Low-level proof of identity related to Weierstrass P-function

Question

A theorem which can be extracted from Theorem V.1.1 of Silverman's "advanced topics in the theory of elliptic curves" is the following. Here $\mathbb{Q}(u)$ denotes rational functions in a variable $u$ and $\mathbb{Q}(u)[[q]]$ denotes the formal power series ring in one variable $q$ over this field.

Make the following definitions (they're all elements of $\mathbb{Q}(u)[[q]]$, and $\sigma_k(n)$ is the sum of the $k$th powers of the positive divisors of $n$);

$$s_k=\sum_{n\geq1}\sigma_k(n)q^n;$$ $$a_4=-5s_3=-5q-45q^2-\cdots;$$ $$a_6=-\frac{5s_3+7s_5}{12}=-q-23q^2-\cdots;$$ $$X=\frac{u}{(1-u)^2}+\sum_{n\geq1}\left(\sum_{0<d\mid n}d(u^d+u^{-d}-2)\right)q^n=\frac{u}{(1-u)^2} + \frac{(1-u)^2}{u}q + \cdots;$$ $$Y=\frac{u^2}{(1-u)^3}+\sum_{n\geq1}\left(\sum_{0<d\mid n}\binom{d}{2}u^d-\binom{d+1}{2}u^{-d}+d\right)q^n=\frac{u^3}{(1-u)^3}+\frac{u-1}{u}q+\cdots.$$

Theorem. $$Y^2+XY=X^3+a_4X+a_6.$$

This is a purely algebraic statement. The proof in Silverman 2 is: first develop the theory of the Weierstrass $\mathcal{P}$-function, prove $\mathcal{P}'^2=4\mathcal{P}^3+g_4\mathcal{P}+g_6$ using complex analysis, and now work out the power series expansions of everything and all the powers of $\pi$ magically cancel and you get the result above. Theorem V.1.1 isn't quite stated like this but it's equivalent after a little elementary manipulation. Here is some pari-gp code which convinced me that I've not made a slip:

q
u
s3=sum(n=1,10,sigma(n,3)*q^n)+O(q^11)
s5=sum(n=1,10,sigma(n,5)*q^n)+O(q^11)
a4=-5*s3
a6=-(5*s3+7*s5)/12
X=u/(1-u)^2+sum(n=1,10,sumdiv(n,d,d*(u^d+u^(-d)-2))*q^n)+O(q^11)
Y=u^2/(1-u)^3+sum(n=1,10,sumdiv(n,d,d*(d-1)/2*u^d-d*(d+1)/2*u^(-d)+d)*q^n)+O(q^11)
Y^2+X*Y-X^3-a4*X-a6

and it duly prints out O(q^11) for the calculation.

I looked through the literature and I couldn't find anyone proving this directly, everyone seems to resort to the argument involving Weierstrass P. Is there a more low-level/elementary (note: I initially said "combinatorial" but see the comments) proof of this?

I'm interested in this identity because one can use it to analyse the $p$-adic points of an elliptic curve with multiplicative reduction, via the theory of the Tate curve, which has this identity as an input.

Answer 1

Here are two other approaches to the formula, one of which may satisfy you, depending on your tastes:

• Roquette's Analytic Theory of Elliptic Functions over Local Fields includes a self-contained discussion of this formula in Section 3 on pages 23 to 29. Roquette's approach, though, is (non-Archimedean) analytic and also uses a bit of the theory of genus-1 function fields--not really low-level.

• In the paper where he introduces the tau function, "On certain arithmetical functions," Ramanujan sketches a direct proof of something related to $(\wp')^2 = 4\wp^3 + a_4\wp + a_6$, namely $\wp'' = 6\wp^2 + \frac{a_4}{2}$ (Equation (21) in that paper). One can find some further details of the computations in Chapter 4 ("Eisenstein Series") of Berndt's Number Theory in the Spirit of Ramanujan, but Berndt leaves part of the work as an exercise for his readers. A more comprehensive treatment, generalizing Ramanujan's approach, is in the first chapter of Development of Elliptic Functions According to Ramanujan by Venkatachaliengar and Cooper. The authors reach $(\wp')^2 = 4\wp^3 + a_4\wp + a_6$ at the bottom of page 11. Perhaps by reducing the generality of Venkatachaliengar and Cooper's intermediate formulas, one could produce a minimal low-level proof that fills in the details of Ramanujan's sketch.