Growth of Mordell-Weil Rank of Elliptic Curves over Field Extensions












3












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I'm a graduate student just checking to make sure that what he's researching isn't already known.



Let $mathbb{F}$ be a number field, and let $E$ be an elliptic curve defined over $mathbb{F}$. Is it already known that, for any integer $rgeq1$, there exists a finite-degree extension $mathbb{K}$ of $mathbb{F}$ so that $textrm{rank}left(Eleft(mathbb{K}right)right)geq r$?






elliptic-curves







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    3












    $begingroup$


    I'm a graduate student just checking to make sure that what he's researching isn't already known.



    Let $mathbb{F}$ be a number field, and let $E$ be an elliptic curve defined over $mathbb{F}$. Is it already known that, for any integer $rgeq1$, there exists a finite-degree extension $mathbb{K}$ of $mathbb{F}$ so that $textrm{rank}left(Eleft(mathbb{K}right)right)geq r$?






    elliptic-curves







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I'm a graduate student just checking to make sure that what he's researching isn't already known.



      Let $mathbb{F}$ be a number field, and let $E$ be an elliptic curve defined over $mathbb{F}$. Is it already known that, for any integer $rgeq1$, there exists a finite-degree extension $mathbb{K}$ of $mathbb{F}$ so that $textrm{rank}left(Eleft(mathbb{K}right)right)geq r$?






      elliptic-curves







      share|cite|improve this question









      $endgroup$




      I'm a graduate student just checking to make sure that what he's researching isn't already known.



      Let $mathbb{F}$ be a number field, and let $E$ be an elliptic curve defined over $mathbb{F}$. Is it already known that, for any integer $rgeq1$, there exists a finite-degree extension $mathbb{K}$ of $mathbb{F}$ so that $textrm{rank}left(Eleft(mathbb{K}right)right)geq r$?






      elliptic-curves




      elliptic-curves






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 6 hours ago









      MCSMCS

      1875




      1875






















          2 Answers
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          $begingroup$

          This is (unfortunately for you) well-known. Theorem 10.1 of this paper shows that over an algebraically closed field of characteristic zero the Mordell-Weil rank of any positive-dimensional abelian variety is infinite, and (as remark 1 below the theorem mentions) this implies that for any abelian variety over field of characteristic zero, there are finite extensions over which the ranks get arbitrarily large.



          The result also holds for fields of positive characteristic which are not algebraic over a finite field.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah well. Thanks for the prompt response.
            $endgroup$
            – MCS
            4 hours ago



















          1












          $begingroup$

          For elliptic curves you can be more specific. For example, it's not hard to show that there are elements $a_1,ldots,a_rinmathbb{F}$ so that for $mathbb{K}=mathbb{F}(sqrt{a_1},ldots,sqrt{a_r})$ one has $operatorname{rank}E(mathbb{K})ge r$. In particular, $E$ has infinite rank over the maximal abelian extension of $mathbb F$ of exponent 2. However, if you ask for the minimal degree of an extension $mathbb K/mathbb F$ such that $E(mathbb K)$ has rank $r$, then I think that you get an interesting question, although that, too, has certainly been studied.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you outline or provide a reference regarding the first claim? I am aware of how to do that for $r=1$, but I'm not so sure how to do that for $r>1$.
            $endgroup$
            – Wojowu
            3 hours ago










          • $begingroup$
            Outline: $E:y^2=f(x)$. Find values $x_1,x_2,ldots,x_r$ so that the fields $mathbb{F}(sqrt{f(x_i)})$ are distinct. To do this, choose the $x_i$ successively so that each $f(x_i)$ is divisible by a prime (to an odd power) that doesn't divide any of the others. There's a similar, but rather more elaborate, argument in my paper with Rosen: On the independence of Heegner points associated to distinct quadratic imaginary fields. J. Number Theory 127 (2007), no. 1, 10–36.
            $endgroup$
            – Joe Silverman
            2 hours ago











          Your Answer





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          2 Answers
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          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          5












          $begingroup$

          This is (unfortunately for you) well-known. Theorem 10.1 of this paper shows that over an algebraically closed field of characteristic zero the Mordell-Weil rank of any positive-dimensional abelian variety is infinite, and (as remark 1 below the theorem mentions) this implies that for any abelian variety over field of characteristic zero, there are finite extensions over which the ranks get arbitrarily large.



          The result also holds for fields of positive characteristic which are not algebraic over a finite field.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah well. Thanks for the prompt response.
            $endgroup$
            – MCS
            4 hours ago
















          5












          $begingroup$

          This is (unfortunately for you) well-known. Theorem 10.1 of this paper shows that over an algebraically closed field of characteristic zero the Mordell-Weil rank of any positive-dimensional abelian variety is infinite, and (as remark 1 below the theorem mentions) this implies that for any abelian variety over field of characteristic zero, there are finite extensions over which the ranks get arbitrarily large.



