If $xR=I$ we can say that $xin I$?
$begingroup$
Probably this is a dumb question, I lack knowledge on abstract algebra. Suppose that $R$ is a ring without unity and $Isubset R$ is a non-trivial ideal.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
abstract-algebra ring-theory ideals
asked Dec 21 '18 at 2:16
MasacrosoMasacroso
13.1k41748
$endgroup$
add a comment |
$begingroup$
Probably this is a dumb question, I lack knowledge on abstract algebra. Suppose that $R$ is a ring without unity and $Isubset R$ is a non-trivial ideal.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
abstract-algebra ring-theory ideals
asked Dec 21 '18 at 2:16
MasacrosoMasacroso
13.1k41748
$endgroup$
1
$begingroup$
It is a reasonable question. If you want to think about it, consider $R$ to be a proper ideal of a your favorite ring with unity. Then $xR$ will be an ideal of $R$, and a suitable choice of $x$ will give you an answer.
$endgroup$
– hardmath
Dec 21 '18 at 2:21
8
$begingroup$
Suppose $R=2mathbb{Z}$, $I=4mathbb{Z}$, and $x=2$.
$endgroup$
– Michael Burr
Dec 21 '18 at 2:21
3
$begingroup$
@hardmath I hope their favorite ring is a nonfield domain! If they only like von Neumann regular rings they won't find a counterexample with that hint.
$endgroup$
– rschwieb
Dec 21 '18 at 16:45
add a comment |
$begingroup$
Probably this is a dumb question, I lack knowledge on abstract algebra. Suppose that $R$ is a ring without unity and $Isubset R$ is a non-trivial ideal.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
abstract-algebra ring-theory ideals
asked Dec 21 '18 at 2:16
MasacrosoMasacroso
13.1k41748
$endgroup$
Probably this is a dumb question, I lack knowledge on abstract algebra. Suppose that $R$ is a ring without unity and $Isubset R$ is a non-trivial ideal.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
asked Dec 21 '18 at 2:16
MasacrosoMasacroso
13.1k41748
asked Dec 21 '18 at 2:16
MasacrosoMasacroso
13.1k41748
edited Dec 21 '18 at 3:37
Masacroso
asked Dec 21 '18 at 2:16
MasacrosoMasacroso
13.1k41748
asked Dec 21 '18 at 2:16
MasacrosoMasacroso
13.1k41748
asked Dec 21 '18 at 2:16
MasacrosoMasacroso
13.1k41748
13.1k41748
1
$begingroup$
It is a reasonable question. If you want to think about it, consider $R$ to be a proper ideal of a your favorite ring with unity. Then $xR$ will be an ideal of $R$, and a suitable choice of $x$ will give you an answer.
$endgroup$
– hardmath
Dec 21 '18 at 2:21
8
$begingroup$
Suppose $R=2mathbb{Z}$, $I=4mathbb{Z}$, and $x=2$.
$endgroup$
– Michael Burr
Dec 21 '18 at 2:21
3
$begingroup$
@hardmath I hope their favorite ring is a nonfield domain! If they only like von Neumann regular rings they won't find a counterexample with that hint.
$endgroup$
– rschwieb
Dec 21 '18 at 16:45
add a comment |
1
$begingroup$
It is a reasonable question. If you want to think about it, consider $R$ to be a proper ideal of a your favorite ring with unity. Then $xR$ will be an ideal of $R$, and a suitable choice of $x$ will give you an answer.
$endgroup$
– hardmath
Dec 21 '18 at 2:21
8
$begingroup$
Suppose $R=2mathbb{Z}$, $I=4mathbb{Z}$, and $x=2$.
$endgroup$
– Michael Burr
Dec 21 '18 at 2:21
3
$begingroup$
@hardmath I hope their favorite ring is a nonfield domain! If they only like von Neumann regular rings they won't find a counterexample with that hint.
$endgroup$
– rschwieb
Dec 21 '18 at 16:45
1
1
$begingroup$
It is a reasonable question. If you want to think about it, consider $R$ to be a proper ideal of a your favorite ring with unity. Then $xR$ will be an ideal of $R$, and a suitable choice of $x$ will give you an answer.
