Let $f$ and $g$ be tow elliptic function with the same periods, then there exists an algebraic relationship of the form $P(f,g)=0$, where $P$ a polynomial of tow variables and constants coefficients.

it's a well known property, as special case we have $P(f, f^{\prime})=0$, which is verified by the Weirstrass elliptic function $\mathfrak{D}$:

the Weierstrass elliptic function $\mathfrak{D}$ is a solution of the differential equation in $\Lambda$

$$(Y^{\prime})^{2}=4(Y)^{3}-g_{2}Y -g_{3}$$

where $\Lambda$ the lattice generated by the two Periods of $\mathfrak{D}$ and $g_{1},g_{2}$ are the invariants of the function $\mathfrak{D}$.

My question is: there exists an algebraic relationship between tow elliptic functions, if they don't have the same periods, and if there exists, under which conditions (between periods).

i have proved the existence of an algebraic relationship between tow elliptic functions $f$ and $g$ in this case: (which generalize the above property):

if the periods of $f$ are $\omega_{1}$ and $\omega_{2}$ and the periods of $g$ are $p\omega_{1}$ and $q\omega_{2}$, where $p,q\in\mathbb{Q}$.

the problem still open for furthermore generalisation, you are welcome if you have any suggestions.

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    $\begingroup$ Presumably the answer is that the there is an algebraic relationship if and only if the two elliptic curves are commensurate... $\endgroup$ – Igor Rivin Mar 14 '12 at 19:55
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    $\begingroup$ Just to confirm Igor's suspicion, the algebraic relation between the functions implies (in fact is equivalent to) the existence of an isogeny between the elliptic curves, which implies that the lattices are commensurate. In the OP's notation, the periods of $g$ are $a_{11}\omega_1+a{12}\omega_2,a_{21}\omega_1+a{22}\omega_2$ where the $a_{ij}$ are rational and $\det (a_{ij}) \ne 0$. $\endgroup$ – Felipe Voloch Mar 14 '12 at 22:46
  • $\begingroup$ from the remarks of Pr. Felipe Voloch, if there exists a algebraic relationship between tow elliptic curves, then this tow elliptic curves there lattices are commensurate, which give as a relation of equivalence between the existence of an algebraic relationship and "lattices are commensurate". thanks for Pr. Igor Rivin and Igor Rivin for there valuable remarks $\endgroup$ – Abdelmajid Khadari Mar 16 '12 at 1:16
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    $\begingroup$ If $f$ has periods $\omega_1$ and $\omega_2$ then $f$ is a fortiori $\langle p \omega_1, q \omega_2 \rangle$-periodic, and thus algebraically related with $g$. The same argument shows that more generally if the intersection of the period lattices of $f$ and $g$ is again a lattice then $f,g$ are algebraically dependent. The converse is true too (assuming as always that neither $f$ nor $g$ is constant) but not quite this easy. $\endgroup$ – Noam D. Elkies May 9 '12 at 5:14

There exists an algebraic relation between two elliptic functions if and only if their lattices are commensurable. This means that every period of one function is a rational multiple of periods of another.

Proof. Let $f$ and $g$ be two elliptic functions, let $F(f,g)=0$ be an algebraic relation. Choose a point $a$ such that for each solution $w$ of the equation $F(a,w)=0$ we have $F_w(a,w)\neq 0$. Then there are finitely many holomorphic germs $w=\phi_j(z)$, that describe all solutions of $F(z,w)=0$ near $z=a$.

Let $T$ be a period of $f$. Let $z_0$ be a point such that $f(a_0)=a$. Consider the sequence $z_0, z_0+T, z_0+2T...$ Near these points we must have the relations $g=\phi_j(f)$. Because there are only finitely many $\phi_j$ we will have that the germs of $g$ at $z_0+mT$ and at $z_0+nT$ coinside. That is $(m-n)T$ will be a period of $g$.


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