this sounds right.
– user26857
Oct 29 at 22:29