          The result also holds for fields of positive characteristic which are not algebraic over a finite field.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah well. Thanks for the prompt response.
            $endgroup$
            – MCS
            4 hours ago














          5












          5








          5





          $begingroup$

          This is (unfortunately for you) well-known. Theorem 10.1 of this paper shows that over an algebraically closed field of characteristic zero the Mordell-Weil rank of any positive-dimensional abelian variety is infinite, and (as remark 1 below the theorem mentions) this implies that for any abelian variety over field of characteristic zero, there are finite extensions over which the ranks get arbitrarily large.



          The result also holds for fields of positive characteristic which are not algebraic over a finite field.






          share|cite|improve this answer









          $endgroup$



          This is (unfortunately for you) well-known. Theorem 10.1 of this paper shows that over an algebraically closed field of characteristic zero the Mordell-Weil rank of any positive-dimensional abelian variety is infinite, and (as remark 1 below the theorem mentions) this implies that for any abelian variety over field of characteristic zero, there are finite extensions over which the ranks get arbitrarily large.



          The result also holds for fields of positive characteristic which are not algebraic over a finite field.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 6 hours ago









          WojowuWojowu

          6,64912851




          6,64912851












          • $begingroup$
            Ah well. Thanks for the prompt response.
            $endgroup$
            – MCS
            4 hours ago


















          • $begingroup$
            Ah well. Thanks for the prompt response.
            $endgroup$
            – MCS
            4 hours ago
















          $begingroup$
          Ah well. Thanks for the prompt response.
          $endgroup$
          – MCS
          4 hours ago




          $begingroup$
          Ah well. Thanks for the prompt response.
          $endgroup$
          – MCS
          4 hours ago











          1












          $begingroup$

          For elliptic curves you can be more specific. For example, it's not hard to show that there are elements $a_1,ldots,a_rinmathbb{F}$ so that for $mathbb{K}=mathbb{F}(sqrt{a_1},ldots,sqrt{a_r})$ one has $operatorname{rank}E(mathbb{K})ge r$. In particular, $E$ has infinite rank over the maximal abelian extension of $mathbb F$ of exponent 2. However, if you ask for the minimal degree of an extension $mathbb K/mathbb F$ such that $E(mathbb K)$ has rank $r$, then I think that you get an interesting question, although that, too, has certainly been studied.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you outline or provide a reference regarding the first claim? I am aware of how to do that for $r=1$, but I'm not so sure how to do that for $r>1$.
            $endgroup$
            – Wojowu
            3 hours ago










          • $begingroup$
            Outline: $E:y^2=f(x)$. Find values $x_1,x_2,ldots,x_r$ so that the fields $mathbb{F}(sqrt{f(x_i)})$ are distinct. To do this, choose the $x_i$ successively so that each $f(x_i)$ is divisible by a prime (to an odd power) that doesn't divide any of the others. There's a similar, but rather more elaborate, argument in my paper with Rosen: On the independence of Heegner points associated to distinct quadratic imaginary fields. J. Number Theory 127 (2007), no. 1, 10–36.
            $endgroup$
            – Joe Silverman
            2 hours ago
















          1












          $begingroup$

          For elliptic curves you can be more specific. For example, it's not hard to show that there are elements $a_1,ldots,a_rinmathbb{F}$ so that for $mathbb{K}=mathbb{F}(sqrt{a_1},ldots,sqrt{a_r})$ one has $operatorname{rank}E(mathbb{K})ge r$. In particular, $E$ has infinite rank over the maximal abelian extension of $mathbb F$ of exponent 2. However, if you ask for the minimal degree of an extension $mathbb K/mathbb F$ such that $E(mathbb K)$ has rank $r$, then I think that you get an interesting question, although that, too, has certainly been studied.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you outline or provide a reference regarding the first claim? I am aware of how to do that for $r=1$, but I'm not so sure how to do that for $r>1$.
            $endgroup$
            – Wojowu
            3 hours ago










          • $begingroup$
            Outline: $E:y^2=f(x)$. Find values $x_1,x_2,ldots,x_r$ so that the fields $mathbb{F}(sqrt{f(x_i)})$ are distinct. To do this, choose the $x_i$ successively so that each $f(x_i)$ is divisible by a prime (to an odd power) that doesn't divide any of the others. There's a similar, but rather more elaborate, argument in my paper with Rosen: On the independence of Heegner points associated to distinct quadratic imaginary fields. J. Number Theory 127 (2007), no. 1, 10–36.
            $endgroup$
            – Joe Silverman
            2 hours ago














