Let $\varphi(u)$ be holomorphic in the neighborhood of the origin of the complex plane. One says that $\varphi(u)$ admits an ** algebraic addition theorem** if it satisfies a functional equation of the form $G[\varphi(u). \varphi(v), \varphi(u+v)]=0,$ where where G(X,Y,Z) is a non vanishing polynomial in the three variables X,Y,Z with complex constant coefficients, while $u,v, u+v$ are in the domain of $\varphi(u)$. Examples are the rational functions, the exponential function, the trigonometric functions, and the elliptic functions. Then it can be proved that $\varphi(u)$ and $\varphi'(u)$ are connected by an algebraic equation $A[\varphi(u)$,$\varphi'(u)]=0$, which defines the ELLIPTIC CURVE parameterized by $\varphi(u)$. It is known that ANY elliptic curve can be realized in this way.

If one finds the greatest common divisor of G(X,Y,Z) and $X'\frac{dG}{d Y}-Y'\frac{dG}{d X},$, where $X'$ means the derivative with respect to $u$, etc., we obtain an irreducible polynomial $D(Z,X,X',Y,Y')$, which, when put equal to zero, gives the GROUP LAW for A(X,X')=0. If the degree of D in Z is equal to one, then the group law is rational in X,X',Y,Y'. Here is the question:

Prove: ** the degree of D in Z is one iff $\varphi(u)$ is uniform** (i.e., has no branch points)

The proof must not use the properties of $\varphi(u)$ as rational, trigonometric, or elliptic functions...rather, only the uniformity of $\varphi(u)$.

I have not seen a proof...it is desirable that one be found. There ARE proofs using the properties of the rational, trigonometric, and elliptic functions.

This gives an elementary algorithm for finding the group law for any elliptic curve without detouring through the Weierstrass $\wp$ function.