Inverse elliptic integral, Weierstrass function, in other fields Take a separable cubic polynomial $4x^3-ax-b = 4 (x-e_1)(x-e_2)(x-e_3)$, let $h'(x) = (4x^3-ax-b)^{-1/2}$ and define its elliptic integral $h(x)= \int h'(x)dx$. Let $P(z) = h^{-1}(z)$ its inverse function. Then $\displaystyle P'(z) = \frac{1}{h'(P(z))}$ and $$P'(z)^2 = P(z)^3-a P(z)-b \tag{1}$$


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*For $z,x \in \mathbb{C}$ this is the definition of the Weierstrass function $\wp$ of the complex elliptic curve $$E(\mathbb{C}) = \{ (x,y) \in \mathbb{C}^2, y^2 = 4x^3-ax-b\}$$ 
Question 1 : How to show easily that $P$ is doubly periodic ?
Take a $c \in\mathbb{C}$ with $P(c) \ne 0$ and a closed-curve $\gamma : P(c) \to P(c)$ enclosing one of the root $e_i$. Then $h \circ \gamma$ is a non-closed curve $c \to c+\omega$ and we find $$0 =  \int_\gamma dx = \int_{h \,\circ\, \gamma} P'(z)dz = P(c+\omega)-P(c)$$
Thanks to $(1)$ it implies $P'(c+\omega) = \pm P'(c)$. We can show the sign is $+$ (if it was not we could double $\omega$) and the differential equation shows that $P$ is $\omega$ periodic. 
Applying the same process with a curve enclosing a different root will produce a different period $\omega_2$ which is $\mathbb{Z}$-linearly independent to $\omega$ (why ?)
So $P$ is doubly periodic and we obtain the (Riemann surface and abelian group) isomorphism with a complex torus $$\varphi : \mathbb{C}/(\omega \mathbb{Z}+ \omega_2\mathbb{Z}) \to E(\mathbb{C}), \qquad \varphi(z) = (P(z),P'(z))$$

*Question 2 : Can we do the same with another (algebraically closed) field $K$ not contained in $\mathbb{C}$ and the corresponding elliptic curve over $K$ ? $K=\overline{\mathbb{F}}_p$ seems out of reach because it doesn't have an absolute value for making sense to analytic functions. 
What about the case $K=\overline{\mathbb{Q}}_p$ ? Algebraically, will $h$ be in some field of formal series, being the anti-derivative of $h' \in \overline{K(x)}$ ? And will its inverse function $P$ be well-defined ? In that case, does it tell us the structure of $E(\overline{\mathbb{Q}}_p)$ ?

One way to do this is to prove there is a lattice $\Lambda$ in $\Bbb C$
whose $\wp$-function satisfies
$$\wp'(z)^2=4\wp(z)^3-a\wp(z)-b.$$
Cox gives a proof in Primes of the form $x^2+ny^2$. This involves
the $j$ modular function. It is a fact that $j$ is surjective, so there
is $\tau$ with
$$j(\tau)=1728\frac{a^3}{a^3-27b^2}.$$
Then a lattice of the form $\alpha\Bbb Z+\alpha\tau\Bbb Z$  works.
As $P$ and $\wp$ satisfy the same differential equation and their
Laurent series at $0$ start off the same, then their Laurent series
are the same, so $P=\wp$ by analytic continuation. As $\wp$
is periodic, so is $P$.
