Does $frac{b}{s} in S^{-1} I$ directly implies that $b in I$?












2












$begingroup$


Let $R$ be a commutative ring with identity $1_R$, $0 not in Ssubseteq R$ be a multiplicative set, and $Isubseteq R$ be an ideal of $R$.Consider the ring of quotients $S^{-1}I$.



I was trying to prove a theorem, and I proved almost everything, but stuck at showing that if $frac{b}{s} in I $, then $b in I$, so my question is




Does $frac{b}{s} in S^{-1} I$ directly implies that $b in I$ ?




The definition only says that for $i in I, s' in S$ satisfying $frac{b}{s} = frac{i}{s'}$, i.e $exists s_1 in S quad s.t quad s_1 (s'b - si) = 0,$, the equivalence class $b/s in S^{-1}I$. but since $R$ is not necessarily an integral domain, I couldn't derive anything useful from this relation.



Edit:



In the original theorem, $I$ and $b$ has some other properties, but what I want to understand in this question is that whether I need to look for other thing, or I'm done but just I haven't realized it.






abstract-algebra ring-theory ideals localization







share|cite|improve this question











$endgroup$












  • $begingroup$
    See wikipedia for the definition of the equivalence class $b/s$.
    $endgroup$
    – Dietrich Burde
    Jan 13 at 12:11






  • 3




    $begingroup$
    As I see it, it only means there exists $s'in S$ such that $s'bin I$.
    $endgroup$
    – Bernard
    Jan 13 at 12:17










  • $begingroup$
    @Bernard I think so too.
    $endgroup$
    – onurcanbektas
    Jan 13 at 12:19










  • $begingroup$
    However, it would be true if $I$ were a prime ideal (not meeting $S$)
    $endgroup$
    – Bernard
    Jan 13 at 12:23










  • $begingroup$
    @Bernard in the original theorem, I'm trying to prove that $I$ is a prime ideal; I assumed it is not the case, and examined every case separately.
    $endgroup$
    – onurcanbektas
    Jan 13 at 12:25
















2












$begingroup$


Let $R$ be a commutative ring with identity $1_R$, $0 not in Ssubseteq R$ be a multiplicative set, and $Isubseteq R$ be an ideal of $R$.Consider the ring of quotients $S^{-1}I$.



I was trying to prove a theorem, and I proved almost everything, but stuck at showing that if $frac{b}{s} in I $, then $b in I$, so my question is




Does $frac{b}{s} in S^{-1} I$ directly implies that $b in I$ ?




The definition only says that for $i in I, s' in S$ satisfying $frac{b}{s} = frac{i}{s'}$, i.e $exists s_1 in S quad s.t quad s_1 (s'b - si) = 0,$, the equivalence class $b/s in S^{-1}I$. but since $R$ is not necessarily an integral domain, I couldn't derive anything useful from this relation.



Edit:



In the original theorem, $I$ and $b$ has some other properties, but what I want to understand in this question is that whether I need to look for other thing, or I'm done but just I haven't realized it.






abstract-algebra ring-theory ideals localization







share|cite|improve this question











$endgroup$












  • $begingroup$
    See wikipedia for the definition of the equivalence class $b/s$.
    $endgroup$
    – Dietrich Burde
    Jan 13 at 12:11






  • 3




    $begingroup$
    As I see it, it only means there exists $s'in S$ such that $s'bin I$.
    $endgroup$
    – Bernard
    Jan 13 at 12:17










  • $begingroup$
    @Bernard I think so too.
    $endgroup$
    – onurcanbektas
    Jan 13 at 12:19










  • $begingroup$
    However, it would be true if $I$ were a prime ideal (not meeting $S$)
    $endgroup$
    – Bernard
    Jan 13 at 12:23










  • $begingroup$
    @Bernard in the original theorem, I'm trying to prove that $I$ is a prime ideal; I assumed it is not the case, and examined every case separately.
    $endgroup$
    – onurcanbektas
    Jan 13 at 12:25














2












2








2





$begingroup$


Let $R$ be a commutative ring with identity $1_R$, $0 not in Ssubseteq R$ be a multiplicative set, and $Isubseteq R$ be an ideal of $R$.Consider the ring of quotients $S^{-1}I$.



