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

7,98461233




7,98461233










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











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    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

        8,7491620




        8,7491620























            -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

            14.3k103554




            14.3k103554










            answered Dec 12 '18 at 17:59









            Tsemo AristideTsemo Aristide

            57.8k11445




            57.8k11445












            • $begingroup$
              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


















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