          1












          1








          1





          $begingroup$

          For elliptic curves you can be more specific. For example, it's not hard to show that there are elements $a_1,ldots,a_rinmathbb{F}$ so that for $mathbb{K}=mathbb{F}(sqrt{a_1},ldots,sqrt{a_r})$ one has $operatorname{rank}E(mathbb{K})ge r$. In particular, $E$ has infinite rank over the maximal abelian extension of $mathbb F$ of exponent 2. However, if you ask for the minimal degree of an extension $mathbb K/mathbb F$ such that $E(mathbb K)$ has rank $r$, then I think that you get an interesting question, although that, too, has certainly been studied.






          share|cite|improve this answer









          $endgroup$



          For elliptic curves you can be more specific. For example, it's not hard to show that there are elements $a_1,ldots,a_rinmathbb{F}$ so that for $mathbb{K}=mathbb{F}(sqrt{a_1},ldots,sqrt{a_r})$ one has $operatorname{rank}E(mathbb{K})ge r$. In particular, $E$ has infinite rank over the maximal abelian extension of $mathbb F$ of exponent 2. However, if you ask for the minimal degree of an extension $mathbb K/mathbb F$ such that $E(mathbb K)$ has rank $r$, then I think that you get an interesting question, although that, too, has certainly been studied.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Joe SilvermanJoe Silverman

          30.9k182158




          30.9k182158












          • $begingroup$
            Could you outline or provide a reference regarding the first claim? I am aware of how to do that for $r=1$, but I'm not so sure how to do that for $r>1$.
            $endgroup$
            – Wojowu
            3 hours ago










          • $begingroup$
            Outline: $E:y^2=f(x)$. Find values $x_1,x_2,ldots,x_r$ so that the fields $mathbb{F}(sqrt{f(x_i)})$ are distinct. To do this, choose the $x_i$ successively so that each $f(x_i)$ is divisible by a prime (to an odd power) that doesn't divide any of the others. There's a similar, but rather more elaborate, argument in my paper with Rosen: On the independence of Heegner points associated to distinct quadratic imaginary fields. J. Number Theory 127 (2007), no. 1, 10–36.
            $endgroup$
            – Joe Silverman
            2 hours ago


















          • $begingroup$
            Could you outline or provide a reference regarding the first claim? I am aware of how to do that for $r=1$, but I'm not so sure how to do that for $r>1$.
            $endgroup$
            – Wojowu
            3 hours ago










          • $begingroup$
            Outline: $E:y^2=f(x)$. Find values $x_1,x_2,ldots,x_r$ so that the fields $mathbb{F}(sqrt{f(x_i)})$ are distinct. To do this, choose the $x_i$ successively so that each $f(x_i)$ is divisible by a prime (to an odd power) that doesn't divide any of the others. There's a similar, but rather more elaborate, argument in my paper with Rosen: On the independence of Heegner points associated to distinct quadratic imaginary fields. J. Number Theory 127 (2007), no. 1, 10–36.
            $endgroup$
            – Joe Silverman
            2 hours ago
















          $begingroup$
          Could you outline or provide a reference regarding the first claim? I am aware of how to do that for $r=1$, but I'm not so sure how to do that for $r>1$.
          $endgroup$
          – Wojowu
          3 hours ago




          $begingroup$
          Could you outline or provide a reference regarding the first claim? I am aware of how to do that for $r=1$, but I'm not so sure how to do that for $r>1$.
          $endgroup$
          – Wojowu
          3 hours ago












          $begingroup$
          Outline: $E:y^2=f(x)$. Find values $x_1,x_2,ldots,x_r$ so that the fields $mathbb{F}(sqrt{f(x_i)})$ are distinct. To do this, choose the $x_i$ successively so that each $f(x_i)$ is divisible by a prime (to an odd power) that doesn't divide any of the others. There's a similar, but rather more elaborate, argument in my paper with Rosen: On the independence of Heegner points associated to distinct quadratic imaginary fields. J. Number Theory 127 (2007), no. 1, 10–36.
          $endgroup$
          – Joe Silverman
          2 hours ago




          $begingroup$
          Outline: $E:y^2=f(x)$. Find values $x_1,x_2,ldots,x_r$ so that the fields $mathbb{F}(sqrt{f(x_i)})$ are distinct. To do this, choose the $x_i$ successively so that each $f(x_i)$ is divisible by a prime (to an odd power) that doesn't divide any of the others. There's a similar, but rather more elaborate, argument in my paper with Rosen: On the independence of Heegner points associated to distinct quadratic imaginary fields. J. Number Theory 127 (2007), no. 1, 10–36.
          $endgroup$
          – Joe Silverman
          2 hours ago


















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