Mixing
Prime ideal containing an ideal and not intersecting a multiplicatively closed set [closed]
0
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Suppose $R$ is a ring and $I$ an ideal of $R$, moreover suppose $M$ is a multiplicatively closed subset of $R$ such that $I$ does not intersect $M$ can you always find a prime ideal $Isubset P$ also not intersecting $M$ ?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
asked Jan 6 at 19:58
sirjoesirjoe
384
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closed as off-topic by Clayton, Dietrich Burde, user26857, egreg, mrtaurho Jan 7 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Clayton, Dietrich Burde, user26857, egreg, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
Suppose $R$ is a ring and $I$ an ideal of $R$, moreover suppose $M$ is a multiplicatively closed subset of $R$ such that $I$ does not intersect $M$ can you always find a prime ideal $Isubset P$ also not intersecting $M$ ?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
asked Jan 6 at 19:58
sirjoesirjoe
384
$endgroup$
closed as off-topic by Clayton, Dietrich Burde, user26857, egreg, mrtaurho Jan 7 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Clayton, Dietrich Burde, user26857, egreg, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
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Hint: Consider $M^{-1}I$ as an ideal in $M^{-1}R$ and prove it is proper; then it is contained in a maximal ideal. Go back to $R$.
$endgroup$
– egreg
Jan 6 at 23:15
add a comment |
0
0
0
$begingroup$
Suppose $R$ is a ring and $I$ an ideal of $R$, moreover suppose $M$ is a multiplicatively closed subset of $R$ such that $I$ does not intersect $M$ can you always find a prime ideal $Isubset P$ also not intersecting $M$ ?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
asked Jan 6 at 19:58
sirjoesirjoe
384
$endgroup$
Suppose $R$ is a ring and $I$ an ideal of $R$, moreover suppose $M$ is a multiplicatively closed subset of $R$ such that $I$ does not intersect $M$ can you always find a prime ideal $Isubset P$ also not intersecting $M$ ?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
abstract-algebra ring-theory ideals maximal-and-prime-ideals
asked Jan 6 at 19:58
sirjoesirjoe
384
asked Jan 6 at 19:58
sirjoesirjoe
384
asked Jan 6 at 19:58
sirjoesirjoe
384
asked Jan 6 at 19:58
sirjoesirjoe
384
asked Jan 6 at 19:58
sirjoesirjoe
384
384
closed as off-topic by Clayton, Dietrich Burde, user26857, egreg, mrtaurho Jan 7 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Clayton, Dietrich Burde, user26857, egreg, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Clayton, Dietrich Burde, user26857, egreg, mrtaurho Jan 7 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Clayton, Dietrich Burde, user26857, egreg, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Hint: Consider $M^{-1}I$ as an ideal in $M^{-1}R$ and prove it is proper; then it is contained in a maximal ideal. Go back to $R$.
$endgroup$
– egreg
Jan 6 at 23:15
add a comment |
$begingroup$
Hint: Consider $M^{-1}I$ as an ideal in $M^{-1}R$ and prove it is proper; then it is contained in a maximal ideal. Go back to $R$.
$endgroup$
– egreg
Jan 6 at 23:15
$begingroup$
Hint: Consider $M^{-1}I$ as an ideal in $M^{-1}R$ and prove it is proper; then it is contained in a maximal ideal. Go back to $R$.
$endgroup$
– egreg
Jan 6 at 23:15
$begingroup$
Hint: Consider $M^{-1}I$ as an ideal in $M^{-1}R$ and prove it is proper; then it is contained in a maximal ideal. Go back to $R$.
$endgroup$
– egreg
Jan 6 at 23:15
add a comment |
1 Answer
1
active
oldest
votes
1
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You haven't shown any of your own efforts, so I am only going to give a vague hint: try Zorn's lemma.
answered Jan 6 at 20:04
A. PongráczA. Pongrácz
5,9631929
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using Zorns Lemma though one can get a maximal ideal $J$ of those containing $I$ such that $J$ does not intersect the set $M$ I suppose the closure of $M$ implies that such an ideal would be prime, but how so?
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– sirjoe
Jan 6 at 20:38
1
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Exactly correct. Simply pick two elements $a,b$ outside the maximal $J$, and show that $abnotin J$. The way to make use of the fact that $a,bnotin J$ is by obsering that $(a)+J$ and $(b)+J$ are ideals containing $I$ that are strictly bigger than the maximal $J$...
$endgroup$
– A. Pongrácz
Jan 6 at 20:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
You haven't shown any of your own efforts, so I am only going to give a vague hint: try Zorn's lemma.
answered Jan 6 at 20:04
A. PongráczA. Pongrácz
5,9631929
$endgroup$
$begingroup$
using Zorns Lemma though one can get a maximal ideal $J$ of those containing $I$ such that $J$ does not intersect the set $M$ I suppose the closure of $M$ implies that such an ideal would be prime, but how so?
