Proof that every codimension 1 subvariety of $mathbbP^n$ is $V((f))$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite
1












I've been told that




Theorem. Every codimension 1 subvariety of $mathbbP^n$ is $V((f))$, where $f$ is some prime homogeneous polynomial.




I'm under the impression that this is true over any algebraically closed field.



However, I'm unable to locate a proof. Does anyone know where a proof of this theorem can be located? A version of this for affine space is here, but a bit light on the details.



Also, in this context, does $V((f))$ mean $V$ of the principal ideal generated by $f$, or is the double bracketing indicative of some kind of a power series construction?










share|cite|improve this question

























    up vote
    2
    down vote

    favorite
    1












    I've been told that




    Theorem. Every codimension 1 subvariety of $mathbbP^n$ is $V((f))$, where $f$ is some prime homogeneous polynomial.




    I'm under the impression that this is true over any algebraically closed field.



    However, I'm unable to locate a proof. Does anyone know where a proof of this theorem can be located? A version of this for affine space is here, but a bit light on the details.



    Also, in this context, does $V((f))$ mean $V$ of the principal ideal generated by $f$, or is the double bracketing indicative of some kind of a power series construction?










    share|cite|improve this question























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      I've been told that




      Theorem. Every codimension 1 subvariety of $mathbbP^n$ is $V((f))$, where $f$ is some prime homogeneous polynomial.




      I'm under the impression that this is true over any algebraically closed field.



      However, I'm unable to locate a proof. Does anyone know where a proof of this theorem can be located? A version of this for affine space is here, but a bit light on the details.



      Also, in this context, does $V((f))$ mean $V$ of the principal ideal generated by $f$, or is the double bracketing indicative of some kind of a power series construction?










      share|cite|improve this question













      I've been told that




      Theorem. Every codimension 1 subvariety of $mathbbP^n$ is $V((f))$, where $f$ is some prime homogeneous polynomial.




      I'm under the impression that this is true over any algebraically closed field.



      However, I'm unable to locate a proof. Does anyone know where a proof of this theorem can be located? A version of this for affine space is here, but a bit light on the details.



      Also, in this context, does $V((f))$ mean $V$ of the principal ideal generated by $f$, or is the double bracketing indicative of some kind of a power series construction?







      algebraic-geometry projective-space






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Sep 10 at 7:15









      goblin

      35.9k1155182




      35.9k1155182




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .



          For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.



          In regards to $V((f))$, your first interpretation is correct.






          share|cite|improve this answer




















            Your Answer




            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2911618%2fproof-that-every-codimension-1-subvariety-of-mathbbpn-is-vf%23new-answer', 'question_page');

            );

            Post as a guest






























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .



            For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.



            In regards to $V((f))$, your first interpretation is correct.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .



              For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.



              In regards to $V((f))$, your first interpretation is correct.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .



                For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.



                In regards to $V((f))$, your first interpretation is correct.






                share|cite|improve this answer












                This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .



                For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.



                In regards to $V((f))$, your first interpretation is correct.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 10 at 10:25









                Gabriel Micolet

                1141




                1141



























                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2911618%2fproof-that-every-codimension-1-subvariety-of-mathbbpn-is-vf%23new-answer', 'question_page');

                    );

                    Post as a guest













































































                    這個網誌中的熱門文章

                    tkz-euclide: tkzDrawCircle[R] not working

                    How to combine Bézier curves to a surface?

                    1st Magritte Awards