If $mathbb{Z}ni underbrace{z}_{ne 0} in I trianglelefteq R:={a+sqrt7b:a,binmathbb{Z}}$ then $[R:I] <...












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Let a ring $R:={a+sqrt7b:a,binmathbb{Z}}$. Let an ideal $Itrianglelefteq R$ such that $zin I$ for some $0ne zinmathbb{Z}$. Prove that $[R:I]<infty$.




I showed that $zin I$ for some $z>0$. I defined $$\ M:={a+sqrt7b+Iin R/I:0leq a,b< z} $$ We want to show $M=R/I$. Let $u:=a+sqrt7b+Iin R/I$. We know that
$$\ a =kq+r,b=kp'+r' $$ for some $0leq r,r'<k$. I showed that $$ \u=r+sqrt7(kq'+r')+I $$ Now I want to show that $u=r+sqrt7r'+I$ but I don't know how.






abstract-algebra ring-theory ideals







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    $begingroup$



    Let a ring $R:={a+sqrt7b:a,binmathbb{Z}}$. Let an ideal $Itrianglelefteq R$ such that $zin I$ for some $0ne zinmathbb{Z}$. Prove that $[R:I]<infty$.




    I showed that $zin I$ for some $z>0$. I defined $$\ M:={a+sqrt7b+Iin R/I:0leq a,b< z} $$ We want to show $M=R/I$. Let $u:=a+sqrt7b+Iin R/I$. We know that
    $$\ a =kq+r,b=kp'+r' $$ for some $0leq r,r'<k$. I showed that $$ \u=r+sqrt7(kq'+r')+I $$ Now I want to show that $u=r+sqrt7r'+I$ but I don't know how.






    abstract-algebra ring-theory ideals







    share|cite|improve this question











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      1








      1





      $begingroup$



      Let a ring $R:={a+sqrt7b:a,binmathbb{Z}}$. Let an ideal $Itrianglelefteq R$ such that $zin I$ for some $0ne zinmathbb{Z}$. Prove that $[R:I]<infty$.




      I showed that $zin I$ for some $z>0$. I defined $$\ M:={a+sqrt7b+Iin R/I:0leq a,b< z} $$ We want to show $M=R/I$. Let $u:=a+sqrt7b+Iin R/I$. We know that
      $$\ a =kq+r,b=kp'+r' $$ for some $0leq r,r'<k$. I showed that $$ \u=r+sqrt7(kq'+r')+I $$ Now I want to show that $u=r+sqrt7r'+I$ but I don't know how.






      abstract-algebra ring-theory ideals







      share|cite|improve this question











      $endgroup$





      Let a ring $R:={a+sqrt7b:a,binmathbb{Z}}$. Let an ideal $Itrianglelefteq R$ such that $zin I$ for some $0ne zinmathbb{Z}$. Prove that $[R:I]<infty$.




      I showed that $zin I$ for some $z>0$. I defined $$\ M:={a+sqrt7b+Iin R/I:0leq a,b< z} $$ We want to show $M=R/I$. Let $u:=a+sqrt7b+Iin R/I$. We know that
      $$\ a =kq+r,b=kp'+r' $$ for some $0leq r,r'<k$. I showed that $$ \u=r+sqrt7(kq'+r')+I $$ Now I want to show that $u=r+sqrt7r'+I$ but I don't know how.






      abstract-algebra ring-theory ideals




      abstract-algebra ring-theory ideals






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      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 23 at 22:05







      J. Doe

















      asked Jan 23 at 22:00









      J. DoeJ. Doe

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      13512






















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          $begingroup$

          You're almost there.



          (I assume $k=z$ in your calculations.)

          Then just use the definition of ideal: $zin Iimpliessqrt7q'zin I$, and we also have $x+I=I$ if $xin I$, thus
          $$u = r+sqrt7r'+sqrt7q'z+I = r+sqrt7r'+I$$
          as required.






          share|cite|improve this answer









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            active

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            1












            $begingroup$

            You're almost there.



            (I assume $k=z$ in your calculations.)

            Then just use the definition of ideal: $zin Iimpliessqrt7q'zin I$, and we also have $x+I=I$ if $xin I$, thus
            $$u = r+sqrt7r'+sqrt7q'z+I = r+sqrt7r'+I$$
            as required.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You're almost there.



              (I assume $k=z$ in your calculations.)

              Then just use the definition of ideal: $zin Iimpliessqrt7q'zin I$, and we also have $x+I=I$ if $xin I$, thus
              $$u = r+sqrt7r'+sqrt7q'z+I = r+sqrt7r'+I$$
              as required.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You're almost there.



                (I assume $k=z$ in your calculations.)

                Then just use the definition of ideal: $zin Iimpliessqrt7q'zin I$, and we also have $x+I=I$ if $xin I$, thus
                $$u = r+sqrt7r'+sqrt7q'z+I = r+sqrt7r'+I$$
                as required.






                share|cite|improve this answer









                $endgroup$



                You're almost there.



                (I assume $k=z$ in your calculations.)

                Then just use the definition of ideal: $zin Iimpliessqrt7q'zin I$, and we also have $x+I=I$ if $xin I$, thus
                $$u = r+sqrt7r'+sqrt7q'z+I = r+sqrt7r'+I$$
                as required.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 23 at 22:34









                BerciBerci

                61.4k23674




                61.4k23674






























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