Finding the associated primes of $A/I$ where $I$ has a primary decomposition (Vakil 5.5.R)
(Vakil 5.5.R.) Let $A=mathbb C[x,y]$. $Q_1=(y-x^2)$, $Q_2=(x-1,y-1)$ and $Q_3=(x-2,y-2)$. Let $$I=Q_1^{3}cap Q_2^{15} cap Q_3.$$
I want to show that the associated primes of $A/I$ are precisely $Q_1, Q_2, Q_3$.
It is easy to verify that the radical of the powers of a prime ideal is the prime itself. But I don't know how to show that this prime ideal is an associated prime of $A/I$ (an annihilator of an element of $A/I$, how to find it?).
On the other hand, I know every associated prime, as prime, must contain some $Q_i$. But why are there no other associated primes?
In respond to comments below:
$Q_1=ann((y-x^2)^2+I)$ doesn't seem to be true, because
$$(y-x^2)((y-x^2)^2+I)neq I, since~ (y-x^2)^3not in I$$
algebraic-geometry commutative-algebra
asked Nov 19 '18 at 4:09
No One
2,0041519
|
show 3 more comments
(Vakil 5.5.R.) Let $A=mathbb C[x,y]$. $Q_1=(y-x^2)$, $Q_2=(x-1,y-1)$ and $Q_3=(x-2,y-2)$. Let $$I=Q_1^{3}cap Q_2^{15} cap Q_3.$$
I want to show that the associated primes of $A/I$ are precisely $Q_1, Q_2, Q_3$.
It is easy to verify that the radical of the powers of a prime ideal is the prime itself. But I don't know how to show that this prime ideal is an associated prime of $A/I$ (an annihilator of an element of $A/I$, how to find it?).
On the other hand, I know every associated prime, as prime, must contain some $Q_i$. But why are there no other associated primes?
In respond to comments below:
$Q_1=ann((y-x^2)^2+I)$ doesn't seem to be true, because
$$(y-x^2)((y-x^2)^2+I)neq I, since~ (y-x^2)^3not in I$$
algebraic-geometry commutative-algebra
asked Nov 19 '18 at 4:09
No One
2,0041519
For your first question, what happens if you take $(y-x^2)^2$ or $(x-1)^{14}$ in this case? Can you see how they're annihilated by an element of $Q_1$ or $Q_2$, respectively?
– KReiser
Nov 19 '18 at 19:36
@KReiser I think it makes no difference whether it is $(y-x^2)^2$ or $(y-x^2)^3$, $(x-1)^{14}$ or $(x-1)^{15}$, is that right?
– No One
Nov 20 '18 at 3:15
It shows that $Q_i$ are associated primes by showing that each of the mentioned nonzero element in $A/I$ is annihilated by the corresponding prime.
– KReiser
Nov 20 '18 at 3:23
When have a primary decomposition of and ideal, then their radicals are the associated primes. (This is a standard fact in Commutative Algebra.) The only thing you have to show is that $Q_1^3$, $Q_2^{15}$, and $Q_3$ are primary.
– user26857
Nov 20 '18 at 10:00
@user26857 I think the radicals are CALLED "associated primes of primary ideals. ". On the other hand, associated primes are defined to be annihilators of $A$ modules ($A/I$ in this case). My question is about why $Q_i$'s are annihilators of elements of $A/I$.
– No One
Nov 20 '18 at 18:16
|
show 3 more comments
(Vakil 5.5.R.) Let $A=mathbb C[x,y]$. $Q_1=(y-x^2)$, $Q_2=(x-1,y-1)$ and $Q_3=(x-2,y-2)$. Let $$I=Q_1^{3}cap Q_2^{15} cap Q_3.$$
I want to show that the associated primes of $A/I$ are precisely $Q_1, Q_2, Q_3$.
It is easy to verify that the radical of the powers of a prime ideal is the prime itself. But I don't know how to show that this prime ideal is an associated prime of $A/I$ (an annihilator of an element of $A/I$, how to find it?).
