Ideal with zero localizations at prime ideals containing it The 2019 Stack Overflow Developer...
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Ideal with zero localizations at prime ideals containing it
The 2019 Stack Overflow Developer Survey Results Are InDo closed immersions preserve stalks?Non-Noetherian rings with an ideal not containing a product of prime idealsDoes every Krull ring have a height 1 prime ideal?Expressing ideals as products of prime ideals in a commutative, Noetherian ring with unityExistence of minimal non-zero prime ideals: Counter examples?Is a graded module over a graded ring zero when all of it's graded localizations at graded primes not containing the irrelevant ideal are zero?Minimal ideals of localizationsCommutative rings with unity over which every non-zero module has an associated primeCommutative Noetherian, local, reduced ring has only one minimal prime ideal?Zero divisors and minimal prime ideals in commutative ringExample of a principal prime ideal containing a proper prime non-zero ideal.
$begingroup$
Let $R$ be a commutative unital ring. I know that if an $R$-module has zero localizations at all prime ideals of $R$, then it is a zero module.
Consider a proper ideal $Isubset R$ as an $R$-module. Is it true that if localizations of this module at all prime ideals containing $I$ are zero, then it is a zero module?
This is true for $R$ a DVR (because there is only one non-zero prime ideal which contains all proper ideals).
commutative-algebra localization
asked yesterday
Stepan BanachStepan Banach
1397
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative unital ring. I know that if an $R$-module has zero localizations at all prime ideals of $R$, then it is a zero module.
Consider a proper ideal $Isubset R$ as an $R$-module. Is it true that if localizations of this module at all prime ideals containing $I$ are zero, then it is a zero module?
This is true for $R$ a DVR (because there is only one non-zero prime ideal which contains all proper ideals).
commutative-algebra localization
asked yesterday
Stepan BanachStepan Banach
1397
$endgroup$
$begingroup$
Is $I$ finitely generated ? Suppose $M$ is a finitely generated $R-module$. Then $M=0 iff M_p=0 forall p in Spec R$
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
@SoumikGhosh No finiteness hypotheses
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
@SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
$endgroup$
– Alex Mathers
yesterday
add a comment |
$begingroup$
Let $R$ be a commutative unital ring. I know that if an $R$-module has zero localizations at all prime ideals of $R$, then it is a zero module.
Consider a proper ideal $Isubset R$ as an $R$-module. Is it true that if localizations of this module at all prime ideals containing $I$ are zero, then it is a zero module?
This is true for $R$ a DVR (because there is only one non-zero prime ideal which contains all proper ideals).
commutative-algebra localization
asked yesterday
Stepan BanachStepan Banach
1397
$endgroup$
Let $R$ be a commutative unital ring. I know that if an $R$-module has zero localizations at all prime ideals of $R$, then it is a zero module.
Consider a proper ideal $Isubset R$ as an $R$-module. Is it true that if localizations of this module at all prime ideals containing $I$ are zero, then it is a zero module?
This is true for $R$ a DVR (because there is only one non-zero prime ideal which contains all proper ideals).
