Doubly periodic meromorphic function with prescribed poles and zeros The field of the meromorphic functions on a complex torus
$\mathbb{C} \mathbin{/} \Lambda$ is $\mathbb{C}(\wp, \wp')$,
where $\wp$ is the weierstrass p-function to the lattice $\Lambda$.
Furthermore, for such a function $f$ and its finite set $U$ of poles and zeros holds: $\sum_{ u \in U } \operatorname{ord}_u(f) = 0$ and $\sum_{ u \in U } u \cdot \operatorname{ord}_u(f) \in \Lambda$, where $\operatorname{ord}_u(f)$ is the order of the pole (if negative) resp. the zero (if positive) of $f$ at $u$.
If now some points $U$ and their orders are given, and fulfill constraints above, I believe (because of the Riemann–Roch theorem) that a corresponding meromorphic function exists and is unique (up to a multiplicative constant), but I cannot figure out how to construct it from $\wp$ and $\wp'$.
Are my claims correct? And if yes, how to construct the meromorphic function in question (with closed-form formula, or recursively)?
 A: We can just ignore the extra constraints, since it is enough to prove every meromorphic function on $\mathbb{C}/\Lambda$ is rational in $\wp$ and $\wp'$. Given a meromorphic function $f$ on $\mathbb{C}/\Lambda$, there is indeed a standard proof that $f \in \mathbb{C}(\wp,\wp')$. It goes something like this. Write $f$ as a sum of an even function and an odd function:
$$
f(z) = \frac{f(z) + f(-z)}{2} + \frac{f(z) - f(-z)}{2}
$$
Using this trick we may assume that $f$ is odd, or that $f$ is even.
In fact we can assume $f$ is an even function, since if $f$ is an odd elliptic function then $\wp' \cdot f$ is an even elliptic function. Thus it is enough to show that if $f$ is an even elliptic function then $f \in \mathbb{C}(\wp)$.
For even elliptic functions $f$, the identity
$$
\operatorname{ord}_w f = \operatorname{ord}_{-w} f
$$
holds for all $w \in \mathbb{C}$. Furthermore, if $2 w \in \Lambda$, then
$\operatorname{ord}_w f$ is even, because the $i$-th derivative satisfies
$$
f^{(i)}(-w) = f^{(i)}(w) = (-1)^i f^{(i)}(-w)
$$ for all odd values of $i$ (the first equality follows because $2 w
\in \Lambda$, and the last equality is achieved by repeatedly
applying the chain rule). Therefore
$$
\operatorname{div}(f) = \sum_{w \in H} n_w ((w) + (-w))
$$
for some set of integers $n_w$, where $H$ is half of a fundamental
parallelogram for $\Lambda$, and the sum has only finitely many
nonzero terms.
Consider the function
$$
g(z) = \prod_{w \in H\setminus \{0\}} (\wp(z) - \wp(w))^{n_w}.
$$
We have $\operatorname{div}(\wp(z)-\wp(w)) = (w) + (-w) - 2(0)$, so $\operatorname{div}(g)$ and
$\operatorname{div}(f)$ are identical except possibly at $(0)$. Since every principal divisor has degree zero, $\operatorname{div}(g)$ and $\operatorname{div}(f)$ must in fact also be identical at $(0)$. Hence $f/g$ is an elliptic function with no poles, so it is constant. But then $f \in \mathbb{C}(\wp)$ since $g \in \mathbb{C}(\wp)$.