On the other hand, I know every associated prime, as prime, must contain some $Q_i$. But why are there no other associated primes?
In respond to comments below:
$Q_1=ann((y-x^2)^2+I)$ doesn't seem to be true, because
$$(y-x^2)((y-x^2)^2+I)neq I, since~ (y-x^2)^3not in I$$
algebraic-geometry commutative-algebra
asked Nov 19 '18 at 4:09
No One
2,0041519
(Vakil 5.5.R.) Let $A=mathbb C[x,y]$. $Q_1=(y-x^2)$, $Q_2=(x-1,y-1)$ and $Q_3=(x-2,y-2)$. Let $$I=Q_1^{3}cap Q_2^{15} cap Q_3.$$
I want to show that the associated primes of $A/I$ are precisely $Q_1, Q_2, Q_3$.
It is easy to verify that the radical of the powers of a prime ideal is the prime itself. But I don't know how to show that this prime ideal is an associated prime of $A/I$ (an annihilator of an element of $A/I$, how to find it?).
On the other hand, I know every associated prime, as prime, must contain some $Q_i$. But why are there no other associated primes?
In respond to comments below:
$Q_1=ann((y-x^2)^2+I)$ doesn't seem to be true, because
$$(y-x^2)((y-x^2)^2+I)neq I, since~ (y-x^2)^3not in I$$
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 19 '18 at 4:09
No One
2,0041519
asked Nov 19 '18 at 4:09
No One
2,0041519
edited Nov 20 '18 at 22:46
asked Nov 19 '18 at 4:09
No One
2,0041519
asked Nov 19 '18 at 4:09
No One
2,0041519
asked Nov 19 '18 at 4:09
No One
2,0041519
2,0041519
For your first question, what happens if you take $(y-x^2)^2$ or $(x-1)^{14}$ in this case? Can you see how they're annihilated by an element of $Q_1$ or $Q_2$, respectively?
– KReiser
Nov 19 '18 at 19:36
@KReiser I think it makes no difference whether it is $(y-x^2)^2$ or $(y-x^2)^3$, $(x-1)^{14}$ or $(x-1)^{15}$, is that right?
– No One
Nov 20 '18 at 3:15
It shows that $Q_i$ are associated primes by showing that each of the mentioned nonzero element in $A/I$ is annihilated by the corresponding prime.
– KReiser
Nov 20 '18 at 3:23
When have a primary decomposition of and ideal, then their radicals are the associated primes. (This is a standard fact in Commutative Algebra.) The only thing you have to show is that $Q_1^3$, $Q_2^{15}$, and $Q_3$ are primary.
– user26857
Nov 20 '18 at 10:00
@user26857 I think the radicals are CALLED "associated primes of primary ideals. ". On the other hand, associated primes are defined to be annihilators of $A$ modules ($A/I$ in this case). My question is about why $Q_i$'s are annihilators of elements of $A/I$.
– No One
Nov 20 '18 at 18:16
|
show 3 more comments
For your first question, what happens if you take $(y-x^2)^2$ or $(x-1)^{14}$ in this case? Can you see how they're annihilated by an element of $Q_1$ or $Q_2$, respectively?
– KReiser
Nov 19 '18 at 19:36
@KReiser I think it makes no difference whether it is $(y-x^2)^2$ or $(y-x^2)^3$, $(x-1)^{14}$ or $(x-1)^{15}$, is that right?
– No One
Nov 20 '18 at 3:15
It shows that $Q_i$ are associated primes by showing that each of the mentioned nonzero element in $A/I$ is annihilated by the corresponding prime.
– KReiser
Nov 20 '18 at 3:23
When have a primary decomposition of and ideal, then their radicals are the associated primes. (This is a standard fact in Commutative Algebra.) The only thing you have to show is that $Q_1^3$, $Q_2^{15}$, and $Q_3$ are primary.
