Polynomials that suffice $S = V(f,g)$ for $S$ finite set in $mathbb{A}^2(mathbb{C})$












0












$begingroup$


Let $ S subset mathbb{A}^2(mathbb{C})$ a finite set of points. Show that exists polynomials $f$ and $g$ such that S = V(f,g).



I know that if I am working in $mathbb{A}^2(mathbb{R})$ I can make horizontal lines and little circles tangent to those lines and make the polynomials $f$ and $g$ the multiplication of such lines and circles, (given that the radius of these circles is less then the minimum of the vertical distance between all points).



My problem is if we consider $mathbb{A}^2(mathbb{C})$, because then I lose control of other intersection points. How can I solve this prroblem considering in $mathbb{A}^2(mathbb{C})$?






algebraic-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you do it for $|S| = 1$?
    $endgroup$
    – Eric Towers
    Dec 12 '18 at 17:54










  • $begingroup$
    Then can you do "something" that takes two varieties and gives you back their union?
    $endgroup$
    – Eric Towers
    Dec 12 '18 at 17:56
















0












$begingroup$


Let $ S subset mathbb{A}^2(mathbb{C})$ a finite set of points. Show that exists polynomials $f$ and $g$ such that S = V(f,g).



I know that if I am working in $mathbb{A}^2(mathbb{R})$ I can make horizontal lines and little circles tangent to those lines and make the polynomials $f$ and $g$ the multiplication of such lines and circles, (given that the radius of these circles is less then the minimum of the vertical distance between all points).



My problem is if we consider $mathbb{A}^2(mathbb{C})$, because then I lose control of other intersection points. How can I solve this prroblem considering in $mathbb{A}^2(mathbb{C})$?






algebraic-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you do it for $|S| = 1$?
    $endgroup$
    – Eric Towers
    Dec 12 '18 at 17:54










  • $begingroup$
    Then can you do "something" that takes two varieties and gives you back their union?
    $endgroup$
    – Eric Towers
    Dec 12 '18 at 17:56














0












0








0





$begingroup$


Let $ S subset mathbb{A}^2(mathbb{C})$ a finite set of points. Show that exists polynomials $f$ and $g$ such that S = V(f,g).



I know that if I am working in $mathbb{A}^2(mathbb{R})$ I can make horizontal lines and little circles tangent to those lines and make the polynomials $f$ and $g$ the multiplication of such lines and circles, (given that the radius of these circles is less then the minimum of the vertical distance between all points).



My problem is if we consider $mathbb{A}^2(mathbb{C})$, because then I lose control of other intersection points. How can I solve this prroblem considering in $mathbb{A}^2(mathbb{C})$?






algebraic-geometry







share|cite|improve this question











$endgroup$




Let $ S subset mathbb{A}^2(mathbb{C})$ a finite set of points. Show that exists polynomials $f$ and $g$ such that S = V(f,g).



I know that if I am working in $mathbb{A}^2(mathbb{R})$ I can make horizontal lines and little circles tangent to those lines and make the polynomials $f$ and $g$ the multiplication of such lines and circles, (given that the radius of these circles is less then the minimum of the vertical distance between all points).



My problem is if we consider $mathbb{A}^2(mathbb{C})$, because then I lose control of other intersection points. How can I solve this prroblem considering in $mathbb{A}^2(mathbb{C})$?






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 12 '18 at 18:36









Key Flex

8,28261233




8,28261233










asked Dec 12 '18 at 17:49









Joao BregunciJoao Bregunci

216




216












  • $begingroup$
    Can you do it for $|S| = 1$?
    $endgroup$
    – Eric Towers
    Dec 12 '18 at 17:54










  • $begingroup$
    Then can you do "something" that takes two varieties and gives you back their union?
    $endgroup$
    – Eric Towers
    Dec 12 '18 at 17:56


















  • $begingroup$
    Can you do it for $|S| = 1$?
    $endgroup$
    – Eric Towers
    Dec 12 '18 at 17:54










  • $begingroup$
    Then can you do "something" that takes two varieties and gives you back their union?
    $endgroup$
    – Eric Towers
    Dec 12 '18 at 17:56
















$begingroup$
Can you do it for $|S| = 1$?
$endgroup$
– Eric Towers
Dec 12 '18 at 17:54




$begingroup$
Can you do it for $|S| = 1$?
$endgroup$
– Eric Towers
Dec 12 '18 at 17:54












$begingroup$
Then can you do "something" that takes two varieties and gives you back their union?
$endgroup$
– Eric Towers
Dec 12 '18 at 17:56




$begingroup$
Then can you do "something" that takes two varieties and gives you back their union?
$endgroup$
– Eric Towers
Dec 12 '18 at 17:56










2 Answers
2






active

oldest

votes


















2












$begingroup$

Suppose $S = { (a_1, b_1), (a_2, b_2), ldots, (a_n, b_n) }$. First, find a solution for the "generic" case in which all $a_i$ are distinct. (Hint to get started: there exists a polynomial function $p$ such that $p(a_1) = b_1, ldots, p(a_n) = b_n$.)



