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When is this statement true $fracas in S^-1I implies a in I$?

2

$begingroup$

If $R$ is a ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset of $R$.

We know that if $a in I$ then $fracas in S^-1I$.
But the converse isn't true always. If $fracas in S^-1P$ then $a in P$ if $P$ is a prime ideal and $P cap S= emptyset$.

My question is that, if we know that $I cap S= emptyset$, is there another property that we can have on the ideal $I$ so that this statement $fracas in S^-1I implies a in I$ is true? (I mean other than $I$ being prime.)

ring-theory ideals localization

share|cite|improve this question

edited Apr 2 at 8:02

Orat

3,04021231

asked Mar 24 at 16:35

Fareed AFFareed AF

822112

$endgroup$

add a comment | 

2

$begingroup$

If $R$ is a ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset of $R$.

We know that if $a in I$ then $fracas in S^-1I$.
But the converse isn't true always. If $fracas in S^-1P$ then $a in P$ if $P$ is a prime ideal and $P cap S= emptyset$.

My question is that, if we know that $I cap S= emptyset$, is there another property that we can have on the ideal $I$ so that this statement $fracas in S^-1I implies a in I$ is true? (I mean other than $I$ being prime.)

ring-theory ideals localization

share|cite|improve this question

edited Apr 2 at 8:02

Orat

3,04021231

asked Mar 24 at 16:35

Fareed AFFareed AF

822112

$endgroup$

add a comment | 

$begingroup$

If $R$ is a ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset of $R$.

We know that if $a in I$ then $fracas in S^-1I$.
But the converse isn't true always. If $fracas in S^-1P$ then $a in P$ if $P$ is a prime ideal and $P cap S= emptyset$.

My question is that, if we know that $I cap S= emptyset$, is there another property that we can have on the ideal $I$ so that this statement $fracas in S^-1I implies a in I$ is true? (I mean other than $I$ being prime.)

ring-theory ideals localization

share|cite|improve this question

edited Apr 2 at 8:02

Orat

3,04021231

asked Mar 24 at 16:35

Fareed AFFareed AF

822112

$endgroup$

share|cite|improve this question

edited Apr 2 at 8:02

Orat

3,04021231

asked Mar 24 at 16:35

Fareed AFFareed AF

822112

share|cite|improve this question

edited Apr 2 at 8:02

Orat

3,04021231

asked Mar 24 at 16:35

Fareed AFFareed AF

822112

edited Apr 2 at 8:02

Orat

3,04021231

edited Apr 2 at 8:02

Orat

3,04021231

asked Mar 24 at 16:35

Fareed AFFareed AF

822112

asked Mar 24 at 16:35

Fareed AFFareed AF

822112

2 Answers
2

active

oldest

votes

2

$begingroup$

$I subset R$ with $Icap S = emptyset$ satisfies your property if and only if whenever $sain I$ with $sin S$ and $ain R$, we have that $ain I$.

To prove it is sufficient is the same proof as in the prime case. To prove it is necessary, suppose there is some $ain R-I$ and $sin S$ with $sa in I$. Then $a/1 = (sa)/s in S^-1I$ but $anotin I$.

share|cite|improve this answer

answered Mar 24 at 19:33

cspruncsprun

2,804211

$endgroup$

add a comment | 

0

$begingroup$

If $q$ is primary and $S cap q=varnothing$ then, $fracas in S^-1q implies a in q$

share|cite|improve this answer

answered Apr 2 at 4:31

Fareed AFFareed AF

822112

$endgroup$

add a comment | 

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2

$begingroup$

$I subset R$ with $Icap S = emptyset$ satisfies your property if and only if whenever $sain I$ with $sin S$ and $ain R$, we have that $ain I$.

To prove it is sufficient is the same proof as in the prime case. To prove it is necessary, suppose there is some $ain R-I$ and $sin S$ with $sa in I$. Then $a/1 = (sa)/s in S^-1I$ but $anotin I$.

share|cite|improve this answer

answered Mar 24 at 19:33

cspruncsprun

2,804211

$endgroup$

add a comment | 

2

$begingroup$

$I subset R$ with $Icap S = emptyset$ satisfies your property if and only if whenever $sain I$ with $sin S$ and $ain R$, we have that $ain I$.

To prove it is sufficient is the same proof as in the prime case. To prove it is necessary, suppose there is some $ain R-I$ and $sin S$ with $sa in I$. Then $a/1 = (sa)/s in S^-1I$ but $anotin I$.

share|cite|improve this answer

answered Mar 24 at 19:33

cspruncsprun

2,804211

$endgroup$

add a comment | 

$begingroup$

$I subset R$ with $Icap S = emptyset$ satisfies your property if and only if whenever $sain I$ with $sin S$ and $ain R$, we have that $ain I$.

To prove it is sufficient is the same proof as in the prime case. To prove it is necessary, suppose there is some $ain R-I$ and $sin S$ with $sa in I$. Then $a/1 = (sa)/s in S^-1I$ but $anotin I$.

share|cite|improve this answer

answered Mar 24 at 19:33

cspruncsprun

2,804211

$endgroup$

share|cite|improve this answer

answered Mar 24 at 19:33

cspruncsprun

2,804211

answered Mar 24 at 19:33

cspruncsprun

2,804211

answered Mar 24 at 19:33

cspruncsprun

2,804211

0

$begingroup$

If $q$ is primary and $S cap q=varnothing$ then, $fracas in S^-1q implies a in q$

share|cite|improve this answer

answered Apr 2 at 4:31

Fareed AFFareed AF

822112

$endgroup$

add a comment | 

0

$begingroup$

If $q$ is primary and $S cap q=varnothing$ then, $fracas in S^-1q implies a in q$

share|cite|improve this answer

answered Apr 2 at 4:31

Fareed AFFareed AF

822112

$endgroup$

add a comment | 

$begingroup$

If $q$ is primary and $S cap q=varnothing$ then, $fracas in S^-1q implies a in q$

share|cite|improve this answer

answered Apr 2 at 4:31

Fareed AFFareed AF

822112

$endgroup$

share|cite|improve this answer

answered Apr 2 at 4:31

Fareed AFFareed AF

822112

answered Apr 2 at 4:31

Fareed AFFareed AF

822112

answered Apr 2 at 4:31

Fareed AFFareed AF

822112