A Rationality Theorem

In the last lecture, I mentioned that it can be surprisingly difficult to say anything precise about $L(E,s)$, even with the above formulas. For example, it is a very deep theorem of Gross and Zagier that for the elliptic curve $y^2 + y = x^3 - 7x + 6$ we have

$$L(E,s) = c(s-1)^3 +$$    higher order terms$$,$$

and nobody has any idea how to prove that there is an elliptic curve with

$$L(E,s) = c(s-1)^4 +$$    higher order terms$$.$$

Fortunately, it is possible to decide whether or not $L(E,1)=0$.

Theorem 1.1   Let $y^2 = x^3 + ax + b$ be an elliptic curve. Let

$$\Omega_E = 2^n\int_{\gamma}^{\infty} \frac{dx}{\sqrt{x^3+ax+b}},$$

where $\gamma$ is the largest real root of $x^3 +ax+b$, and $n=0$ if $\Delta(E)<0$, $n=1$ if $\Delta(E)>0$. Then

$$\frac{L(E,1)}{\Omega_E} \in \mathbb{Q},$$

and the denominator is $\leq 24$.

In practice, one computes $\Omega_E$ using the ``Arithmetic-Geometric Mean'', NOT numerical integration. In PARI, $\Omega_E$ is approximated by E.omega[1]*2^(E.disc>0).

Remark 1.2   I don't know if the denominator is ever really as big as $24$. It would be a fun student project to either find an example, or to understand the proof that the quotient is rational and prove that $24$ can be replaced by something smaller.

Example 1.3   Let $E$ be the elliptic curve $y^2 = x^3 - 43x + 166$. We compute $L(E,1)$ using the above formula and observe that $L(E,1)/\Omega_E$ appears to be a rational number, as predicted by the theorem.

? E = ellinit([0,0,0,-43,166]);
? E = ellchangecurve(E, ellglobalred(E)[2]);
? eps = ellrootno(E)
%77 = 1
? N = ellglobalred(E)[1]
%78 = 26
? L = (1+eps) * sum(n=1,100, ellak(E,n)/n * exp(-2*Pi*n/sqrt(N)))
%79 = 0.6209653495490554663758626727
? Om = E.omega[1]*2^(E.disc>0)
%80 = 4.346757446843388264631038710
? L/Om
%81 = 0.1428571428571428571428571427
? contfrac(L/Om)
%84 = [0, 7]
? 1/7.0
%85 = 0.1428571428571428571428571428
? elltors(E)
%86 = [7, [7], [[1, 0]]]

Notice that in this example, $L(E,1)/\Omega_E = 1/7 = 1/\char93 E(\mathbb{Q})$. This is shadow of a more refined conjecture of Birch and Swinnerton-Dyer.

Example 1.4   In this example, we verify that $L(E,1)=0$ computationally.

? E=ellinit([0, 1, 1, -2, 0]);
? L1 = elllseries(E,1)
%4 = -6.881235151133426545894712438 E-29
? Omega = E.omega[1]*2^(E.disc>0)
%5 = 4.980425121710110150642715583
? L1/Omega
%6 = 1.795732353252503036074927634 E-20