Mixing
Primary ideals are those whose quotient has irreducible prime spectrum. [duplicate]
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This question already has an answer here:
Irreducible Spectrum implies Zero-Divisors are Nilpotent
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I'm currently trying to figure out primary ideals and I think I proved the following geometric statement, but I'm not sure if I did it right and would like to have some feedback.
Lemma: An ideal $mathfrak{a}$ of a commutative ring $A$ is primary if and only if $text{Spec}(A/mathfrak{a})$ is irreducible.
Proof: First note that $text{Spec}(A/mathfrak{a})$ is irreducible if and only if $sqrt{(0)} subset A / mathfrak{a}$ is a prime ideal. Now suppose $mathfrak{a}$ is primary, and $bar x cdot bar y insqrt{0} subset A / mathfrak{a}$. Then $x^ncdot y^n = (xy)^n in mathfrak{a}$ for some $n in mathbb{N}$, so either $x^n in mathfrak{a}$ or $(y^n)^m in mathfrak{a}$ for some $m in mathbb{N}$. I.e. either $x in sqrt{(0)}$ or $y in sqrt{(0)}$.
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$.
I'm still not 100% sure that I'm done here, because I only have $x^n in mathfrak{a}$, not $xin mathfrak{a}$. Am I missing something, or is the claim wrong? If so, I would like to see a counterexample.
abstract-algebra commutative-algebra ideals primary-decomposition
asked Jan 29 at 18:13
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Jan 30 at 20:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Irreducible Spectrum implies Zero-Divisors are Nilpotent
1 answer
I'm currently trying to figure out primary ideals and I think I proved the following geometric statement, but I'm not sure if I did it right and would like to have some feedback.
Lemma: An ideal $mathfrak{a}$ of a commutative ring $A$ is primary if and only if $text{Spec}(A/mathfrak{a})$ is irreducible.
Proof: First note that $text{Spec}(A/mathfrak{a})$ is irreducible if and only if $sqrt{(0)} subset A / mathfrak{a}$ is a prime ideal. Now suppose $mathfrak{a}$ is primary, and $bar x cdot bar y insqrt{0} subset A / mathfrak{a}$. Then $x^ncdot y^n = (xy)^n in mathfrak{a}$ for some $n in mathbb{N}$, so either $x^n in mathfrak{a}$ or $(y^n)^m in mathfrak{a}$ for some $m in mathbb{N}$. I.e. either $x in sqrt{(0)}$ or $y in sqrt{(0)}$.
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$.
I'm still not 100% sure that I'm done here, because I only have $x^n in mathfrak{a}$, not $xin mathfrak{a}$. Am I missing something, or is the claim wrong? If so, I would like to see a counterexample.
abstract-algebra commutative-algebra ideals primary-decomposition
asked Jan 29 at 18:13
red_trumpetred_trumpet
1,025319
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marked as duplicate by rschwieb abstract-algebra
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Jan 30 at 20:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$. . Are you aware that this condition ("$xyin I$ implies $x^nin I$ or $y^nin I$") is strictly weaker than being primary? You seem to be using it as if you thought it was equivalent. That's what catches my eye. The converse proof seems to fall short in this manner. I guess that's what you're getting at in the last sentence.
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– rschwieb
Jan 29 at 18:21
1
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Yeah exactly. Do you know a counterexample of an ideal where $xy in I Rightarrow (x^n in I$ or $y^n in I)$, but I is not primary?
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– red_trumpet
Jan 29 at 18:30
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Of course I do! :)
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– rschwieb
Jan 29 at 18:31
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I am not handy at all with the topology on Spec. It feels to me like it should somehow be very straightforward to say that the spectrum is irreducible since $A/mathfrak{a}$ only has one minimal prime.,
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– rschwieb
Jan 29 at 18:37
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Thanks! As far as I understand up to now, the spectrum is irreducible if and only if $mathfrak{a}$ satisfies the weaker condition above, as shown by my proof. So in particular if $mathfrak{a}$ is primary, this holds.
$endgroup$
– red_trumpet
Jan 29 at 18:48
|
show 6 more comments
$begingroup$
This question already has an answer here:
Irreducible Spectrum implies Zero-Divisors are Nilpotent
1 answer
I'm currently trying to figure out primary ideals and I think I proved the following geometric statement, but I'm not sure if I did it right and would like to have some feedback.
Lemma: An ideal $mathfrak{a}$ of a commutative ring $A$ is primary if and only if $text{Spec}(A/mathfrak{a})$ is irreducible.
