Computation of tensor product












2












$begingroup$


Let $k$ be a field.



Consider the map $k[X,X^{-1},Y, Y^{-1}] to k[T, T^{-1}] $ where $X to T$ and
$Y to T$.



I'm interested in computing the tensor product
$ k[T,T^{-1}] otimes_{k[X,X^{-1},Y, Y^{-1}]} k[X^{1/n},Y^{1/n},X^{-1/n},Y^{-1/n}] $.



I'm not sure how to proceed with this computation. Any help would be appreciated






abstract-algebra tensor-products







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$endgroup$

















    2












    $begingroup$


    Let $k$ be a field.



    Consider the map $k[X,X^{-1},Y, Y^{-1}] to k[T, T^{-1}] $ where $X to T$ and
    $Y to T$.



    I'm interested in computing the tensor product
    $ k[T,T^{-1}] otimes_{k[X,X^{-1},Y, Y^{-1}]} k[X^{1/n},Y^{1/n},X^{-1/n},Y^{-1/n}] $.



    I'm not sure how to proceed with this computation. Any help would be appreciated






    abstract-algebra tensor-products







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $k$ be a field.



      Consider the map $k[X,X^{-1},Y, Y^{-1}] to k[T, T^{-1}] $ where $X to T$ and
      $Y to T$.



      I'm interested in computing the tensor product
      $ k[T,T^{-1}] otimes_{k[X,X^{-1},Y, Y^{-1}]} k[X^{1/n},Y^{1/n},X^{-1/n},Y^{-1/n}] $.



      I'm not sure how to proceed with this computation. Any help would be appreciated






      abstract-algebra tensor-products







      share|cite|improve this question









      $endgroup$




      Let $k$ be a field.



      Consider the map $k[X,X^{-1},Y, Y^{-1}] to k[T, T^{-1}] $ where $X to T$ and
      $Y to T$.



      I'm interested in computing the tensor product
      $ k[T,T^{-1}] otimes_{k[X,X^{-1},Y, Y^{-1}]} k[X^{1/n},Y^{1/n},X^{-1/n},Y^{-1/n}] $.



      I'm not sure how to proceed with this computation. Any help would be appreciated






      abstract-algebra tensor-products




      abstract-algebra tensor-products






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 12:15









      grontimgrontim

      310110




      310110






















          1 Answer
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          $begingroup$

          Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.



          Capital variables are very easy to mistype, so I'll use lower case.



          Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.



          Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
          $$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
          $$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
          $$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
          $$cong k[a,b,u]/(a^n-b^n,au-1)$$






          share|cite|improve this answer









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            1 Answer
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            1 Answer
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            active

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            1












            $begingroup$

            Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.



            Capital variables are very easy to mistype, so I'll use lower case.



            Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.



            Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
            $$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
            $$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
            $$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
            $$cong k[a,b,u]/(a^n-b^n,au-1)$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.



              Capital variables are very easy to mistype, so I'll use lower case.



              Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.



              Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
              $$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
              $$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
              $$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
              $$cong k[a,b,u]/(a^n-b^n,au-1)$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.



                Capital variables are very easy to mistype, so I'll use lower case.



                Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.



                Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
                $$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
                $$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
                $$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
                $$cong k[a,b,u]/(a^n-b^n,au-1)$$






                share|cite|improve this answer









                $endgroup$



                Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.



                Capital variables are very easy to mistype, so I'll use lower case.



                Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.



                Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
                $$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
                $$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
                $$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
                $$cong k[a,b,u]/(a^n-b^n,au-1)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 19:07









                jgonjgon

                13.5k22041




                13.5k22041






























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