commutative-algebra localization
commutative-algebra localization
asked yesterday
Stepan BanachStepan Banach
1397
asked yesterday
Stepan BanachStepan Banach
1397
edited yesterday
Stepan Banach
asked yesterday
Stepan BanachStepan Banach
1397
asked yesterday
Stepan BanachStepan Banach
1397
asked yesterday
Stepan BanachStepan Banach
1397
1397
$begingroup$
Is $I$ finitely generated ? Suppose $M$ is a finitely generated $R-module$. Then $M=0 iff M_p=0 forall p in Spec R$
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
@SoumikGhosh No finiteness hypotheses
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
@SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
$endgroup$
– Alex Mathers
yesterday
add a comment |
$begingroup$
Is $I$ finitely generated ? Suppose $M$ is a finitely generated $R-module$. Then $M=0 iff M_p=0 forall p in Spec R$
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
@SoumikGhosh No finiteness hypotheses
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
@SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
$endgroup$
– Alex Mathers
yesterday
$begingroup$
Is $I$ finitely generated ? Suppose $M$ is a finitely generated $R-module$. Then $M=0 iff M_p=0 forall p in Spec R$
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
Is $I$ finitely generated ? Suppose $M$ is a finitely generated $R-module$. Then $M=0 iff M_p=0 forall p in Spec R$
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
@SoumikGhosh No finiteness hypotheses
$endgroup$
– Stepan Banach
yesterday
$begingroup$
@SoumikGhosh No finiteness hypotheses
$endgroup$
– Stepan Banach
yesterday
1
1
$begingroup$
@SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
$endgroup$
– Alex Mathers
yesterday
$begingroup$
@SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
$endgroup$
– Alex Mathers
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Counterexamples exist. In fact:
Proposition. TFAE:
$I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$
- $forall x in I : Ann(x) + I = R$
- $forall x in I : exists y in I : xy=x$
In particular, a nonzero proper ideal in a boolean ring is a counterexample.
Proof. $1 implies 2$: Take $x in I$. Suppose there is a prime ideal $mathfrak p$ containing $Ann(x) + I$. Localizing at $mathfrak p$, we find $r in R - mathfrak p$ with $rx= 0$. This contradicts $Ann(x) subset mathfrak p$.
$2 implies 3$: Take $a in R$ and $y in I$ with $a+y = 1$ and $ax = 0$. Then $xy = x$.
$3 implies 1$: Take $x in I$, $y in I$ with $xy = x$ and take $mathfrak p supset I$. Because $(1-y)x = 0$ and $1-y notin mathfrak p$, $x$ becomes $0$ in $I_{mathfrak p}$. $square$
Milking this, we find:
- such $I$ consists of zero divisors.
- There are no counterexamples when $R$ is an integral domain.
- If $I$ is nonzero, proper and finitely generated, it contains a nontrivial idempotent. Proof: From 2. it follows that $Ann(I) + I = R$. Write $a + x = 1$ with $aI = 0$ and $x in I$. Then $x^2=x$.
Since you asked about $Spec(R)$ connected and reduced, here's an example (necessarily not Noetherian). Take a field $k$, let $R = k[X_1, X_2, ldots]$ modulo the ideal generated by the $X_iX_j-X_i$ for $j>i$, and let $I$ be the ideal generated by the $X_i$.
answered yesterday
rabotarabota
14.6k32885
$endgroup$
$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I do not completely understand this. All this proves is that tensoring with $mathbb{Z}[x]/(x^2)$ is not going to work. There still might be an irreducible, non-reduced example constructed in another way. Or did you mean something else?
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced)
$endgroup$
– rabota
yesterday
add a comment |
$begingroup$
Consider the ring $$R= mathbb Z_2 times mathbb Z_2 times... times mathbb Z_2 times.....$$ countable number of times .
Look at the ideal $$ I= bigoplus _{mathbb N} mathbb Z_2$$
I claim $I_p=0 forall p in Spec R$ containing $I$.
Say $I subset p$ Consider $ain I$. Observe that $a^2=a$ and hence $a= 0 $ in $A_p$ since $1-a $ is not in $p$ ( since $ain I subset p$)
Thus every element of $I$ becomes $0$ in $A_p$ and hence $I_p=0 forall p in Spec A $
answered yesterday
Soumik GhoshSoumik Ghosh
1,298112
$endgroup$
$begingroup$
do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
yesterday
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Counterexamples exist. In fact:
Proposition. TFAE:
$I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$
- $forall x in I : Ann(x) + I = R$
- $forall x in I : exists y in I : xy=x$
In particular, a nonzero proper ideal in a boolean ring is a counterexample.
Proof. $1 implies 2$: Take $x in I$. Suppose there is a prime ideal $mathfrak p$ containing $Ann(x) + I$. Localizing at $mathfrak p$, we find $r in R - mathfrak p$ with $rx= 0$. This contradicts $Ann(x) subset mathfrak p$.
