Hypersurfaces of degree $d$ in $mathbb{P}^n_k$ that contain a given closed $X$












6












$begingroup$


Let $k$ be an algebraically closed field and consider $mathbb{P}^n_k$, the be the $n-$dimensional projective space over $k$.



It is known that, for any integer $d>0$, there is a bijection between the hypersurfaces of degree $d$ and $mathbb{P} H^0(mathbb{P}^n_k, mathcal{O}_{mathbb{P}^n_k}(d))$.



Now, let $Xsubseteq mathbb{P}^n_k$ be a closed subscheme and let $mathcal I_X$ be the corresponding sheaf of ideals. Intuitively I can say that there is a bijection between the hypersurfaces of degree $d$ containing $X$ and $mathbb{P} H^0(mathbb{P}^n_k, mathcal{I}_X(d))$. How can I show this fact formally?






algebraic-geometry schemes projective-schemes







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    6












    $begingroup$


    Let $k$ be an algebraically closed field and consider $mathbb{P}^n_k$, the be the $n-$dimensional projective space over $k$.



    It is known that, for any integer $d>0$, there is a bijection between the hypersurfaces of degree $d$ and $mathbb{P} H^0(mathbb{P}^n_k, mathcal{O}_{mathbb{P}^n_k}(d))$.



    Now, let $Xsubseteq mathbb{P}^n_k$ be a closed subscheme and let $mathcal I_X$ be the corresponding sheaf of ideals. Intuitively I can say that there is a bijection between the hypersurfaces of degree $d$ containing $X$ and $mathbb{P} H^0(mathbb{P}^n_k, mathcal{I}_X(d))$. How can I show this fact formally?






    algebraic-geometry schemes projective-schemes







    share|cite|improve this question









    $endgroup$















      6












      6








      6


      1



      $begingroup$


      Let $k$ be an algebraically closed field and consider $mathbb{P}^n_k$, the be the $n-$dimensional projective space over $k$.



      It is known that, for any integer $d>0$, there is a bijection between the hypersurfaces of degree $d$ and $mathbb{P} H^0(mathbb{P}^n_k, mathcal{O}_{mathbb{P}^n_k}(d))$.



      Now, let $Xsubseteq mathbb{P}^n_k$ be a closed subscheme and let $mathcal I_X$ be the corresponding sheaf of ideals. Intuitively I can say that there is a bijection between the hypersurfaces of degree $d$ containing $X$ and $mathbb{P} H^0(mathbb{P}^n_k, mathcal{I}_X(d))$. How can I show this fact formally?






      algebraic-geometry schemes projective-schemes







      share|cite|improve this question









      $endgroup$




      Let $k$ be an algebraically closed field and consider $mathbb{P}^n_k$, the be the $n-$dimensional projective space over $k$.



      It is known that, for any integer $d>0$, there is a bijection between the hypersurfaces of degree $d$ and $mathbb{P} H^0(mathbb{P}^n_k, mathcal{O}_{mathbb{P}^n_k}(d))$.



      Now, let $Xsubseteq mathbb{P}^n_k$ be a closed subscheme and let $mathcal I_X$ be the corresponding sheaf of ideals. Intuitively I can say that there is a bijection between the hypersurfaces of degree $d$ containing $X$ and $mathbb{P} H^0(mathbb{P}^n_k, mathcal{I}_X(d))$. How can I show this fact formally?






      algebraic-geometry schemes projective-schemes




      algebraic-geometry schemes projective-schemes






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 17 at 0:16









      baobabbaobab

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          $begingroup$

          Consider the exact sequence $$ 0 to mathcal{I}_X to mathcal{O}_{Bbb P^n} to mathcal{O}_X to 0.$$



          After twisting by $mathcal{O}(d)$, this sequence becomes $$ 0 to mathcal{I}_X(d) to mathcal{O}_{Bbb P^n}(d) to mathcal{O}_X(d) to 0$$ and remains exact (we're tensoring with an invertible sheaf).



          Taking cohomology/global sections, we see that the exact sequence becomes $$ 0 to H^0(mathcal{I}_X(d)) to H^0(mathcal{O}_{Bbb P^n}(d)) to H^0(mathcal{O}_X(d))$$
          and now we may notice several things. First, the global sections of $mathcal{O}(d)$ are the degree $d$ homogeneous polynomials. Second, those degree $d$ homogeneous polynomials in $H^0(mathcal{O}(d))$ coming from $H^0(mathcal{I}_X(d))$ must vanish on $X$, by the exactness of the above sequence. So we may conclude that $H^0(mathcal{I}_X(d))$ is precisely the vector space of degree $d$ polynomials vanishing on $X$, and after taking the projectivization, we may say that $Bbb PH^0(mathcal{I}_X(d))$ is exactly the set of hypersurfaces of degree $d$ containing $X$.






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            $begingroup$

            Consider the exact sequence $$ 0 to mathcal{I}_X to mathcal{O}_{Bbb P^n} to mathcal{O}_X to 0.$$



            After twisting by $mathcal{O}(d)$, this sequence becomes $$ 0 to mathcal{I}_X(d) to mathcal{O}_{Bbb P^n}(d) to mathcal{O}_X(d) to 0$$ and remains exact (we're tensoring with an invertible sheaf).



