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?






abstract-algebra ring-theory commutative-algebra 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
    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?






abstract-algebra ring-theory commutative-algebra 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
    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





$begingroup$


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?






abstract-algebra ring-theory commutative-algebra 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 $m_1 in M_1$ 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




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











$endgroup$





















    0












    $begingroup$

    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 $$(1,1)-(2,0)=(-1,1)notin M_2$$



    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 $a,b$.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      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























              0












              $begingroup$

              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 $$(1,1)-(2,0)=(-1,1)notin M_2$$



              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 $a,b$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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 $$(1,1)-(2,0)=(-1,1)notin M_2$$



                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 $a,b$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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 $$(1,1)-(2,0)=(-1,1)notin M_2$$



                  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 $a,b$.






                  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 $$(1,1)-(2,0)=(-1,1)notin M_2$$



                  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 $a,b$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 37 mins ago









                  user647486user647486

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