$endgroup$
– hardmath
Dec 21 '18 at 2:21
$begingroup$
It is a reasonable question. If you want to think about it, consider $R$ to be a proper ideal of a your favorite ring with unity. Then $xR$ will be an ideal of $R$, and a suitable choice of $x$ will give you an answer.
$endgroup$
– hardmath
Dec 21 '18 at 2:21
8
8
$begingroup$
Suppose $R=2mathbb{Z}$, $I=4mathbb{Z}$, and $x=2$.
$endgroup$
– Michael Burr
Dec 21 '18 at 2:21
$begingroup$
Suppose $R=2mathbb{Z}$, $I=4mathbb{Z}$, and $x=2$.
$endgroup$
– Michael Burr
Dec 21 '18 at 2:21
3
3
$begingroup$
@hardmath I hope their favorite ring is a nonfield domain! If they only like von Neumann regular rings they won't find a counterexample with that hint.
$endgroup$
– rschwieb
Dec 21 '18 at 16:45
$begingroup$
@hardmath I hope their favorite ring is a nonfield domain! If they only like von Neumann regular rings they won't find a counterexample with that hint.
$endgroup$
– rschwieb
Dec 21 '18 at 16:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The request for a counterexample has already been amply covered by existing answers/comments mentioning subrngs of $mathbb Z$.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
Trivially $R$ having identity ensures this, but I understand you're probably interested in weaker conditions.
One interesting condition that jumps to mind for me is for $I$ to be a modular ideal. If $I$ is left modular, that means that there exists $ein R$ such that $re-rin I$ for all $rin R$. (Said another way, $R/I$ has a right identity.)
But now look at what that would mean for $I=xR=Rx$: you'd have $xe-xin I$, but you already know $xein I$, so that would mean $xin I$ as well.
Obviously being right modular would suffice as well, given your assumption that $Rx=I$.
answered Dec 21 '18 at 14:31
rschwiebrschwieb
108k12103253
$endgroup$
add a comment |
$begingroup$
Per Michael Burr's suggestion in the comments, look at $R=2mathbb Z$, $I=4mathbb Z$, and $x=2$. $xR=Rx=4mathbb Z=I$, but obviously, $2notin4mathbb Z$.
answered Dec 21 '18 at 7:37
YiFanYiFan
5,2882728
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The request for a counterexample has already been amply covered by existing answers/comments mentioning subrngs of $mathbb Z$.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
Trivially $R$ having identity ensures this, but I understand you're probably interested in weaker conditions.
One interesting condition that jumps to mind for me is for $I$ to be a modular ideal. If $I$ is left modular, that means that there exists $ein R$ such that $re-rin I$ for all $rin R$. (Said another way, $R/I$ has a right identity.)
But now look at what that would mean for $I=xR=Rx$: you'd have $xe-xin I$, but you already know $xein I$, so that would mean $xin I$ as well.
Obviously being right modular would suffice as well, given your assumption that $Rx=I$.
answered Dec 21 '18 at 14:31
rschwiebrschwieb
108k12103253
$endgroup$
add a comment |
$begingroup$
The request for a counterexample has already been amply covered by existing answers/comments mentioning subrngs of $mathbb Z$.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
Trivially $R$ having identity ensures this, but I understand you're probably interested in weaker conditions.
One interesting condition that jumps to mind for me is for $I$ to be a modular ideal. If $I$ is left modular, that means that there exists $ein R$ such that $re-rin I$ for all $rin R$. (Said another way, $R/I$ has a right identity.)
But now look at what that would mean for $I=xR=Rx$: you'd have $xe-xin I$, but you already know $xein I$, so that would mean $xin I$ as well.
Obviously being right modular would suffice as well, given your assumption that $Rx=I$.
answered Dec 21 '18 at 14:31
rschwiebrschwieb
108k12103253
$endgroup$
add a comment |
$begingroup$
The request for a counterexample has already been amply covered by existing answers/comments mentioning subrngs of $mathbb Z$.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
Trivially $R$ having identity ensures this, but I understand you're probably interested in weaker conditions.
