Algebraic addition theorem

Given a function $f:\mathbb C\to\mathbb C$ (or, more generally, on another field $\mathbb k$) we say that $f$ has an algebraic addition theorem (following Hancock's terminology) if there exists a polynomial $G(\xi,\eta,\zeta)$ of three variables such that $$G(f(u)~ , ~ f(v) ~, ~ f(u+v)) = 0$$ for all $u,v\in\mathbb C$ (or $u,v\in\mathbb k$). To give a few examples:

• The identity function $f(u)=u$ has the trivial algebraic addition theorem given by the polynomial $$G(\xi,\eta,\zeta)=\xi+\eta-\zeta$$

• The exponential function $f(u)=e^u$ has the algebraic addition theorem given by $$G(\xi,\eta,\zeta)=\xi\eta-\zeta$$

• The function $f(u)=u^2$ has the algebraic addition theorem given by $$G(\xi,\eta,\zeta)=(\xi+\eta-\zeta)^2-4\xi\eta$$

• The function $f(u)=u^3$ has the algebraic addition theorem given by $$G(\xi,\eta,\zeta)=(\xi+\eta-\zeta)^3-27\xi\eta\zeta$$

Hancock writes that for Weierstrass, the central problem of the theory of elliptic functions is to determine all complex-valued meromorphic functions with an algebraic addition theorem.

Proposition 1. If the function $\phi$ has an algebraic addition theorem and $R$ is a rational function, then $R(\phi)$ has an algebraic addition theorem.

Proof. Let $\xi,\eta,\zeta$ represent the quantities $\phi(u),\phi(v),\phi(u+v)$ for arbitrary arguments $u,v\in\mathbb k$. If $\phi$ has an algebraic addition theorem as postulated, we have that $$\mathbb k(\xi,\eta,\zeta)\supset\mathbb k(\xi,\eta)$$ is an algebraic extension if the above are viewed as fields of rational functions over the field $\mathbb k$. If $R\in \mathbb k(x)$ is a rational function, then we have that $\mathbb k(R(\xi),R(\eta),R(\zeta))$ is a subfield of $\mathbb k(\xi,\eta,\zeta)$. Further, $\mathbb k(R(\xi),R(\eta))$ is a subfield of $\mathbb k(\xi,\eta)$, and the latter is certainly an algebraic extension of the former. Hence, we have the following situation: $$\mathbb k(R(\xi),R(\eta))\subset \mathbb k(R(\xi),R(\eta),R(\zeta))\subset \mathbb k(\xi,\eta,\zeta)\supset \mathbb k(\xi,\eta)\supset \mathbb k(R(\xi),R(\eta))$$ with all above extensions being algebraic. This means that $\mathbb k(\xi,\eta,\zeta)$ is algebraic over $\mathbb k(R(\xi),R(\eta))$, and since $\mathbb k(R(\xi),R(\eta),R(\zeta))$ is an intermediate extension between these two fields, it must also be an algebraic extension over $\mathbb k(R(\xi),R(\eta))$. An algebraic dependence between $R(\xi),R(\eta),R(\zeta)$ amounts to an algebraic addition theorem, as desired. $\blacksquare$

This gives rise to the following corollary:

Proposition 2. If $\mathbb k =\mathbb C$ and $\phi$ is a polynomial function of $u$, or a polynomial function of $e^{\alpha u}$ for any $\alpha\in\mathbb C$, then $\phi$ has an algebraic addition theorem.

complex-analysis

elliptic-functions

abstract-algebra

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