$endgroup$
– sirjoe
Jan 6 at 20:38
1
$begingroup$
Exactly correct. Simply pick two elements $a,b$ outside the maximal $J$, and show that $abnotin J$. The way to make use of the fact that $a,bnotin J$ is by obsering that $(a)+J$ and $(b)+J$ are ideals containing $I$ that are strictly bigger than the maximal $J$...
$endgroup$
– A. Pongrácz
Jan 6 at 20:44
add a comment |
1
$begingroup$
You haven't shown any of your own efforts, so I am only going to give a vague hint: try Zorn's lemma.
answered Jan 6 at 20:04
A. PongráczA. Pongrácz
5,9631929
$endgroup$
$begingroup$
using Zorns Lemma though one can get a maximal ideal $J$ of those containing $I$ such that $J$ does not intersect the set $M$ I suppose the closure of $M$ implies that such an ideal would be prime, but how so?
$endgroup$
– sirjoe
Jan 6 at 20:38
1
$begingroup$
Exactly correct. Simply pick two elements $a,b$ outside the maximal $J$, and show that $abnotin J$. The way to make use of the fact that $a,bnotin J$ is by obsering that $(a)+J$ and $(b)+J$ are ideals containing $I$ that are strictly bigger than the maximal $J$...
$endgroup$
– A. Pongrácz
Jan 6 at 20:44
add a comment |
1
1
1
$begingroup$
You haven't shown any of your own efforts, so I am only going to give a vague hint: try Zorn's lemma.
answered Jan 6 at 20:04
A. PongráczA. Pongrácz
5,9631929
$endgroup$
You haven't shown any of your own efforts, so I am only going to give a vague hint: try Zorn's lemma.
answered Jan 6 at 20:04
A. PongráczA. Pongrácz
5,9631929
answered Jan 6 at 20:04
A. PongráczA. Pongrácz
5,9631929
answered Jan 6 at 20:04
A. PongráczA. Pongrácz
5,9631929
answered Jan 6 at 20:04
A. PongráczA. Pongrácz
5,9631929
5,9631929
$begingroup$
using Zorns Lemma though one can get a maximal ideal $J$ of those containing $I$ such that $J$ does not intersect the set $M$ I suppose the closure of $M$ implies that such an ideal would be prime, but how so?
$endgroup$
– sirjoe
Jan 6 at 20:38
1
$begingroup$
Exactly correct. Simply pick two elements $a,b$ outside the maximal $J$, and show that $abnotin J$. The way to make use of the fact that $a,bnotin J$ is by obsering that $(a)+J$ and $(b)+J$ are ideals containing $I$ that are strictly bigger than the maximal $J$...
$endgroup$
– A. Pongrácz
Jan 6 at 20:44
add a comment |
$begingroup$
using Zorns Lemma though one can get a maximal ideal $J$ of those containing $I$ such that $J$ does not intersect the set $M$ I suppose the closure of $M$ implies that such an ideal would be prime, but how so?
$endgroup$
– sirjoe
Jan 6 at 20:38
1
$begingroup$
Exactly correct. Simply pick two elements $a,b$ outside the maximal $J$, and show that $abnotin J$. The way to make use of the fact that $a,bnotin J$ is by obsering that $(a)+J$ and $(b)+J$ are ideals containing $I$ that are strictly bigger than the maximal $J$...
$endgroup$
– A. Pongrácz
Jan 6 at 20:44
$begingroup$
using Zorns Lemma though one can get a maximal ideal $J$ of those containing $I$ such that $J$ does not intersect the set $M$ I suppose the closure of $M$ implies that such an ideal would be prime, but how so?
$endgroup$
– sirjoe
Jan 6 at 20:38
$begingroup$
using Zorns Lemma though one can get a maximal ideal $J$ of those containing $I$ such that $J$ does not intersect the set $M$ I suppose the closure of $M$ implies that such an ideal would be prime, but how so?
$endgroup$
– sirjoe
Jan 6 at 20:38
1
1
$begingroup$
Exactly correct. Simply pick two elements $a,b$ outside the maximal $J$, and show that $abnotin J$. The way to make use of the fact that $a,bnotin J$ is by obsering that $(a)+J$ and $(b)+J$ are ideals containing $I$ that are strictly bigger than the maximal $J$...
$endgroup$
– A. Pongrácz
Jan 6 at 20:44
$begingroup$
Exactly correct. Simply pick two elements $a,b$ outside the maximal $J$, and show that $abnotin J$. The way to make use of the fact that $a,bnotin J$ is by obsering that $(a)+J$ and $(b)+J$ are ideals containing $I$ that are strictly bigger than the maximal $J$...
$endgroup$
– A. Pongrácz
Jan 6 at 20:44
add a comment |
$begingroup$
Hint: Consider $M^{-1}I$ as an ideal in $M^{-1}R$ and prove it is proper; then it is contained in a maximal ideal. Go back to $R$.
$endgroup$
– egreg
Jan 6 at 23:15