            Taking cohomology/global sections, we see that the exact sequence becomes $$ 0 to H^0(mathcal{I}_X(d)) to H^0(mathcal{O}_{Bbb P^n}(d)) to H^0(mathcal{O}_X(d))$$
            and now we may notice several things. First, the global sections of $mathcal{O}(d)$ are the degree $d$ homogeneous polynomials. Second, those degree $d$ homogeneous polynomials in $H^0(mathcal{O}(d))$ coming from $H^0(mathcal{I}_X(d))$ must vanish on $X$, by the exactness of the above sequence. So we may conclude that $H^0(mathcal{I}_X(d))$ is precisely the vector space of degree $d$ polynomials vanishing on $X$, and after taking the projectivization, we may say that $Bbb PH^0(mathcal{I}_X(d))$ is exactly the set of hypersurfaces of degree $d$ containing $X$.






            share|cite|improve this answer









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              0





              +100







              $begingroup$

              Consider the exact sequence $$ 0 to mathcal{I}_X to mathcal{O}_{Bbb P^n} to mathcal{O}_X to 0.$$



              After twisting by $mathcal{O}(d)$, this sequence becomes $$ 0 to mathcal{I}_X(d) to mathcal{O}_{Bbb P^n}(d) to mathcal{O}_X(d) to 0$$ and remains exact (we're tensoring with an invertible sheaf).



              Taking cohomology/global sections, we see that the exact sequence becomes $$ 0 to H^0(mathcal{I}_X(d)) to H^0(mathcal{O}_{Bbb P^n}(d)) to H^0(mathcal{O}_X(d))$$
              and now we may notice several things. First, the global sections of $mathcal{O}(d)$ are the degree $d$ homogeneous polynomials. Second, those degree $d$ homogeneous polynomials in $H^0(mathcal{O}(d))$ coming from $H^0(mathcal{I}_X(d))$ must vanish on $X$, by the exactness of the above sequence. So we may conclude that $H^0(mathcal{I}_X(d))$ is precisely the vector space of degree $d$ polynomials vanishing on $X$, and after taking the projectivization, we may say that $Bbb PH^0(mathcal{I}_X(d))$ is exactly the set of hypersurfaces of degree $d$ containing $X$.






              share|cite|improve this answer









              $endgroup$
















                0





                +100







                0





                +100



                0




                +100



                $begingroup$

                Consider the exact sequence $$ 0 to mathcal{I}_X to mathcal{O}_{Bbb P^n} to mathcal{O}_X to 0.$$



                After twisting by $mathcal{O}(d)$, this sequence becomes $$ 0 to mathcal{I}_X(d) to mathcal{O}_{Bbb P^n}(d) to mathcal{O}_X(d) to 0$$ and remains exact (we're tensoring with an invertible sheaf).



                Taking cohomology/global sections, we see that the exact sequence becomes $$ 0 to H^0(mathcal{I}_X(d)) to H^0(mathcal{O}_{Bbb P^n}(d)) to H^0(mathcal{O}_X(d))$$
                and now we may notice several things. First, the global sections of $mathcal{O}(d)$ are the degree $d$ homogeneous polynomials. Second, those degree $d$ homogeneous polynomials in $H^0(mathcal{O}(d))$ coming from $H^0(mathcal{I}_X(d))$ must vanish on $X$, by the exactness of the above sequence. So we may conclude that $H^0(mathcal{I}_X(d))$ is precisely the vector space of degree $d$ polynomials vanishing on $X$, and after taking the projectivization, we may say that $Bbb PH^0(mathcal{I}_X(d))$ is exactly the set of hypersurfaces of degree $d$ containing $X$.






                share|cite|improve this answer









                $endgroup$



                Consider the exact sequence $$ 0 to mathcal{I}_X to mathcal{O}_{Bbb P^n} to mathcal{O}_X to 0.$$



                After twisting by $mathcal{O}(d)$, this sequence becomes $$ 0 to mathcal{I}_X(d) to mathcal{O}_{Bbb P^n}(d) to mathcal{O}_X(d) to 0$$ and remains exact (we're tensoring with an invertible sheaf).



                Taking cohomology/global sections, we see that the exact sequence becomes $$ 0 to H^0(mathcal{I}_X(d)) to H^0(mathcal{O}_{Bbb P^n}(d)) to H^0(mathcal{O}_X(d))$$
                and now we may notice several things. First, the global sections of $mathcal{O}(d)$ are the degree $d$ homogeneous polynomials. Second, those degree $d$ homogeneous polynomials in $H^0(mathcal{O}(d))$ coming from $H^0(mathcal{I}_X(d))$ must vanish on $X$, by the exactness of the above sequence. So we may conclude that $H^0(mathcal{I}_X(d))$ is precisely the vector space of degree $d$ polynomials vanishing on $X$, and after taking the projectivization, we may say that $Bbb PH^0(mathcal{I}_X(d))$ is exactly the set of hypersurfaces of degree $d$ containing $X$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 4:35









                KReiserKReiser

                9,72721435




                9,72721435






























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