$R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=Icap J$

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If $R$ is a commutative ring and $I$ and $J$ are ideals s.t. $I+J=R$ then show that $IJ=Icap J$.



I've already shown that $IJ subset Icap J$, now I need to show the reverse inclusion.



I'm a bit lost, so far i'm just figuring out what pieces I have to work with.



Such as:



$forall rin R$ $exists iin I ,jin J$ s.t. $i+j=r$



$forall ijin IJ$, $ij=i_1$ and $ij=j_1$ for some $i_1in I$, $j_1in J$.



Also, if I let $xin Icap J$, then $x=i_2=j_2=i+j$ for some $i_2in I$, $j_2in J$



Anyone, having problem getting to the conclusion here, thanks in advance






abstract-algebra ring-theory







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  • Is $Isubset IJ$? How about $J$?
    – John Douma
    Sep 1 at 1:50










  • @JohnDouma It's very doubtful that $I$ is a subset of $IJ$. Take for instance, $R = mathbbZ$, $I = 2 mathbbZ$ and $J = 3 mathbbZ$.
    – Quasicoherent
    Sep 1 at 3:18










  • @Quasicoherent Yes. Thank you.
    – John Douma
    Sep 1 at 3:22














up vote
4
down vote

favorite
1












If $R$ is a commutative ring and $I$ and $J$ are ideals s.t. $I+J=R$ then show that $IJ=Icap J$.



I've already shown that $IJ subset Icap J$, now I need to show the reverse inclusion.



I'm a bit lost, so far i'm just figuring out what pieces I have to work with.



Such as:



$forall rin R$ $exists iin I ,jin J$ s.t. $i+j=r$



$forall ijin IJ$, $ij=i_1$ and $ij=j_1$ for some $i_1in I$, $j_1in J$.



Also, if I let $xin Icap J$, then $x=i_2=j_2=i+j$ for some $i_2in I$, $j_2in J$



Anyone, having problem getting to the conclusion here, thanks in advance






abstract-algebra ring-theory







share|cite|improve this question





















  • Is $Isubset IJ$? How about $J$?
    – John Douma
    Sep 1 at 1:50










  • @JohnDouma It's very doubtful that $I$ is a subset of $IJ$. Take for instance, $R = mathbbZ$, $I = 2 mathbbZ$ and $J = 3 mathbbZ$.
    – Quasicoherent
    Sep 1 at 3:18










  • @Quasicoherent Yes. Thank you.
    – John Douma
    Sep 1 at 3:22












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





If $R$ is a commutative ring and $I$ and $J$ are ideals s.t. $I+J=R$ then show that $IJ=Icap J$.



I've already shown that $IJ subset Icap J$, now I need to show the reverse inclusion.



I'm a bit lost, so far i'm just figuring out what pieces I have to work with.



Such as:



$forall rin R$ $exists iin I ,jin J$ s.t. $i+j=r$



$forall ijin IJ$, $ij=i_1$ and $ij=j_1$ for some $i_1in I$, $j_1in J$.



Also, if I let $xin Icap J$, then $x=i_2=j_2=i+j$ for some $i_2in I$, $j_2in J$



Anyone, having problem getting to the conclusion here, thanks in advance






abstract-algebra ring-theory







share|cite|improve this question













If $R$ is a commutative ring and $I$ and $J$ are ideals s.t. $I+J=R$ then show that $IJ=Icap J$.



I've already shown that $IJ subset Icap J$, now I need to show the reverse inclusion.



I'm a bit lost, so far i'm just figuring out what pieces I have to work with.



Such as:



$forall rin R$ $exists iin I ,jin J$ s.t. $i+j=r$



$forall ijin IJ$, $ij=i_1$ and $ij=j_1$ for some $i_1in I$, $j_1in J$.



