Can this argument about homogenous polynomials over $mathbb{R}$ in two variables having graphs that are...












2












$begingroup$


If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.



Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.



(This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.



Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?






algebraic-geometry







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.



    Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.



    (This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.



    Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?






    algebraic-geometry







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.



      Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.



      (This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.



      Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?






      algebraic-geometry







      share|cite|improve this question











      $endgroup$




      If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.



      Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.



      (This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.



      Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?






      algebraic-geometry




      algebraic-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 2:18







      Michael Barz

















      asked Dec 31 '18 at 2:10









      Michael BarzMichael Barz

      915




      915






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
            $endgroup$
            – Michael Barz
            Dec 31 '18 at 2:26






          • 3




            $begingroup$
            I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
            $endgroup$
            – KReiser
            Dec 31 '18 at 2:34













          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',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          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%2f3057379%2fcan-this-argument-about-homogenous-polynomials-over-mathbbr-in-two-variable%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
            $endgroup$
            – Michael Barz
            Dec 31 '18 at 2:26






          • 3




            $begingroup$
            I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
            $endgroup$
            – KReiser
            Dec 31 '18 at 2:34


















          4












          $begingroup$

          It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
            $endgroup$
            – Michael Barz
            Dec 31 '18 at 2:26






          • 3




            $begingroup$
            I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
            $endgroup$
            – KReiser
            Dec 31 '18 at 2:34
















          4












          4








          4





          $begingroup$

          It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.






          share|cite|improve this answer









          $endgroup$



          It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 2:21









          KReiserKReiser

          9,94021435




          9,94021435












          • $begingroup$
            I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
            $endgroup$
            – Michael Barz
            Dec 31 '18 at 2:26






          • 3




            $begingroup$
            I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
            $endgroup$
            – KReiser
            Dec 31 '18 at 2:34




















          • $begingroup$
            I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
            $endgroup$
            – Michael Barz
            Dec 31 '18 at 2:26






          • 3




            $begingroup$
            I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
            $endgroup$
            – KReiser
            Dec 31 '18 at 2:34


















          $begingroup$
          I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
          $endgroup$
          – Michael Barz
          Dec 31 '18 at 2:26




          $begingroup$
          I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
          $endgroup$
          – Michael Barz
          Dec 31 '18 at 2:26




          3




          3




          $begingroup$
          I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
          $endgroup$
          – KReiser
          Dec 31 '18 at 2:34






          $begingroup$
          I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
          $endgroup$
          – KReiser
          Dec 31 '18 at 2:34




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057379%2fcan-this-argument-about-homogenous-polynomials-over-mathbbr-in-two-variable%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more