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Let $E$ be an elliptic curve defined over a fixed number field $F$. Note that there exists a unique complex number $\tau$ in the upper half plane so that

$E(\mathbb{C})\simeq\mathbb{C}/L_\tau$

where $L_\tau=\mathbb{Z}+\mathbb{Z}\tau$. From this, we can find an equation for $E$ of the form

$y^2=4x^3-g_2(\tau)x-g_3(\tau)$.

My question is, are $g_2(\tau)$ and $g_3(\tau)$ contained in $F$? If not, does $g_2(\tau)$ and $g_3(\tau)$ have something to do with $F$?

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First, $\tau$ is certainly not unique, it is only well-defined up to linear fractional conjugation by elements of $\text{SL}_2(\mathbb Z)$. Second, to answer your question, no, $g_2(\tau)$ and $g_3(\tau)$ need not be in $F$. What is true is that there exists a complex number $u\in\mathbb C^*$ such that $u^2g_2(\tau)\in F$ and $u^3g_3(\tau)\in F$.

A good way to think about this is that $g_2$ and $g_3$ are really functions on the space of lattices, $g_2=60G_4$ and $g_3=140G_6$, where $$ G_k(L) = \sum_{\substack{\omega\in L\\\omega\ne0\\}} \frac{1}{\omega^k}. $$ The uniformization theorem for elliptic curves implies that if $E$ is an elliptic curve given by a Weierstrass equation $$ y^2 = 4x^3 + ax + b, $$ (with $a,b\in\mathbb C$), then there is a lattice $L$ such that $$ g_2(L)=a\quad\text{and}\quad g_3(L)=b. $$

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  • $\begingroup$ The idea is that $j= j(\tau) \in K$ and $E_j : y^2 = x^3-\frac{27j}{1728-j}x-\frac{27j}{1728-j}$ is $\simeq \mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z})$ with the map $z \mapsto (\wp(z)u^{-2},\wp'(z) u^{-3}), u = (g_3/g_2)^{1/2}$ ? $\endgroup$
    – reuns
    Commented Aug 7, 2017 at 3:13
  • $\begingroup$ @reuns Sure, that will give a model defined over $\mathbb Q(j(E))$, unless $j(E)=0$ or $j(E)=1728$. $\endgroup$
    – Joe Silverman
    Commented Aug 7, 2017 at 13:07

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