Does $frac{b}{s} in S^{-1} I$ directly implies that $b in I$?
$begingroup$
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
edited Jan 13 at 20:17
Eric Wofsey
193k14221352
asked Jan 13 at 12:04
onurcanbektasonurcanbektas
3,48011037
$endgroup$
|
show 1 more comment
$begingroup$
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
edited Jan 13 at 20:17
Eric Wofsey
193k14221352
asked Jan 13 at 12:04
onurcanbektasonurcanbektas
3,48011037
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
edited Jan 13 at 20:17
Eric Wofsey
193k14221352
asked Jan 13 at 12:04
onurcanbektasonurcanbektas
3,48011037
$endgroup$
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
edited Jan 13 at 20:17
Eric Wofsey
193k14221352
asked Jan 13 at 12:04
onurcanbektasonurcanbektas
3,48011037
edited Jan 13 at 20:17
Eric Wofsey
193k14221352
asked Jan 13 at 12:04
onurcanbektasonurcanbektas
3,48011037
edited Jan 13 at 20:17
Eric Wofsey
193k14221352
edited Jan 13 at 20:17
Eric Wofsey
193k14221352
edited Jan 13 at 20:17
Eric Wofsey
193k14221352
193k14221352
asked Jan 13 at 12:04
onurcanbektasonurcanbektas
3,48011037
asked Jan 13 at 12:04
onurcanbektasonurcanbektas
3,48011037
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
1 Answer
1
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votes
$begingroup$
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 $3 notin I$.
answered Jan 13 at 14:40
André 3000André 3000
12.8k22243
$endgroup$
1
$begingroup$
Thanks for your answer Andre.
$endgroup$
– onurcanbektas
Jan 13 at 15:13
add a comment |
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$begingroup$
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 $3 notin I$.
answered Jan 13 at 14:40
André 3000André 3000
12.8k22243
$endgroup$
1
$begingroup$
Thanks for your answer Andre.
$endgroup$
– onurcanbektas
Jan 13 at 15:13
add a comment |
$begingroup$
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 $3 notin I$.
answered Jan 13 at 14:40
André 3000André 3000
12.8k22243
$endgroup$
1
$begingroup$
Thanks for your answer Andre.
$endgroup$
– onurcanbektas
Jan 13 at 15:13
add a comment |
$begingroup$
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 $3 notin I$.
answered Jan 13 at 14:40
André 3000André 3000
12.8k22243
$endgroup$
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 $3 notin I$.
answered Jan 13 at 14:40
André 3000André 3000
12.8k22243
answered Jan 13 at 14:40
André 3000André 3000
12.8k22243
answered Jan 13 at 14:40
André 3000André 3000
12.8k22243
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
add a comment |
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
add a comment |
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$begingroup$
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