Also, if I let $xin Icap J$, then $x=i_2=j_2=i+j$ for some $i_2in I$, $j_2in J$



Anyone, having problem getting to the conclusion here, thanks in advance






abstract-algebra ring-theory




abstract-algebra ring-theory






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asked Sep 1 at 1:13









Michael Vaughan

39911




39911











  • Is $Isubset IJ$? How about $J$?
    – John Douma
    Sep 1 at 1:50










  • @JohnDouma It's very doubtful that $I$ is a subset of $IJ$. Take for instance, $R = mathbbZ$, $I = 2 mathbbZ$ and $J = 3 mathbbZ$.
    – Quasicoherent
    Sep 1 at 3:18










  • @Quasicoherent Yes. Thank you.
    – John Douma
    Sep 1 at 3:22
















  • Is $Isubset IJ$? How about $J$?
    – John Douma
    Sep 1 at 1:50










  • @JohnDouma It's very doubtful that $I$ is a subset of $IJ$. Take for instance, $R = mathbbZ$, $I = 2 mathbbZ$ and $J = 3 mathbbZ$.
    – Quasicoherent
    Sep 1 at 3:18










  • @Quasicoherent Yes. Thank you.
    – John Douma
    Sep 1 at 3:22















Is $Isubset IJ$? How about $J$?
– John Douma
Sep 1 at 1:50




Is $Isubset IJ$? How about $J$?
– John Douma
Sep 1 at 1:50












@JohnDouma It's very doubtful that $I$ is a subset of $IJ$. Take for instance, $R = mathbbZ$, $I = 2 mathbbZ$ and $J = 3 mathbbZ$.
– Quasicoherent
Sep 1 at 3:18




@JohnDouma It's very doubtful that $I$ is a subset of $IJ$. Take for instance, $R = mathbbZ$, $I = 2 mathbbZ$ and $J = 3 mathbbZ$.
– Quasicoherent
Sep 1 at 3:18












@Quasicoherent Yes. Thank you.
– John Douma
Sep 1 at 3:22




@Quasicoherent Yes. Thank you.
– John Douma
Sep 1 at 3:22










2 Answers
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accepted










$I cap J = (I cap J) cdot R \ = (I cap J) cdot (I + J) \ = (I cap J) cdot I + (I cap J) cdot J \ subseteq IJ + IJ \ = IJ.$






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    There are $iin I, jin J$ such that $i+j=1$. Then, for all $ain Icap J$,
    $$a=a1=a(i+j)=ai+ajin JI+IJ=IJ.$$






    share|cite|improve this answer




















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      2 Answers
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      active

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      2 Answers
      2






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      active

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      up vote
      1
      down vote



      accepted










      $I cap J = (I cap J) cdot R \ = (I cap J) cdot (I + J) \ = (I cap J) cdot I + (I cap J) cdot J \ subseteq IJ + IJ \ = IJ.$






      share|cite|improve this answer
























        up vote
        1
        down vote



        accepted










        $I cap J = (I cap J) cdot R \ = (I cap J) cdot (I + J) \ = (I cap J) cdot I + (I cap J) cdot J \ subseteq IJ + IJ \ = IJ.$






        share|cite|improve this answer






















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          $I cap J = (I cap J) cdot R \ = (I cap J) cdot (I + J) \ = (I cap J) cdot I + (I cap J) cdot J \ subseteq IJ + IJ \ = IJ.$






          share|cite|improve this answer












          $I cap J = (I cap J) cdot R \ = (I cap J) cdot (I + J) \ = (I cap J) cdot I + (I cap J) cdot J \ subseteq IJ + IJ \ = IJ.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 1 at 1:51









          qu binggang

          1287




          1287




















              up vote
              1
              down vote













              There are $iin I, jin J$ such that $i+j=1$. Then, for all $ain Icap J$,
              $$a=a1=a(i+j)=ai+ajin JI+IJ=IJ.$$






              share|cite|improve this answer
























                up vote
                1
                down vote













                There are $iin I, jin J$ such that $i+j=1$. Then, for all $ain Icap J$,
                $$a=a1=a(i+j)=ai+ajin JI+IJ=IJ.$$






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  There are $iin I, jin J$ such that $i+j=1$. Then, for all $ain Icap J$,
                  $$a=a1=a(i+j)=ai+ajin JI+IJ=IJ.$$






                  share|cite|improve this answer












                  There are $iin I, jin J$ such that $i+j=1$. Then, for all $ain Icap J$,
                  $$a=a1=a(i+j)=ai+ajin JI+IJ=IJ.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 1 at 3:24









                  Rafael Holanda

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