This is essentially a reference request (apologies if it is a duplicate): it is known that every lattice $\Lambda$ in $\mathbb{C}$ is isomorphic to one of the form $\mathbb{Z} \oplus \mathbb{Z}[\tau]$ for some $\tau \in \mathbb{H}$, the upper half-plane. Let $\Lambda_1 = \mathbb{Z} \oplus \mathbb{Z}[\tau_1]$, $\Lambda_2 = \mathbb{Z} \oplus \mathbb{Z}[\tau_2]$
It can be shown that $\mathbb{C}/\Lambda_1 \cong \mathbb{C}/\Lambda_2$ if and only if there exists $\gamma \in \text{SL}(2, \mathbb{Z})$ such that $\tau_1 = A \cdot \tau_2$. Can someone point me to a resource which contains a full proof of this result? Thank you.
I've worked out the details: proof below.
We use the fact that $\mathbb{C}/\Lambda_1 \cong \mathbb{C}/\Lambda_2$ if and only if $\Lambda_1, \Lambda_2$ are homothetic, i.e. there exists $\alpha \in \mathbb{C}^{*}$ such that $\Lambda_1 = \alpha \Lambda_2$. Due to this, an arbitrary lattice generated by $\langle \omega_1, \omega_2 \rangle$ is isomorphic to $\langle 1, \omega_1/\omega_2 \rangle$. Then without loss of generality we can take $\Lambda_1 = \langle 1, \tau_1 \rangle$ for $\tau_1 \in \mathbb{H}$ and similarly for $\Lambda_2$.
The group $\text{SL}(2, \mathbb{Z})$ acts on $\mathbb{H}$ via fractional linear transformations: $$\gamma = \begin{pmatrix} a & b\\ c & d\\ \end{pmatrix} \in \text{SL}(2, \mathbb{Z})$$ acts via $$\gamma \cdot \tau = \dfrac{a \tau + b}{c \tau + d}$$ and we can compute that $$\text{Im}(\gamma \cdot \tau) = \dfrac{\text{det}(\gamma)}{|c \tau + d|^2} \text{Im}(\tau)$$
Now we show that $\Lambda_1 \cong \Lambda_2$ if and only if there exists $\gamma \in \text{SL}(2, \mathbb{Z})$ such that $\tau_2 = \gamma \cdot \tau_1$.
Suppose $\tau_2 = \gamma \cdot \tau_1$ for some $\gamma$. Then $\Lambda_2 = \langle a\tau_1 + b, c\tau_1 + d \rangle$, but since $ad - bc = 1$ this is the same lattice as $\langle 1, \tau_1 \rangle = \Lambda_1$.
Conversely, suppose $\Lambda_1 = \alpha \Lambda_2$ for some $\alpha \in \mathbb{C}$. On generators this means $\langle \alpha, \alpha \tau_2 \rangle = \langle 1, \tau_1 \rangle.$ In this case $$\alpha = c \tau_1 + d$$ $$\alpha \tau_2 = a \tau_1 + b$$ for some $a, b, c, d \in \mathbb{Z}$.
Similarly, there exist $e, f, g, h \in \mathbb{Z}$ such that $$\tau_1 = e(a \tau_1 + b) + f(c \tau_1 + d)$$ $$1 = g(a \tau_1 + b) + h(c \tau_1 + d)$$ Since $1, \tau_1$ are $\mathbb{Z}$-linearly independent, we can rephrase this as $$\begin{pmatrix} e & f\\ g & h\\ \end{pmatrix} \begin{pmatrix} a & b\\ c & d\\ \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1\\ \end{pmatrix}$$
Taking determinants yields $\text{det}(\gamma) \in \{\pm 1\}$, but $$\text{Im}(\tau_2) = \text{Im}\dfrac{a \tau_1 + b}{c \tau_1 + d} = \dfrac{\text{det}(\gamma)}{|c \tau + d|^2} \text{Im}(\tau_1)$$ and since $\tau_1, \tau_2 \in \mathbb{H}$, we have $\text{Im}(\tau_1), \text{Im}(\tau_2) > 0$, and so $\text{det}(\gamma) = 1$. Therefore, such a $\gamma$ exists in $\text{SL}(2, \mathbb{Z})$, and we are done.
As a consequence, we can identify the space of isomorphism classes of complex tori with the quotient $\mathbb{H} \backslash \text{SL}(2, \mathbb{Z})$, or in other words, the modular curve $Y(1)$, whose points can be identified with the fundamental domain for the $\text{SL}(2, \mathbb{Z})$-action on the upper half-plane: $$\mathcal{F} = \{\tau \in \mathbb{H} \mid |\text{Re}(\tau)| \leq 1/2, |\tau| \geq 1\}$$