$2 implies 3$: Take $a in R$ and $y in I$ with $a+y = 1$ and $ax = 0$. Then $xy = x$.
$3 implies 1$: Take $x in I$, $y in I$ with $xy = x$ and take $mathfrak p supset I$. Because $(1-y)x = 0$ and $1-y notin mathfrak p$, $x$ becomes $0$ in $I_{mathfrak p}$. $square$
Milking this, we find:
- such $I$ consists of zero divisors.
- There are no counterexamples when $R$ is an integral domain.
- If $I$ is nonzero, proper and finitely generated, it contains a nontrivial idempotent. Proof: From 2. it follows that $Ann(I) + I = R$. Write $a + x = 1$ with $aI = 0$ and $x in I$. Then $x^2=x$.
Since you asked about $Spec(R)$ connected and reduced, here's an example (necessarily not Noetherian). Take a field $k$, let $R = k[X_1, X_2, ldots]$ modulo the ideal generated by the $X_iX_j-X_i$ for $j>i$, and let $I$ be the ideal generated by the $X_i$.
answered yesterday
rabotarabota
14.6k32885
$endgroup$
$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I do not completely understand this. All this proves is that tensoring with $mathbb{Z}[x]/(x^2)$ is not going to work. There still might be an irreducible, non-reduced example constructed in another way. Or did you mean something else?
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced)
$endgroup$
– rabota
yesterday
add a comment |
$begingroup$
Counterexamples exist. In fact:
Proposition. TFAE:
$I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$
- $forall x in I : Ann(x) + I = R$
- $forall x in I : exists y in I : xy=x$
In particular, a nonzero proper ideal in a boolean ring is a counterexample.
Proof. $1 implies 2$: Take $x in I$. Suppose there is a prime ideal $mathfrak p$ containing $Ann(x) + I$. Localizing at $mathfrak p$, we find $r in R - mathfrak p$ with $rx= 0$. This contradicts $Ann(x) subset mathfrak p$.
$2 implies 3$: Take $a in R$ and $y in I$ with $a+y = 1$ and $ax = 0$. Then $xy = x$.
$3 implies 1$: Take $x in I$, $y in I$ with $xy = x$ and take $mathfrak p supset I$. Because $(1-y)x = 0$ and $1-y notin mathfrak p$, $x$ becomes $0$ in $I_{mathfrak p}$. $square$
Milking this, we find:
- such $I$ consists of zero divisors.
- There are no counterexamples when $R$ is an integral domain.
- If $I$ is nonzero, proper and finitely generated, it contains a nontrivial idempotent. Proof: From 2. it follows that $Ann(I) + I = R$. Write $a + x = 1$ with $aI = 0$ and $x in I$. Then $x^2=x$.
Since you asked about $Spec(R)$ connected and reduced, here's an example (necessarily not Noetherian). Take a field $k$, let $R = k[X_1, X_2, ldots]$ modulo the ideal generated by the $X_iX_j-X_i$ for $j>i$, and let $I$ be the ideal generated by the $X_i$.
answered yesterday
rabotarabota
14.6k32885
$endgroup$
$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I do not completely understand this. All this proves is that tensoring with $mathbb{Z}[x]/(x^2)$ is not going to work. There still might be an irreducible, non-reduced example constructed in another way. Or did you mean something else?
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced)
$endgroup$
– rabota
yesterday
add a comment |
$begingroup$
Counterexamples exist. In fact:
Proposition. TFAE:
$I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$
- $forall x in I : Ann(x) + I = R$
- $forall x in I : exists y in I : xy=x$
In particular, a nonzero proper ideal in a boolean ring is a counterexample.
Proof. $1 implies 2$: Take $x in I$. Suppose there is a prime ideal $mathfrak p$ containing $Ann(x) + I$. Localizing at $mathfrak p$, we find $r in R - mathfrak p$ with $rx= 0$. This contradicts $Ann(x) subset mathfrak p$.