One interesting condition that jumps to mind for me is for $I$ to be a modular ideal. If $I$ is left modular, that means that there exists $ein R$ such that $re-rin I$ for all $rin R$. (Said another way, $R/I$ has a right identity.)
But now look at what that would mean for $I=xR=Rx$: you'd have $xe-xin I$, but you already know $xein I$, so that would mean $xin I$ as well.
Obviously being right modular would suffice as well, given your assumption that $Rx=I$.
answered Dec 21 '18 at 14:31
rschwiebrschwieb
108k12103253
$endgroup$
The request for a counterexample has already been amply covered by existing answers/comments mentioning subrngs of $mathbb Z$.
If $xR=Rx=I$ for some $xin R$, we can says that $xin I$? If not, there is some other conditions that ensures that $xin I$?
Trivially $R$ having identity ensures this, but I understand you're probably interested in weaker conditions.
One interesting condition that jumps to mind for me is for $I$ to be a modular ideal. If $I$ is left modular, that means that there exists $ein R$ such that $re-rin I$ for all $rin R$. (Said another way, $R/I$ has a right identity.)
But now look at what that would mean for $I=xR=Rx$: you'd have $xe-xin I$, but you already know $xein I$, so that would mean $xin I$ as well.
Obviously being right modular would suffice as well, given your assumption that $Rx=I$.
answered Dec 21 '18 at 14:31
rschwiebrschwieb
108k12103253
edited Dec 21 '18 at 16:41
answered Dec 21 '18 at 14:31
rschwiebrschwieb
108k12103253
answered Dec 21 '18 at 14:31
rschwiebrschwieb
108k12103253
answered Dec 21 '18 at 14:31
rschwiebrschwieb
108k12103253
108k12103253
add a comment |
add a comment |
$begingroup$
Per Michael Burr's suggestion in the comments, look at $R=2mathbb Z$, $I=4mathbb Z$, and $x=2$. $xR=Rx=4mathbb Z=I$, but obviously, $2notin4mathbb Z$.
answered Dec 21 '18 at 7:37
YiFanYiFan
5,2882728
$endgroup$
add a comment |
$begingroup$
Per Michael Burr's suggestion in the comments, look at $R=2mathbb Z$, $I=4mathbb Z$, and $x=2$. $xR=Rx=4mathbb Z=I$, but obviously, $2notin4mathbb Z$.
answered Dec 21 '18 at 7:37
YiFanYiFan
5,2882728
$endgroup$
add a comment |
$begingroup$
Per Michael Burr's suggestion in the comments, look at $R=2mathbb Z$, $I=4mathbb Z$, and $x=2$. $xR=Rx=4mathbb Z=I$, but obviously, $2notin4mathbb Z$.
answered Dec 21 '18 at 7:37
YiFanYiFan
5,2882728
$endgroup$
Per Michael Burr's suggestion in the comments, look at $R=2mathbb Z$, $I=4mathbb Z$, and $x=2$. $xR=Rx=4mathbb Z=I$, but obviously, $2notin4mathbb Z$.
answered Dec 21 '18 at 7:37
YiFanYiFan
5,2882728
answered Dec 21 '18 at 7:37
YiFanYiFan
5,2882728
answered Dec 21 '18 at 7:37
YiFanYiFan
5,2882728
answered Dec 21 '18 at 7:37
YiFanYiFan
5,2882728
5,2882728
add a comment |
add a comment |
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1
$begingroup$
It is a reasonable question. If you want to think about it, consider $R$ to be a proper ideal of a your favorite ring with unity. Then $xR$ will be an ideal of $R$, and a suitable choice of $x$ will give you an answer.
$endgroup$
– hardmath
Dec 21 '18 at 2:21
8
$begingroup$
Suppose $R=2mathbb{Z}$, $I=4mathbb{Z}$, and $x=2$.
$endgroup$
– Michael Burr
Dec 21 '18 at 2:21
3
$begingroup$
@hardmath I hope their favorite ring is a nonfield domain! If they only like von Neumann regular rings they won't find a counterexample with that hint.
$endgroup$
– rschwieb
Dec 21 '18 at 16:45