If $xR=I$ we can say that $xin I$?












5












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















5












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















5












5








5





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 3:37







Masacroso

















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
















  • 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












2 Answers
2






active

oldest

votes


















4












$begingroup$

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 $Rx=I$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    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, $2notin4mathbb Z$.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      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 $Rx=I$.






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        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 $Rx=I$.






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          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 $Rx=I$.






          share|cite|improve this answer











          $endgroup$



          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 $Rx=I$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 21 '18 at 16:41

























          answered Dec 21 '18 at 14:31









          rschwiebrschwieb

          108k12103253




          108k12103253























              2












              $begingroup$

              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, $2notin4mathbb Z$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                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, $2notin4mathbb Z$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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, $2notin4mathbb Z$.






                  share|cite|improve this answer









                  $endgroup$



                  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, $2notin4mathbb Z$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 7:37









                  YiFanYiFan

                  5,2882728




                  5,2882728






























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