Proof: First note that $text{Spec}(A/mathfrak{a})$ is irreducible if and only if $sqrt{(0)} subset A / mathfrak{a}$ is a prime ideal. Now suppose $mathfrak{a}$ is primary, and $bar x cdot bar y insqrt{0} subset A / mathfrak{a}$. Then $x^ncdot y^n = (xy)^n in mathfrak{a}$ for some $n in mathbb{N}$, so either $x^n in mathfrak{a}$ or $(y^n)^m in mathfrak{a}$ for some $m in mathbb{N}$. I.e. either $x in sqrt{(0)}$ or $y in sqrt{(0)}$.
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$.
I'm still not 100% sure that I'm done here, because I only have $x^n in mathfrak{a}$, not $xin mathfrak{a}$. Am I missing something, or is the claim wrong? If so, I would like to see a counterexample.
abstract-algebra commutative-algebra ideals primary-decomposition
asked Jan 29 at 18:13
red_trumpetred_trumpet
1,025319
$endgroup$
This question already has an answer here:
Irreducible Spectrum implies Zero-Divisors are Nilpotent
1 answer
I'm currently trying to figure out primary ideals and I think I proved the following geometric statement, but I'm not sure if I did it right and would like to have some feedback.
Lemma: An ideal $mathfrak{a}$ of a commutative ring $A$ is primary if and only if $text{Spec}(A/mathfrak{a})$ is irreducible.
Proof: First note that $text{Spec}(A/mathfrak{a})$ is irreducible if and only if $sqrt{(0)} subset A / mathfrak{a}$ is a prime ideal. Now suppose $mathfrak{a}$ is primary, and $bar x cdot bar y insqrt{0} subset A / mathfrak{a}$. Then $x^ncdot y^n = (xy)^n in mathfrak{a}$ for some $n in mathbb{N}$, so either $x^n in mathfrak{a}$ or $(y^n)^m in mathfrak{a}$ for some $m in mathbb{N}$. I.e. either $x in sqrt{(0)}$ or $y in sqrt{(0)}$.
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$.
I'm still not 100% sure that I'm done here, because I only have $x^n in mathfrak{a}$, not $xin mathfrak{a}$. Am I missing something, or is the claim wrong? If so, I would like to see a counterexample.
This question already has an answer here:
Irreducible Spectrum implies Zero-Divisors are Nilpotent
1 answer
abstract-algebra commutative-algebra ideals primary-decomposition
abstract-algebra commutative-algebra ideals primary-decomposition
asked Jan 29 at 18:13
red_trumpetred_trumpet
1,025319
asked Jan 29 at 18:13
red_trumpetred_trumpet
1,025319
asked Jan 29 at 18:13
red_trumpetred_trumpet
1,025319
asked Jan 29 at 18:13
red_trumpetred_trumpet
1,025319
asked Jan 29 at 18:13
red_trumpetred_trumpet
1,025319
1,025319
marked as duplicate by rschwieb abstract-algebra
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Jan 30 at 20:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by rschwieb abstract-algebra
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Jan 30 at 20:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$. . Are you aware that this condition ("$xyin I$ implies $x^nin I$ or $y^nin I$") is strictly weaker than being primary? You seem to be using it as if you thought it was equivalent. That's what catches my eye. The converse proof seems to fall short in this manner. I guess that's what you're getting at in the last sentence.
$endgroup$
– rschwieb
Jan 29 at 18:21
1
$begingroup$
Yeah exactly. Do you know a counterexample of an ideal where $xy in I Rightarrow (x^n in I$ or $y^n in I)$, but I is not primary?
$endgroup$
– red_trumpet
Jan 29 at 18:30
$begingroup$
Of course I do! :)
$endgroup$
– rschwieb
Jan 29 at 18:31
$begingroup$
I am not handy at all with the topology on Spec. It feels to me like it should somehow be very straightforward to say that the spectrum is irreducible since $A/mathfrak{a}$ only has one minimal prime.,
$endgroup$
– rschwieb
Jan 29 at 18:37
$begingroup$
Thanks! As far as I understand up to now, the spectrum is irreducible if and only if $mathfrak{a}$ satisfies the weaker condition above, as shown by my proof. So in particular if $mathfrak{a}$ is primary, this holds.
$endgroup$
– red_trumpet
Jan 29 at 18:48
|
show 6 more comments
1
$begingroup$
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$. . Are you aware that this condition ("$xyin I$ implies $x^nin I$ or $y^nin I$") is strictly weaker than being primary? You seem to be using it as if you thought it was equivalent. That's what catches my eye. The converse proof seems to fall short in this manner. I guess that's what you're getting at in the last sentence.
