Does this property of comaximal ideals always holds?












4












$begingroup$


I am reading a paper in which the following result is used but I can’t see the proof of this.




let $R$ be a commutative ring with only two maximal ideals say $M_1$ and $M_2$. Suppose $m_1 in M_1$ be such that $m_1 notin M_2$ then can be always find $m_2 in M_2$ such that $m_1+m_2=1$




Any ideas?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
    $endgroup$
    – B.Swan
    1 hour ago










  • $begingroup$
    @B.Swan this approach doesn't work, to see why try writing out the details
    $endgroup$
    – Alex Mathers
    1 hour ago






  • 1




    $begingroup$
    Set $I=(M_2 cup {m_1}) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
    $endgroup$
    – B.Swan
    1 hour ago


















4












$begingroup$


I am reading a paper in which the following result is used but I can’t see the proof of this.




let $R$ be a commutative ring with only two maximal ideals say $M_1$ and $M_2$. Suppose $m_1 in M_1$ be such that $m_1 notin M_2$ then can be always find $m_2 in M_2$ such that $m_1+m_2=1$




Any ideas?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
    $endgroup$
    – B.Swan
    1 hour ago










  • $begingroup$
    @B.Swan this approach doesn't work, to see why try writing out the details
    $endgroup$
    – Alex Mathers
    1 hour ago






  • 1




    $begingroup$
    Set $I=(M_2 cup {m_1}) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
    $endgroup$
    – B.Swan
    1 hour ago
















4












4








4


1



$begingroup$


I am reading a paper in which the following result is used but I can’t see the proof of this.




let $R$ be a commutative ring with only two maximal ideals say $M_1$ and $M_2$. Suppose $m_1 in M_1$ be such that $m_1 notin M_2$ then can be always find $m_2 in M_2$ such that $m_1+m_2=1$




Any ideas?










share|cite|improve this question









$endgroup$




I am reading a paper in which the following result is used but I can’t see the proof of this.




let $R$ be a commutative ring with only two maximal ideals say $M_1$ and $M_2$. Suppose $m_1 in M_1$ be such that $m_1 notin M_2$ then can be always find $m_2 in M_2$ such that $m_1+m_2=1$




Any ideas?







abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









Math LoverMath Lover

1,024315




1,024315












  • $begingroup$
    Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
    $endgroup$
    – B.Swan
    1 hour ago










  • $begingroup$
    @B.Swan this approach doesn't work, to see why try writing out the details
    $endgroup$
    – Alex Mathers
    1 hour ago






  • 1




    $begingroup$
    Set $I=(M_2 cup {m_1}) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
    $endgroup$
    – B.Swan
    1 hour ago




















  • $begingroup$
    Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
    $endgroup$
    – B.Swan
    1 hour ago










  • $begingroup$
    @B.Swan this approach doesn't work, to see why try writing out the details
    $endgroup$
    – Alex Mathers
    1 hour ago






  • 1




    $begingroup$
    Set $I=(M_2 cup {m_1}) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
    $endgroup$
    – B.Swan
    1 hour ago


















$begingroup$
Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
$endgroup$
– B.Swan
1 hour ago




$begingroup$
Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
$endgroup$
– B.Swan
1 hour ago












$begingroup$
@B.Swan this approach doesn't work, to see why try writing out the details
$endgroup$
– Alex Mathers
1 hour ago




$begingroup$
@B.Swan this approach doesn't work, to see why try writing out the details
$endgroup$
– Alex Mathers
1 hour ago




1




1




$begingroup$
Set $I=(M_2 cup {m_1}) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
$endgroup$
– B.Swan
1 hour ago






$begingroup$
Set $I=(M_2 cup {m_1}) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
$endgroup$
– B.Swan
1 hour ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.



Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.





Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$



    Therefore, that property is not satisfied in general.



    Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.






    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3148803%2fdoes-this-property-of-comaximal-ideals-always-holds%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.



      Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.





      Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.



        Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.





        Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.



          Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.





          Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.






          share|cite|improve this answer











          $endgroup$



          First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.



          Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.





          Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          Alex MathersAlex Mathers

          11k21344




          11k21344























              2












              $begingroup$

              Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$



              Therefore, that property is not satisfied in general.



              Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$



                Therefore, that property is not satisfied in general.



                Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$



                  Therefore, that property is not satisfied in general.



                  Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.






                  share|cite|improve this answer











                  $endgroup$



                  Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$



                  Therefore, that property is not satisfied in general.



                  Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 7 mins ago

























                  answered 1 hour ago









                  user647486user647486

                  413




                  413






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3148803%2fdoes-this-property-of-comaximal-ideals-always-holds%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

                      Calculate evaluation metrics using cross_val_predict sklearn

                      Insert data from modal to MySQL (multiple modal on website)