I will now give the idea of how to treat the general case. The idea will be to find some perturbation of $mathbb{A}^2(mathbb{C})$ which maps our problem into a problem of the generic case which was solved in the previous paragraph. As it turns out, instead of considering the whole class of automorphisms of $mathbb{A}^2(mathbb{C})$, it will suffice to consider a class of shear transformations $phi_lambda$, $(x, y) mapsto (x + lambda y, y)$. So, what you will need to prove is:




  • There exists $lambda in mathbb{C}$ such that $phi_lambda(S)$ has all $x$-coordinates distinct.

  • There exist $f,g$ such that $S = V(f,g)$ if and only if there exist $f', g'$ such that $phi_lambda(S) = V(f', g')$.






share|cite|improve this answer









$endgroup$





















    -2












    $begingroup$

    $S={(a_1,b_1),dotsc,(a_n,b_n)}$



    $P=(z_1-a_1)dotsb(z_n-a_n)$, $Q=(z_2-b_1)dotsb(z-b_n)$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If you mean $P = (z_1 - a_1) (z_1 - a_2) cdots (z_1 - a_n)$ and $Q = (z_2 - b_1) (z_2 - b_2) cdots (z_2 - b_n)$ - then wouldn't $Z(P,Q)$ also contain points of the form $(a_i, b_j)$ with $i ne j$?
      $endgroup$
      – Daniel Schepler
      Dec 12 '18 at 18:07











    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%2f3037008%2fpolynomials-that-suffice-s-vf-g-for-s-finite-set-in-mathbba2-mathb%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Suppose $S = { (a_1, b_1), (a_2, b_2), ldots, (a_n, b_n) }$. First, find a solution for the "generic" case in which all $a_i$ are distinct. (Hint to get started: there exists a polynomial function $p$ such that $p(a_1) = b_1, ldots, p(a_n) = b_n$.)



    I will now give the idea of how to treat the general case. The idea will be to find some perturbation of $mathbb{A}^2(mathbb{C})$ which maps our problem into a problem of the generic case which was solved in the previous paragraph. As it turns out, instead of considering the whole class of automorphisms of $mathbb{A}^2(mathbb{C})$, it will suffice to consider a class of shear transformations $phi_lambda$, $(x, y) mapsto (x + lambda y, y)$. So, what you will need to prove is:




    • There exists $lambda in mathbb{C}$ such that $phi_lambda(S)$ has all $x$-coordinates distinct.

    • There exist $f,g$ such that $S = V(f,g)$ if and only if there exist $f', g'$ such that $phi_lambda(S) = V(f', g')$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Suppose $S = { (a_1, b_1), (a_2, b_2), ldots, (a_n, b_n) }$. First, find a solution for the "generic" case in which all $a_i$ are distinct. (Hint to get started: there exists a polynomial function $p$ such that $p(a_1) = b_1, ldots, p(a_n) = b_n$.)



      I will now give the idea of how to treat the general case. The idea will be to find some perturbation of $mathbb{A}^2(mathbb{C})$ which maps our problem into a problem of the generic case which was solved in the previous paragraph. As it turns out, instead of considering the whole class of automorphisms of $mathbb{A}^2(mathbb{C})$, it will suffice to consider a class of shear transformations $phi_lambda$, $(x, y) mapsto (x + lambda y, y)$. So, what you will need to prove is:




      • There exists $lambda in mathbb{C}$ such that $phi_lambda(S)$ has all $x$-coordinates distinct.

      • There exist $f,g$ such that $S = V(f,g)$ if and only if there exist $f', g'$ such that $phi_lambda(S) = V(f', g')$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Suppose $S = { (a_1, b_1), (a_2, b_2), ldots, (a_n, b_n) }$. First, find a solution for the "generic" case in which all $a_i$ are distinct. (Hint to get started: there exists a polynomial function $p$ such that $p(a_1) = b_1, ldots, p(a_n) = b_n$.)



        I will now give the idea of how to treat the general case. The idea will be to find some perturbation of $mathbb{A}^2(mathbb{C})$ which maps our problem into a problem of the generic case which was solved in the previous paragraph. As it turns out, instead of considering the whole class of automorphisms of $mathbb{A}^2(mathbb{C})$, it will suffice to consider a class of shear transformations $phi_lambda$, $(x, y) mapsto (x + lambda y, y)$. So, what you will need to prove is:




        • There exists $lambda in mathbb{C}$ such that $phi_lambda(S)$ has all $x$-coordinates distinct.

        • There exist $f,g$ such that $S = V(f,g)$ if and only if there exist $f', g'$ such that $phi_lambda(S) = V(f', g')$.






        share|cite|improve this answer









        $endgroup$



        Suppose $S = { (a_1, b_1), (a_2, b_2), ldots, (a_n, b_n) }$. First, find a solution for the "generic" case in which all $a_i$ are distinct. (Hint to get started: there exists a polynomial function $p$ such that $p(a_1) = b_1, ldots, p(a_n) = b_n$.)