– user26857
Nov 20 '18 at 10:00
@user26857 I think the radicals are CALLED "associated primes of primary ideals. ". On the other hand, associated primes are defined to be annihilators of $A$ modules ($A/I$ in this case). My question is about why $Q_i$'s are annihilators of elements of $A/I$.
– No One
Nov 20 '18 at 18:16
For your first question, what happens if you take $(y-x^2)^2$ or $(x-1)^{14}$ in this case? Can you see how they're annihilated by an element of $Q_1$ or $Q_2$, respectively?
– KReiser
Nov 19 '18 at 19:36
For your first question, what happens if you take $(y-x^2)^2$ or $(x-1)^{14}$ in this case? Can you see how they're annihilated by an element of $Q_1$ or $Q_2$, respectively?
– KReiser
Nov 19 '18 at 19:36
@KReiser I think it makes no difference whether it is $(y-x^2)^2$ or $(y-x^2)^3$, $(x-1)^{14}$ or $(x-1)^{15}$, is that right?
– No One
Nov 20 '18 at 3:15
@KReiser I think it makes no difference whether it is $(y-x^2)^2$ or $(y-x^2)^3$, $(x-1)^{14}$ or $(x-1)^{15}$, is that right?
– No One
Nov 20 '18 at 3:15
It shows that $Q_i$ are associated primes by showing that each of the mentioned nonzero element in $A/I$ is annihilated by the corresponding prime.
– KReiser
Nov 20 '18 at 3:23
It shows that $Q_i$ are associated primes by showing that each of the mentioned nonzero element in $A/I$ is annihilated by the corresponding prime.
– KReiser
Nov 20 '18 at 3:23
When have a primary decomposition of and ideal, then their radicals are the associated primes. (This is a standard fact in Commutative Algebra.) The only thing you have to show is that $Q_1^3$, $Q_2^{15}$, and $Q_3$ are primary.
– user26857
Nov 20 '18 at 10:00
When have a primary decomposition of and ideal, then their radicals are the associated primes. (This is a standard fact in Commutative Algebra.) The only thing you have to show is that $Q_1^3$, $Q_2^{15}$, and $Q_3$ are primary.
– user26857
Nov 20 '18 at 10:00
@user26857 I think the radicals are CALLED "associated primes of primary ideals. ". On the other hand, associated primes are defined to be annihilators of $A$ modules ($A/I$ in this case). My question is about why $Q_i$'s are annihilators of elements of $A/I$.
– No One
Nov 20 '18 at 18:16
@user26857 I think the radicals are CALLED "associated primes of primary ideals. ". On the other hand, associated primes are defined to be annihilators of $A$ modules ($A/I$ in this case). My question is about why $Q_i$'s are annihilators of elements of $A/I$.
– No One
Nov 20 '18 at 18:16
|
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For your first question, what happens if you take $(y-x^2)^2$ or $(x-1)^{14}$ in this case? Can you see how they're annihilated by an element of $Q_1$ or $Q_2$, respectively?
– KReiser
Nov 19 '18 at 19:36
@KReiser I think it makes no difference whether it is $(y-x^2)^2$ or $(y-x^2)^3$, $(x-1)^{14}$ or $(x-1)^{15}$, is that right?
– No One
Nov 20 '18 at 3:15
It shows that $Q_i$ are associated primes by showing that each of the mentioned nonzero element in $A/I$ is annihilated by the corresponding prime.
– KReiser
Nov 20 '18 at 3:23
When have a primary decomposition of and ideal, then their radicals are the associated primes. (This is a standard fact in Commutative Algebra.) The only thing you have to show is that $Q_1^3$, $Q_2^{15}$, and $Q_3$ are primary.
– user26857
Nov 20 '18 at 10:00
@user26857 I think the radicals are CALLED "associated primes of primary ideals. ". On the other hand, associated primes are defined to be annihilators of $A$ modules ($A/I$ in this case). My question is about why $Q_i$'s are annihilators of elements of $A/I$.
– No One
Nov 20 '18 at 18:16