$I$ and $J$ are coprime ideals iff $x to (x + I, x + J)$ is surjective.












1












$begingroup$


I'm stuck on this exercise and any help would be well appreciated:




Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I times R/J$.




Assuming surjectivity I got as far as to realize that I need to show $Rsubset I+J$ since $I+Jsubset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.



Also, I don't even know where to begin in the direction $R=I+J$ implies $phi$ is onto.



Again, any help, pointers or references are much appreciated.



Thanks in advance.






abstract-algebra ring-theory commutative-algebra ideals







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$endgroup$

















    1












    $begingroup$


    I'm stuck on this exercise and any help would be well appreciated:




    Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I times R/J$.




    Assuming surjectivity I got as far as to realize that I need to show $Rsubset I+J$ since $I+Jsubset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.



    Also, I don't even know where to begin in the direction $R=I+J$ implies $phi$ is onto.



    Again, any help, pointers or references are much appreciated.



    Thanks in advance.






    abstract-algebra ring-theory commutative-algebra ideals







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I'm stuck on this exercise and any help would be well appreciated:




      Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I times R/J$.




      Assuming surjectivity I got as far as to realize that I need to show $Rsubset I+J$ since $I+Jsubset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.



      Also, I don't even know where to begin in the direction $R=I+J$ implies $phi$ is onto.



      Again, any help, pointers or references are much appreciated.



      Thanks in advance.






      abstract-algebra ring-theory commutative-algebra ideals







      share|cite|improve this question











      $endgroup$




      I'm stuck on this exercise and any help would be well appreciated:




      Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I times R/J$.




      Assuming surjectivity I got as far as to realize that I need to show $Rsubset I+J$ since $I+Jsubset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.



      Also, I don't even know where to begin in the direction $R=I+J$ implies $phi$ is onto.



      Again, any help, pointers or references are much appreciated.



      Thanks in advance.






      abstract-algebra ring-theory commutative-algebra ideals




      abstract-algebra ring-theory commutative-algebra ideals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '14 at 17:56









      user26857

      39.3k124183




      39.3k124183










      asked Dec 15 '14 at 11:02









      WinstonWinston

      369313




      369313






















          1 Answer
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          $begingroup$

          If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.

          Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
            $endgroup$
            – Jyrki Lahtonen
            Dec 15 '14 at 11:47













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          $begingroup$

          If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.

          Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
            $endgroup$
            – Jyrki Lahtonen
            Dec 15 '14 at 11:47


















          2












          $begingroup$

          If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.

          Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
            $endgroup$
            – Jyrki Lahtonen
            Dec 15 '14 at 11:47
















          2












          2








          2





          $begingroup$

          If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.

          Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.






          share|cite|improve this answer











          $endgroup$



          If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.

          Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '14 at 12:13

























          answered Dec 15 '14 at 11:39









          user26857user26857

          39.3k124183




          39.3k124183








          • 3




            $begingroup$
            You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
            $endgroup$
            – Jyrki Lahtonen
            Dec 15 '14 at 11:47
















          • 3




            $begingroup$
            You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
            $endgroup$
            – Jyrki Lahtonen
            Dec 15 '14 at 11:47










          3




          3




          $begingroup$
          You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
          $endgroup$
          – Jyrki Lahtonen
          Dec 15 '14 at 11:47






          $begingroup$
          You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
          $endgroup$
          – Jyrki Lahtonen
          Dec 15 '14 at 11:47




















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