I was trying to prove a theorem, and I proved almost everything, but stuck at showing that if $frac{b}{s} in I $, then $b in I$, so my question is




Does $frac{b}{s} in S^{-1} I$ directly implies that $b in I$ ?




The definition only says that for $i in I, s' in S$ satisfying $frac{b}{s} = frac{i}{s'}$, i.e $exists s_1 in S quad s.t quad s_1 (s'b - si) = 0,$, the equivalence class $b/s in S^{-1}I$. but since $R$ is not necessarily an integral domain, I couldn't derive anything useful from this relation.



Edit:



In the original theorem, $I$ and $b$ has some other properties, but what I want to understand in this question is that whether I need to look for other thing, or I'm done but just I haven't realized it.






abstract-algebra ring-theory ideals localization







share|cite|improve this question











$endgroup$




Let $R$ be a commutative ring with identity $1_R$, $0 not in Ssubseteq R$ be a multiplicative set, and $Isubseteq R$ be an ideal of $R$.Consider the ring of quotients $S^{-1}I$.



I was trying to prove a theorem, and I proved almost everything, but stuck at showing that if $frac{b}{s} in I $, then $b in I$, so my question is




Does $frac{b}{s} in S^{-1} I$ directly implies that $b in I$ ?




The definition only says that for $i in I, s' in S$ satisfying $frac{b}{s} = frac{i}{s'}$, i.e $exists s_1 in S quad s.t quad s_1 (s'b - si) = 0,$, the equivalence class $b/s in S^{-1}I$. but since $R$ is not necessarily an integral domain, I couldn't derive anything useful from this relation.



Edit:



In the original theorem, $I$ and $b$ has some other properties, but what I want to understand in this question is that whether I need to look for other thing, or I'm done but just I haven't realized it.






abstract-algebra ring-theory ideals localization




abstract-algebra ring-theory ideals localization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 20:17









Eric Wofsey

193k14221352




193k14221352










asked Jan 13 at 12:04









onurcanbektasonurcanbektas

3,48011037




3,48011037












  • $begingroup$
    See wikipedia for the definition of the equivalence class $b/s$.
    $endgroup$
    – Dietrich Burde
    Jan 13 at 12:11






  • 3




    $begingroup$
    As I see it, it only means there exists $s'in S$ such that $s'bin I$.
    $endgroup$
    – Bernard
    Jan 13 at 12:17










  • $begingroup$
    @Bernard I think so too.
    $endgroup$
    – onurcanbektas
    Jan 13 at 12:19










  • $begingroup$
    However, it would be true if $I$ were a prime ideal (not meeting $S$)
    $endgroup$
    – Bernard
    Jan 13 at 12:23










  • $begingroup$
    @Bernard in the original theorem, I'm trying to prove that $I$ is a prime ideal; I assumed it is not the case, and examined every case separately.
    $endgroup$
    – onurcanbektas
    Jan 13 at 12:25


















  • $begingroup$
    See wikipedia for the definition of the equivalence class $b/s$.
    $endgroup$
    – Dietrich Burde
    Jan 13 at 12:11






  • 3




    $begingroup$
    As I see it, it only means there exists $s'in S$ such that $s'bin I$.
    $endgroup$
    – Bernard
    Jan 13 at 12:17










  • $begingroup$
    @Bernard I think so too.
    $endgroup$
    – onurcanbektas
    Jan 13 at 12:19










  • $begingroup$
    However, it would be true if $I$ were a prime ideal (not meeting $S$)
    $endgroup$
    – Bernard
    Jan 13 at 12:23










  • $begingroup$
    @Bernard in the original theorem, I'm trying to prove that $I$ is a prime ideal; I assumed it is not the case, and examined every case separately.
    $endgroup$
    – onurcanbektas
    Jan 13 at 12:25
















$begingroup$
See wikipedia for the definition of the equivalence class $b/s$.
$endgroup$
– Dietrich Burde
Jan 13 at 12:11




$begingroup$
See wikipedia for the definition of the equivalence class $b/s$.
$endgroup$
– Dietrich Burde
Jan 13 at 12:11




3




3




$begingroup$
As I see it, it only means there exists $s'in S$ such that $s'bin I$.
$endgroup$
– Bernard
Jan 13 at 12:17




