Brumer's upper bound for the rank

Theorem. Let $L_E(s)$ be the L-function associated with an elliptic curve of conductor $N$. Let $r$ be the order of vanishing at the center of the critical strip. If RH holds for $L_E$, then

$$r\le(1/2+o(1))\frac{\log N}{\log \log N}$$

as $N\to \infty$.

Proof. Let $F(x)=1-\vert x\vert$ be supported on $[-1,1]$. The Fourier transform of this is $G(t)=(2/t^2)(1-\cos t)$, which is everywhere nonnegative. By the explicit formula for this L-function (see page 219 of Mestre), we have

$$\sum_\rho \Phi_\lambda(\rho)+ 2\sum_{p^m} b(p^m) F_\lambda(m... ...2\int_0^\infty {F_\lambda(x)\over e^x-1}-{e^{-x}\over x}\, dx,$$

where $b(p^m)$ is the Frobenius trace (i.e. minus the coefficient of the logarithmic derivative of $L_E(s)$. The coefficient is not normalized and so, by Hasse or Weil, $b(m)$ has size around $\sqrt{m}$ for $m$ a prime power.), $\Phi(s)=G(-i(s-1/2))$ and $F_\lambda(x)=F(x/\lambda)$ so that $G_\lambda(t)=\lambda G(\lambda t)$ where $\lambda$ is a parameter. The integral tends to $\gamma$ as $\lambda\rightarrow\infty$. The positivity of $\Phi$ (under GRH) and the Weil bound for $b(p^m)$ give us that $\lambda r G(0)\le \log N + 4\sum_{p^m\le e^\lambda} {\log p\over p^{m/2}}+O(1)$; the sum is easily bounded by $2e^{\lambda/2}\log 3$. By taking $\lambda=2\log\log N-\log\log\log N$, and noting that $G(0)=1$, we thus get that $r\le (1/2+o(1)) {\log N\over\log\log N}$.

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