$endgroup$
– rschwieb
Jan 29 at 18:21
1
$begingroup$
Yeah exactly. Do you know a counterexample of an ideal where $xy in I Rightarrow (x^n in I$ or $y^n in I)$, but I is not primary?
$endgroup$
– red_trumpet
Jan 29 at 18:30
$begingroup$
Of course I do! :)
$endgroup$
– rschwieb
Jan 29 at 18:31
$begingroup$
I am not handy at all with the topology on Spec. It feels to me like it should somehow be very straightforward to say that the spectrum is irreducible since $A/mathfrak{a}$ only has one minimal prime.,
$endgroup$
– rschwieb
Jan 29 at 18:37
$begingroup$
Thanks! As far as I understand up to now, the spectrum is irreducible if and only if $mathfrak{a}$ satisfies the weaker condition above, as shown by my proof. So in particular if $mathfrak{a}$ is primary, this holds.
$endgroup$
– red_trumpet
Jan 29 at 18:48
1
1
$begingroup$
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$. . Are you aware that this condition ("$xyin I$ implies $x^nin I$ or $y^nin I$") is strictly weaker than being primary? You seem to be using it as if you thought it was equivalent. That's what catches my eye. The converse proof seems to fall short in this manner. I guess that's what you're getting at in the last sentence.
$endgroup$
– rschwieb
Jan 29 at 18:21
$begingroup$
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$. . Are you aware that this condition ("$xyin I$ implies $x^nin I$ or $y^nin I$") is strictly weaker than being primary? You seem to be using it as if you thought it was equivalent. That's what catches my eye. The converse proof seems to fall short in this manner. I guess that's what you're getting at in the last sentence.
$endgroup$
– rschwieb
Jan 29 at 18:21
1
1
$begingroup$
Yeah exactly. Do you know a counterexample of an ideal where $xy in I Rightarrow (x^n in I$ or $y^n in I)$, but I is not primary?
$endgroup$
– red_trumpet
Jan 29 at 18:30
$begingroup$
Yeah exactly. Do you know a counterexample of an ideal where $xy in I Rightarrow (x^n in I$ or $y^n in I)$, but I is not primary?
$endgroup$
– red_trumpet
Jan 29 at 18:30
$begingroup$
Of course I do! :)
$endgroup$
– rschwieb
Jan 29 at 18:31
$begingroup$
Of course I do! :)
$endgroup$
– rschwieb
Jan 29 at 18:31
$begingroup$
I am not handy at all with the topology on Spec. It feels to me like it should somehow be very straightforward to say that the spectrum is irreducible since $A/mathfrak{a}$ only has one minimal prime.,
$endgroup$
– rschwieb
Jan 29 at 18:37
$begingroup$
I am not handy at all with the topology on Spec. It feels to me like it should somehow be very straightforward to say that the spectrum is irreducible since $A/mathfrak{a}$ only has one minimal prime.,
$endgroup$
– rschwieb
Jan 29 at 18:37
$begingroup$
Thanks! As far as I understand up to now, the spectrum is irreducible if and only if $mathfrak{a}$ satisfies the weaker condition above, as shown by my proof. So in particular if $mathfrak{a}$ is primary, this holds.
$endgroup$
– red_trumpet
Jan 29 at 18:48
$begingroup$
Thanks! As far as I understand up to now, the spectrum is irreducible if and only if $mathfrak{a}$ satisfies the weaker condition above, as shown by my proof. So in particular if $mathfrak{a}$ is primary, this holds.
$endgroup$
– red_trumpet
Jan 29 at 18:48
|
show 6 more comments
1 Answer
1
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$begingroup$
The lemma I wrote above is wrong. The proof actually shows
The quotient $A/mathfrak{a}$ has irreducible prime spectrum if and only if $mathfrak{a}$ satisfies: $xy in mathfrak{a} implies (x^n in mathfrak{a} text{ or }y^n in mathfrak{a})$.
Clearly, any primary ideal satisfies this, so quotients of primary ideals have irreducible prime spectrum, but this condition is strictly weaker. Here is a counterexample.
A similar characterisation of primary ideals which can be found
here would be:
$mathfrak{a}$ is primary if and only if all zerodivisors of $A/mathfrak{a}$ are nilpotent.
answered Jan 30 at 13:57
red_trumpetred_trumpet
1,025319
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The lemma I wrote above is wrong. The proof actually shows
The quotient $A/mathfrak{a}$ has irreducible prime spectrum if and only if $mathfrak{a}$ satisfies: $xy in mathfrak{a} implies (x^n in mathfrak{a} text{ or }y^n in mathfrak{a})$.