$begingroup$
As I see it, it only means there exists $s'in S$ such that $s'bin I$.
$endgroup$
– Bernard
Jan 13 at 12:17












$begingroup$
@Bernard I think so too.
$endgroup$
– onurcanbektas
Jan 13 at 12:19




$begingroup$
@Bernard I think so too.
$endgroup$
– onurcanbektas
Jan 13 at 12:19












$begingroup$
However, it would be true if $I$ were a prime ideal (not meeting $S$)
$endgroup$
– Bernard
Jan 13 at 12:23




$begingroup$
However, it would be true if $I$ were a prime ideal (not meeting $S$)
$endgroup$
– Bernard
Jan 13 at 12:23












$begingroup$
@Bernard in the original theorem, I'm trying to prove that $I$ is a prime ideal; I assumed it is not the case, and examined every case separately.
$endgroup$
– onurcanbektas
Jan 13 at 12:25




$begingroup$
@Bernard in the original theorem, I'm trying to prove that $I$ is a prime ideal; I assumed it is not the case, and examined every case separately.
$endgroup$
– onurcanbektas
Jan 13 at 12:25










1 Answer
1






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oldest

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2












$begingroup$

No, this is false. For example, let $newcommand{Z}{mathbb{Z}}R = Z$, $P = 3Z$, so $S = Z setminus 3Z$, and $I = 6Z$. Then
$$
I_P = S^{-1} I = left{frac{6m}{n} : m,n in Z, 3 nmid nright} , .
$$
Since $2 in S$ then $frac{3}{1} = frac{6}{2} in I_P$, but $3 notin I$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks for your answer Andre.
    $endgroup$
    – onurcanbektas
    Jan 13 at 15:13












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1 Answer
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1 Answer
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2












$begingroup$

No, this is false. For example, let $newcommand{Z}{mathbb{Z}}R = Z$, $P = 3Z$, so $S = Z setminus 3Z$, and $I = 6Z$. Then
$$
I_P = S^{-1} I = left{frac{6m}{n} : m,n in Z, 3 nmid nright} , .
$$
Since $2 in S$ then $frac{3}{1} = frac{6}{2} in I_P$, but $3 notin I$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks for your answer Andre.
    $endgroup$
    – onurcanbektas
    Jan 13 at 15:13
















2












$begingroup$

No, this is false. For example, let $newcommand{Z}{mathbb{Z}}R = Z$, $P = 3Z$, so $S = Z setminus 3Z$, and $I = 6Z$. Then
$$
I_P = S^{-1} I = left{frac{6m}{n} : m,n in Z, 3 nmid nright} , .
$$
Since $2 in S$ then $frac{3}{1} = frac{6}{2} in I_P$, but $3 notin I$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks for your answer Andre.
    $endgroup$
    – onurcanbektas
    Jan 13 at 15:13














2












2








2





$begingroup$

No, this is false. For example, let $newcommand{Z}{mathbb{Z}}R = Z$, $P = 3Z$, so $S = Z setminus 3Z$, and $I = 6Z$. Then
$$
I_P = S^{-1} I = left{frac{6m}{n} : m,n in Z, 3 nmid nright} , .
$$
Since $2 in S$ then $frac{3}{1} = frac{6}{2} in I_P$, but $3 notin I$.






share|cite|improve this answer









$endgroup$



No, this is false. For example, let $newcommand{Z}{mathbb{Z}}R = Z$, $P = 3Z$, so $S = Z setminus 3Z$, and $I = 6Z$. Then
$$
I_P = S^{-1} I = left{frac{6m}{n} : m,n in Z, 3 nmid nright} , .
$$
Since $2 in S$ then $frac{3}{1} = frac{6}{2} in I_P$, but $3 notin I$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 14:40









André 3000André 3000

12.8k22243




12.8k22243








  • 1




    $begingroup$
    Thanks for your answer Andre.
    $endgroup$
    – onurcanbektas
    Jan 13 at 15:13














  • 1




    $begingroup$
    Thanks for your answer Andre.
    $endgroup$
    – onurcanbektas
    Jan 13 at 15:13








1




1




$begingroup$
Thanks for your answer Andre.
$endgroup$
– onurcanbektas
Jan 13 at 15:13




$begingroup$
Thanks for your answer Andre.
$endgroup$
– onurcanbektas
Jan 13 at 15:13


















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