        I will now give the idea of how to treat the general case. The idea will be to find some perturbation of $mathbb{A}^2(mathbb{C})$ which maps our problem into a problem of the generic case which was solved in the previous paragraph. As it turns out, instead of considering the whole class of automorphisms of $mathbb{A}^2(mathbb{C})$, it will suffice to consider a class of shear transformations $phi_lambda$, $(x, y) mapsto (x + lambda y, y)$. So, what you will need to prove is:




        • There exists $lambda in mathbb{C}$ such that $phi_lambda(S)$ has all $x$-coordinates distinct.

        • There exist $f,g$ such that $S = V(f,g)$ if and only if there exist $f', g'$ such that $phi_lambda(S) = V(f', g')$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 18:37









        Daniel ScheplerDaniel Schepler

        8,9491620




        8,9491620























            -2












            $begingroup$

            $S={(a_1,b_1),dotsc,(a_n,b_n)}$



            $P=(z_1-a_1)dotsb(z_n-a_n)$, $Q=(z_2-b_1)dotsb(z-b_n)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              If you mean $P = (z_1 - a_1) (z_1 - a_2) cdots (z_1 - a_n)$ and $Q = (z_2 - b_1) (z_2 - b_2) cdots (z_2 - b_n)$ - then wouldn't $Z(P,Q)$ also contain points of the form $(a_i, b_j)$ with $i ne j$?
              $endgroup$
              – Daniel Schepler
              Dec 12 '18 at 18:07
















            -2












            $begingroup$

            $S={(a_1,b_1),dotsc,(a_n,b_n)}$



            $P=(z_1-a_1)dotsb(z_n-a_n)$, $Q=(z_2-b_1)dotsb(z-b_n)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              If you mean $P = (z_1 - a_1) (z_1 - a_2) cdots (z_1 - a_n)$ and $Q = (z_2 - b_1) (z_2 - b_2) cdots (z_2 - b_n)$ - then wouldn't $Z(P,Q)$ also contain points of the form $(a_i, b_j)$ with $i ne j$?
              $endgroup$
              – Daniel Schepler
              Dec 12 '18 at 18:07














            -2












            -2








            -2





            $begingroup$

            $S={(a_1,b_1),dotsc,(a_n,b_n)}$



            $P=(z_1-a_1)dotsb(z_n-a_n)$, $Q=(z_2-b_1)dotsb(z-b_n)$.






            share|cite|improve this answer











            $endgroup$



            $S={(a_1,b_1),dotsc,(a_n,b_n)}$



            $P=(z_1-a_1)dotsb(z_n-a_n)$, $Q=(z_2-b_1)dotsb(z-b_n)$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 12 '18 at 18:20









            Xander Henderson

            14.6k103555




            14.6k103555










            answered Dec 12 '18 at 17:59









            Tsemo AristideTsemo Aristide

            58.7k11445




            58.7k11445












            • $begingroup$
              If you mean $P = (z_1 - a_1) (z_1 - a_2) cdots (z_1 - a_n)$ and $Q = (z_2 - b_1) (z_2 - b_2) cdots (z_2 - b_n)$ - then wouldn't $Z(P,Q)$ also contain points of the form $(a_i, b_j)$ with $i ne j$?
              $endgroup$
              – Daniel Schepler
              Dec 12 '18 at 18:07


















            • $begingroup$
              If you mean $P = (z_1 - a_1) (z_1 - a_2) cdots (z_1 - a_n)$ and $Q = (z_2 - b_1) (z_2 - b_2) cdots (z_2 - b_n)$ - then wouldn't $Z(P,Q)$ also contain points of the form $(a_i, b_j)$ with $i ne j$?
              $endgroup$
              – Daniel Schepler
              Dec 12 '18 at 18:07
















            $begingroup$
            If you mean $P = (z_1 - a_1) (z_1 - a_2) cdots (z_1 - a_n)$ and $Q = (z_2 - b_1) (z_2 - b_2) cdots (z_2 - b_n)$ - then wouldn't $Z(P,Q)$ also contain points of the form $(a_i, b_j)$ with $i ne j$?
            $endgroup$
            – Daniel Schepler
            Dec 12 '18 at 18:07




            $begingroup$
            If you mean $P = (z_1 - a_1) (z_1 - a_2) cdots (z_1 - a_n)$ and $Q = (z_2 - b_1) (z_2 - b_2) cdots (z_2 - b_n)$ - then wouldn't $Z(P,Q)$ also contain points of the form $(a_i, b_j)$ with $i ne j$?
            $endgroup$
            – Daniel Schepler
            Dec 12 '18 at 18:07


















            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%2f3037008%2fpolynomials-that-suffice-s-vf-g-for-s-finite-set-in-mathbba2-mathb%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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more