5
$\begingroup$

Goldfeld proved the following result. Let $E$ be an elliptic curve (with conductor $N$) over $\mathbb{Q}$ whose Hasse-Weil L-function has a zero at $s = 1$ with multiplicity $g$ then for sufficiently large $D > 0$ then the class number $h(D)$ of an imaginary quadratic field with discriminant $-D$ satisfies the following inequality if $(N, D) = 1$, $$h(D) > c(g, N) {\log(D)}^{g - \mu - 1} \exp(-21\sqrt{g\log(\log(D))})$$ where $\mu = 1, 2$ depending on $\chi_{D}(-N) = {(-1)}^{g - \mu}$, $\chi_{D}$ being the quadratic character with conductor $-D$.

Now, Goldfeld claims that if there is an elliptic curve with $g = 3$ (which later Gross - Zagier came up with), then we can effectively find bounds for $D$ that satisfy $h(D) = n$ for any fixed $n$. What I am unable to understand is the fact that suppose with $g = 3$ it happened to be that $\mu = 2$, the above stated lower bound on $h(D)$ would we decreasing with $D$ and we wouldn't be able to bound $D$ (from above). For a fixed elliptic curve $\mu = \mu(D)$ is a function of $D$. So my question is, in that case how is it possible to conclude anything useful when $\mu = 2$ i.e. when $\chi_D(-N) = -1$ with $g = 3$.

$\endgroup$
2
  • 3
    $\begingroup$ Yeah so usually the auxiliary rank 3 curve that’s used has conductor equal to some small prime p times a square (see e.g. Prop. 7.4 of Gross-Zagier’s people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/BF01388809/… where p = 37). I see you assumed that $(D,N)= 1$ —- phew :p. (Otherwise see 4.3 of Oesterle’s numdam.org/article/SB_1983-1984__26__309_0.pdf .). Now $\chi_D(-1) = -1$ since we’re talking about im. q. fields, and if $\chi_D(N) = \chi_D(p) = 1$, then p would split, so p^h would be a nontrivial norm, so $h\gg \log{|D|}$. Otherwise $\mu = 1$ so use Goldfeld. $\endgroup$
    – alpoge
    Apr 25, 2021 at 0:26
  • 1
    $\begingroup$ @alpoge: Can you turn your comment to a response? So that this question can be closed. $\endgroup$
    – GH from MO
    Apr 25, 2021 at 6:50

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy