Does this property of comaximal ideals always hold?












5












$begingroup$


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 $m_1 in M_1$ is such that $m_1 notin M_2$. Then can we always find $m_2 in M_2$ such that $m_1+m_2=1$?




Any ideas?






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







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
    yesterday






  • 1




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






  • 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
    yesterday


















5












$begingroup$


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 $m_1 in M_1$ is such that $m_1 notin M_2$. Then can we always find $m_2 in M_2$ such that $m_1+m_2=1$?




Any ideas?






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







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
    yesterday






  • 1




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






  • 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
    yesterday
















5












5








5


2



$begingroup$


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 $m_1 in M_1$ is such that $m_1 notin M_2$. Then can we always find $m_2 in M_2$ such that $m_1+m_2=1$?




Any ideas?






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







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 $m_1 in M_1$ is such that $m_1 notin M_2$. Then can we always find $m_2 in M_2$ such that $m_1+m_2=1$?




Any ideas?






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




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago









Peter Mortensen

561310




561310










asked yesterday









Math LoverMath Lover

1,029315




1,029315












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






  • 1




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






  • 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
    yesterday




















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






  • 1




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






  • 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
    yesterday


















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




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




1




1




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




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




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
yesterday






$begingroup$
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
yesterday












2 Answers
2






active

oldest

votes


















8












$begingroup$

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











$endgroup$





















    5












    $begingroup$

    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











    $endgroup$













      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-hold%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









      8












      $begingroup$

      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











      $endgroup$


















        8












        $begingroup$

        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











        $endgroup$
















          8












          8








          8





          $begingroup$

          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 yesterday

























          answered yesterday









          user647486user647486

          34616




          34616























              5












              $begingroup$

              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











              $endgroup$


















                5












                $begingroup$

                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











                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  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 yesterday

























                  answered yesterday









                  Alex MathersAlex Mathers

                  11.1k21344




                  11.1k21344






























                      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-hold%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

                      If I really need a card on my start hand, how many mulligans make sense? [duplicate]

                      Image
                      5 This question already has an answer here: How do you calculate the likelihood of drawing certain cards in your opening hand? 3 answers There is a funny modern deck where you play Treasure Hunt, but you really need to have it on the starting hand. How many mulligans should I take, if my deck contains 60 cards and contains 4 copies of Treasure Hunt? magic-the-gathering probability share | improve this question asked Nov 23 '18 at 12:12 dan dan 152 4
                      Read more

                      Alcedinidae

                      Image
                      Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
                      Read more

                      Can an atomic nucleus contain both particles and antiparticles? [duplicate]

                      Image
                      9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
                      Read more