$2 implies 3$: Take $a in R$ and $y in I$ with $a+y = 1$ and $ax = 0$. Then $xy = x$.
$3 implies 1$: Take $x in I$, $y in I$ with $xy = x$ and take $mathfrak p supset I$. Because $(1-y)x = 0$ and $1-y notin mathfrak p$, $x$ becomes $0$ in $I_{mathfrak p}$. $square$
Milking this, we find:
- such $I$ consists of zero divisors.
- There are no counterexamples when $R$ is an integral domain.
- If $I$ is nonzero, proper and finitely generated, it contains a nontrivial idempotent. Proof: From 2. it follows that $Ann(I) + I = R$. Write $a + x = 1$ with $aI = 0$ and $x in I$. Then $x^2=x$.
Since you asked about $Spec(R)$ connected and reduced, here's an example (necessarily not Noetherian). Take a field $k$, let $R = k[X_1, X_2, ldots]$ modulo the ideal generated by the $X_iX_j-X_i$ for $j>i$, and let $I$ be the ideal generated by the $X_i$.
answered yesterday
rabotarabota
14.6k32885
$endgroup$
Counterexamples exist. In fact:
Proposition. TFAE:
$I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$
- $forall x in I : Ann(x) + I = R$
- $forall x in I : exists y in I : xy=x$
In particular, a nonzero proper ideal in a boolean ring is a counterexample.
Proof. $1 implies 2$: Take $x in I$. Suppose there is a prime ideal $mathfrak p$ containing $Ann(x) + I$. Localizing at $mathfrak p$, we find $r in R - mathfrak p$ with $rx= 0$. This contradicts $Ann(x) subset mathfrak p$.
$2 implies 3$: Take $a in R$ and $y in I$ with $a+y = 1$ and $ax = 0$. Then $xy = x$.
$3 implies 1$: Take $x in I$, $y in I$ with $xy = x$ and take $mathfrak p supset I$. Because $(1-y)x = 0$ and $1-y notin mathfrak p$, $x$ becomes $0$ in $I_{mathfrak p}$. $square$
Milking this, we find:
- such $I$ consists of zero divisors.
- There are no counterexamples when $R$ is an integral domain.
- If $I$ is nonzero, proper and finitely generated, it contains a nontrivial idempotent. Proof: From 2. it follows that $Ann(I) + I = R$. Write $a + x = 1$ with $aI = 0$ and $x in I$. Then $x^2=x$.
Since you asked about $Spec(R)$ connected and reduced, here's an example (necessarily not Noetherian). Take a field $k$, let $R = k[X_1, X_2, ldots]$ modulo the ideal generated by the $X_iX_j-X_i$ for $j>i$, and let $I$ be the ideal generated by the $X_i$.
answered yesterday
rabotarabota
14.6k32885
edited yesterday
answered yesterday
rabotarabota
14.6k32885
answered yesterday
rabotarabota
14.6k32885
answered yesterday
rabotarabota
14.6k32885
14.6k32885
$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I do not completely understand this. All this proves is that tensoring with $mathbb{Z}[x]/(x^2)$ is not going to work. There still might be an irreducible, non-reduced example constructed in another way. Or did you mean something else?
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced)
$endgroup$
– rabota
yesterday
add a comment |
$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I do not completely understand this. All this proves is that tensoring with $mathbb{Z}[x]/(x^2)$ is not going to work. There still might be an irreducible, non-reduced example constructed in another way. Or did you mean something else?
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced)
$endgroup$
– rabota
yesterday
$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
yesterday
$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I do not completely understand this. All this proves is that tensoring with $mathbb{Z}[x]/(x^2)$ is not going to work. There still might be an irreducible, non-reduced example constructed in another way. Or did you mean something else?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I do not completely understand this. All this proves is that tensoring with $mathbb{Z}[x]/(x^2)$ is not going to work. There still might be an irreducible, non-reduced example constructed in another way. Or did you mean something else?
