Ideal of regular functions in $Z(xy-z^2)$ vanishing on $(x,y)$ is not principal












2














Let $H = Z(xy-z^2) subset Bbb A ^ {3}$.



And let $L = Z(x,z)$.



I need to show that $L subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.



So showing $Lsubset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.



Im not sure how to show the last part.



The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) cap (x,z)$ is not principal.



Any ideas how?






algebraic-geometry affine-geometry







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    2














    Let $H = Z(xy-z^2) subset Bbb A ^ {3}$.



    And let $L = Z(x,z)$.



    I need to show that $L subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.



    So showing $Lsubset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.



    Im not sure how to show the last part.



    The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) cap (x,z)$ is not principal.



    Any ideas how?






    algebraic-geometry affine-geometry







    share|cite|improve this question

























      2












      2








      2


      2





      Let $H = Z(xy-z^2) subset Bbb A ^ {3}$.



      And let $L = Z(x,z)$.



      I need to show that $L subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.



      So showing $Lsubset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.



      Im not sure how to show the last part.



      The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) cap (x,z)$ is not principal.



      Any ideas how?






      algebraic-geometry affine-geometry







      share|cite|improve this question













      Let $H = Z(xy-z^2) subset Bbb A ^ {3}$.



      And let $L = Z(x,z)$.



      I need to show that $L subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.



      So showing $Lsubset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.



      Im not sure how to show the last part.



      The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) cap (x,z)$ is not principal.



      Any ideas how?






      algebraic-geometry affine-geometry




      algebraic-geometry affine-geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      user123

      1,294316




      1,294316






















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          One approach is to consider the grading.



          $$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$



          So the dimensions are $2,4,6,dots$.



          On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be



          $$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$



          Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.






          share|cite|improve this answer





















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            1 Answer
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            0














            One approach is to consider the grading.



            $$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$



            So the dimensions are $2,4,6,dots$.



            On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be



            $$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$



            Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.






            share|cite|improve this answer


























              0














              One approach is to consider the grading.



              $$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$



              So the dimensions are $2,4,6,dots$.



              On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be



              $$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$



              Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.






              share|cite|improve this answer
























                0












                0








                0






                One approach is to consider the grading.



                $$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$



                So the dimensions are $2,4,6,dots$.



                On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be



                $$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$



                Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.






                share|cite|improve this answer












                One approach is to consider the grading.



                $$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$



                So the dimensions are $2,4,6,dots$.



                On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be



                $$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$



                Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                Trevor Gunn

                14.2k32046




                14.2k32046






























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