Ideal with zero localizations at prime ideals containing it












4












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
    16 hours ago












  • $begingroup$
    @SoumikGhosh No finiteness hypotheses
    $endgroup$
    – Stepan Banach
    16 hours ago






  • 1




    $begingroup$
    @SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
    $endgroup$
    – Alex Mathers
    15 hours ago


















4












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
    16 hours ago












  • $begingroup$
    @SoumikGhosh No finiteness hypotheses
    $endgroup$
    – Stepan Banach
    16 hours ago






  • 1




    $begingroup$
    @SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
    $endgroup$
    – Alex Mathers
    15 hours ago
















4












4








4


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago







Stepan Banach

















asked 16 hours ago









Stepan BanachStepan Banach

995




995












  • $begingroup$
    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
    16 hours ago












  • $begingroup$
    @SoumikGhosh No finiteness hypotheses
    $endgroup$
    – Stepan Banach
    16 hours ago






  • 1




    $begingroup$
    @SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
    $endgroup$
    – Alex Mathers
    15 hours ago




















  • $begingroup$
    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
    16 hours ago












  • $begingroup$
    @SoumikGhosh No finiteness hypotheses
    $endgroup$
    – Stepan Banach
    16 hours ago






  • 1




    $begingroup$
    @SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
    $endgroup$
    – Alex Mathers
    15 hours ago


















$begingroup$
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
16 hours ago






$begingroup$
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
16 hours ago














$begingroup$
@SoumikGhosh No finiteness hypotheses
$endgroup$
– Stepan Banach
16 hours ago




$begingroup$
@SoumikGhosh No finiteness hypotheses
$endgroup$
– Stepan Banach
16 hours ago




1




1




$begingroup$
@SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
$endgroup$
– Alex Mathers
15 hours ago






$begingroup$
@SoumikGhosh your statement about $M$ here is true without finitely generated hypothesis
$endgroup$
– Alex Mathers
15 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

Counterexamples exist. In fact:




Proposition. TFAE:





  1. $I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$

  2. $forall x in I : Ann(x) + I = R$

  3. $forall x in I : exists y in I : xy=x$




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 $X_i$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
    $endgroup$
    – Stepan Banach
    14 hours ago










  • $begingroup$
    but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
    $endgroup$
    – Stepan Banach
    12 hours ago










  • $begingroup$
    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
    12 hours ago






  • 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
    12 hours ago





















3












$begingroup$

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 $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    16 hours ago










  • $begingroup$
    localization in integral domains can never give you $0$ unless you invert $0$ itself.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
    $endgroup$
    – Stepan Banach
    16 hours ago










  • $begingroup$
    My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    but your example is not connected
    $endgroup$
    – Stepan Banach
    16 hours ago












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Counterexamples exist. In fact:




Proposition. TFAE:





  1. $I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$

  2. $forall x in I : Ann(x) + I = R$

  3. $forall x in I : exists y in I : xy=x$




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 $X_i$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
    $endgroup$
    – Stepan Banach
    14 hours ago










  • $begingroup$
    but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
    $endgroup$
    – Stepan Banach
    12 hours ago










  • $begingroup$
    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
    12 hours ago






  • 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
    12 hours ago


















4












$begingroup$

Counterexamples exist. In fact:




Proposition. TFAE:





  1. $I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$

  2. $forall x in I : Ann(x) + I = R$

  3. $forall x in I : exists y in I : xy=x$




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 $X_i$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
    $endgroup$
    – Stepan Banach
    14 hours ago










  • $begingroup$
    but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
    $endgroup$
    – Stepan Banach
    12 hours ago










  • $begingroup$
    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
    12 hours ago






  • 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
    12 hours ago
















4












4








4





$begingroup$

Counterexamples exist. In fact:




Proposition. TFAE:





  1. $I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$

  2. $forall x in I : Ann(x) + I = R$

  3. $forall x in I : exists y in I : xy=x$




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 $X_i$.






share|cite|improve this answer











$endgroup$



Counterexamples exist. In fact:




Proposition. TFAE:





  1. $I_{mathfrak p} = 0$ for all primes $mathfrak p supset I$

  2. $forall x in I : Ann(x) + I = R$

  3. $forall x in I : exists y in I : xy=x$




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 $X_i$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 15 hours ago

























answered 16 hours ago









rabotarabota

14.6k32885




14.6k32885












  • $begingroup$
    is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
    $endgroup$
    – Stepan Banach
    14 hours ago










  • $begingroup$
    but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
    $endgroup$
    – Stepan Banach
    12 hours ago










  • $begingroup$
    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
    12 hours ago






  • 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
    12 hours ago




















  • $begingroup$
    is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
    $endgroup$
    – Stepan Banach
    14 hours ago










  • $begingroup$
    but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
    $endgroup$
    – Stepan Banach
    12 hours ago










  • $begingroup$
    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
    12 hours ago






  • 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
    12 hours ago


















$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
14 hours ago




$begingroup$
is there an example when $mathrm{Spec},R$ is irreducible, but not reduced?
$endgroup$
– Stepan Banach
14 hours ago












$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
12 hours ago




$begingroup$
but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
$endgroup$
– Stepan Banach
12 hours ago












$begingroup$
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
12 hours ago




$begingroup$
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
12 hours ago




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
12 hours ago






$begingroup$
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
12 hours ago













3












$begingroup$

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 $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    16 hours ago










  • $begingroup$
    localization in integral domains can never give you $0$ unless you invert $0$ itself.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
    $endgroup$
    – Stepan Banach
    16 hours ago










  • $begingroup$
    My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    but your example is not connected
    $endgroup$
    – Stepan Banach
    16 hours ago
















3












$begingroup$

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 $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    16 hours ago










  • $begingroup$
    localization in integral domains can never give you $0$ unless you invert $0$ itself.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
    $endgroup$
    – Stepan Banach
    16 hours ago










  • $begingroup$
    My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    but your example is not connected
    $endgroup$
    – Stepan Banach
    16 hours ago














3












3








3





$begingroup$

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 $






share|cite|improve this answer









$endgroup$



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 $







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 16 hours ago









Soumik GhoshSoumik Ghosh

1,300112




1,300112












  • $begingroup$
    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
    16 hours ago










  • $begingroup$
    localization in integral domains can never give you $0$ unless you invert $0$ itself.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
    $endgroup$
    – Stepan Banach
    16 hours ago










  • $begingroup$
    My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    but your example is not connected
    $endgroup$
    – Stepan Banach
    16 hours ago


















  • $begingroup$
    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
    16 hours ago










  • $begingroup$
    localization in integral domains can never give you $0$ unless you invert $0$ itself.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
    $endgroup$
    – Stepan Banach
    16 hours ago










  • $begingroup$
    My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
    $endgroup$
    – Soumik Ghosh
    16 hours ago










  • $begingroup$
    but your example is not connected
    $endgroup$
    – Stepan Banach
    16 hours ago
















$begingroup$
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
16 hours ago




$begingroup$
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
16 hours ago












$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
16 hours ago




$begingroup$
localization in integral domains can never give you $0$ unless you invert $0$ itself.
$endgroup$
– Soumik Ghosh
16 hours ago












$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
16 hours ago




$begingroup$
then what about an example with $mathrm{Spec},R$ connected reduced? or is that also impossible?
$endgroup$
– Stepan Banach
16 hours ago












$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
16 hours ago




$begingroup$
My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring.
$endgroup$
– Soumik Ghosh
16 hours ago












$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
16 hours ago




$begingroup$
but your example is not connected
$endgroup$
– Stepan Banach
16 hours ago


















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