$endgroup$
– Stepan Banach
yesterday
1
1
$begingroup$
I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced)
$endgroup$
– rabota
yesterday
$begingroup$
I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced)
$endgroup$
– rabota
yesterday
add a comment |
$begingroup$
Consider the ring $$R= mathbb Z_2 times mathbb Z_2 times... times mathbb Z_2 times.....$$ countable number of times .
Look at the ideal $$ I= bigoplus _{mathbb N} mathbb Z_2$$
I claim $I_p=0 forall p in Spec R$ containing $I$.
Say $I subset p$ Consider $ain I$. Observe that $a^2=a$ and hence $a= 0 $ in $A_p$ since $1-a $ is not in $p$ ( since $ain I subset p$)
Thus every element of $I$ becomes $0$ in $A_p$ and hence $I_p=0 forall p in Spec A $
answered yesterday
Soumik GhoshSoumik Ghosh
1,298112
$endgroup$
$begingroup$
do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
yesterday
add a comment |
$begingroup$
Consider the ring $$R= mathbb Z_2 times mathbb Z_2 times... times mathbb Z_2 times.....$$ countable number of times .
Look at the ideal $$ I= bigoplus _{mathbb N} mathbb Z_2$$
I claim $I_p=0 forall p in Spec R$ containing $I$.
Say $I subset p$ Consider $ain I$. Observe that $a^2=a$ and hence $a= 0 $ in $A_p$ since $1-a $ is not in $p$ ( since $ain I subset p$)
Thus every element of $I$ becomes $0$ in $A_p$ and hence $I_p=0 forall p in Spec A $
answered yesterday
Soumik GhoshSoumik Ghosh
1,298112
$endgroup$
$begingroup$
do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
yesterday
add a comment |
$begingroup$
Consider the ring $$R= mathbb Z_2 times mathbb Z_2 times... times mathbb Z_2 times.....$$ countable number of times .
Look at the ideal $$ I= bigoplus _{mathbb N} mathbb Z_2$$
I claim $I_p=0 forall p in Spec R$ containing $I$.
Say $I subset p$ Consider $ain I$. Observe that $a^2=a$ and hence $a= 0 $ in $A_p$ since $1-a $ is not in $p$ ( since $ain I subset p$)
Thus every element of $I$ becomes $0$ in $A_p$ and hence $I_p=0 forall p in Spec A $
answered yesterday
Soumik GhoshSoumik Ghosh
1,298112
$endgroup$
Consider the ring $$R= mathbb Z_2 times mathbb Z_2 times... times mathbb Z_2 times.....$$ countable number of times .
Look at the ideal $$ I= bigoplus _{mathbb N} mathbb Z_2$$
I claim $I_p=0 forall p in Spec R$ containing $I$.
Say $I subset p$ Consider $ain I$. Observe that $a^2=a$ and hence $a= 0 $ in $A_p$ since $1-a $ is not in $p$ ( since $ain I subset p$)
Thus every element of $I$ becomes $0$ in $A_p$ and hence $I_p=0 forall p in Spec A $
answered yesterday
Soumik GhoshSoumik Ghosh
1,298112
answered yesterday
Soumik GhoshSoumik Ghosh
1,298112
answered yesterday
Soumik GhoshSoumik Ghosh
1,298112
answered yesterday
Soumik GhoshSoumik Ghosh
1,298112
1,298112
$begingroup$
do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
yesterday
add a comment |
$begingroup$
do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
yesterday
$begingroup$
do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
yesterday
$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
yesterday
add a comment |
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$begingroup$
Is $I$ finitely generated ? Suppose $M$ is a finitely generated $R-module$. Then $M=0 iff M_p=0 forall p in Spec R$
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
@SoumikGhosh No finiteness hypotheses
$endgroup$
– Stepan Banach
yesterday
1
$begingroup$
@SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
$endgroup$
– Alex Mathers
yesterday