Clearly, any primary ideal satisfies this, so quotients of primary ideals have irreducible prime spectrum, but this condition is strictly weaker. Here is a counterexample.
A similar characterisation of primary ideals which can be found
here would be:
$mathfrak{a}$ is primary if and only if all zerodivisors of $A/mathfrak{a}$ are nilpotent.
answered Jan 30 at 13:57
red_trumpetred_trumpet
1,025319
$endgroup$
add a comment |
$begingroup$
The lemma I wrote above is wrong. The proof actually shows
The quotient $A/mathfrak{a}$ has irreducible prime spectrum if and only if $mathfrak{a}$ satisfies: $xy in mathfrak{a} implies (x^n in mathfrak{a} text{ or }y^n in mathfrak{a})$.
Clearly, any primary ideal satisfies this, so quotients of primary ideals have irreducible prime spectrum, but this condition is strictly weaker. Here is a counterexample.
A similar characterisation of primary ideals which can be found
here would be:
$mathfrak{a}$ is primary if and only if all zerodivisors of $A/mathfrak{a}$ are nilpotent.
answered Jan 30 at 13:57
red_trumpetred_trumpet
1,025319
$endgroup$
add a comment |
$begingroup$
The lemma I wrote above is wrong. The proof actually shows
The quotient $A/mathfrak{a}$ has irreducible prime spectrum if and only if $mathfrak{a}$ satisfies: $xy in mathfrak{a} implies (x^n in mathfrak{a} text{ or }y^n in mathfrak{a})$.
Clearly, any primary ideal satisfies this, so quotients of primary ideals have irreducible prime spectrum, but this condition is strictly weaker. Here is a counterexample.
A similar characterisation of primary ideals which can be found
here would be:
$mathfrak{a}$ is primary if and only if all zerodivisors of $A/mathfrak{a}$ are nilpotent.
answered Jan 30 at 13:57
red_trumpetred_trumpet
1,025319
$endgroup$
The lemma I wrote above is wrong. The proof actually shows
The quotient $A/mathfrak{a}$ has irreducible prime spectrum if and only if $mathfrak{a}$ satisfies: $xy in mathfrak{a} implies (x^n in mathfrak{a} text{ or }y^n in mathfrak{a})$.
Clearly, any primary ideal satisfies this, so quotients of primary ideals have irreducible prime spectrum, but this condition is strictly weaker. Here is a counterexample.
A similar characterisation of primary ideals which can be found
here would be:
$mathfrak{a}$ is primary if and only if all zerodivisors of $A/mathfrak{a}$ are nilpotent.
answered Jan 30 at 13:57
red_trumpetred_trumpet
1,025319
answered Jan 30 at 13:57
red_trumpetred_trumpet
1,025319
answered Jan 30 at 13:57
red_trumpetred_trumpet
1,025319
answered Jan 30 at 13:57
red_trumpetred_trumpet
1,025319
1,025319
add a comment |
add a comment |
1
$begingroup$
Conversely, suppose $x cdot y in mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $mathfrak{a}$, i.e. $x^n in mathfrak{a}$ or $y^n in mathfrak{a}$. . Are you aware that this condition ("$xyin I$ implies $x^nin I$ or $y^nin I$") is strictly weaker than being primary? You seem to be using it as if you thought it was equivalent. That's what catches my eye. The converse proof seems to fall short in this manner. I guess that's what you're getting at in the last sentence.
$endgroup$
– rschwieb
Jan 29 at 18:21
1
$begingroup$
Yeah exactly. Do you know a counterexample of an ideal where $xy in I Rightarrow (x^n in I$ or $y^n in I)$, but I is not primary?
$endgroup$
– red_trumpet
Jan 29 at 18:30
$begingroup$
Of course I do! :)
$endgroup$
– rschwieb
Jan 29 at 18:31
$begingroup$
I am not handy at all with the topology on Spec. It feels to me like it should somehow be very straightforward to say that the spectrum is irreducible since $A/mathfrak{a}$ only has one minimal prime.,
$endgroup$
– rschwieb
Jan 29 at 18:37
$begingroup$
Thanks! As far as I understand up to now, the spectrum is irreducible if and only if $mathfrak{a}$ satisfies the weaker condition above, as shown by my proof. So in particular if $mathfrak{a}$ is primary, this holds.
$endgroup$
– red_trumpet
Jan 29 at 18:48