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







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$endgroup$












  • $begingroup$
    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











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      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






























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