UPDATE: My question is equivalent to finding a nice way of writing down (all or some) maps $\{ \alpha \in \mathbb C: \alpha \Lambda_1 \subseteq \Lambda_2\}$ for two complex lattices $\Lambda_1, \Lambda_2 \in \mathbb C$.
Some theory and definitions.
Let $\omega_1, \omega_2\in\mathbb C$ be a set of generators of the complex lattice $\Lambda=\{n_1\omega_1 + n_2\omega_2: n_1,n_2 \in \mathbb Z\} \subset \mathbb C$.
Two lattices $\Lambda_1, \Lambda_2$ are homothetic if $\Lambda_1 = \alpha \Lambda_2$ for some $\alpha \in \mathbb C$
Now, elliptic curves over $\mathbb C$ up to isomorphism correspond to lattices $\Lambda$ up to homothety and isogenies between elliptic curves correspond to maps $\{\alpha\in\mathbb C: \alpha \Lambda_1 \subset \Lambda_2\}$.
Let $E_1(\mathbb C) \cong \mathbb C / \Lambda_1$ and $E_2(\mathbb C) \cong \mathbb C / \Lambda_2$.
I gather that $E_1$ is isomorphic to $E_2$ if and only if $\Lambda_1$ and $\Lambda_2$ are homothetic.
Also, there exists an isogeny between $E_1$ and $E_2$ if and only if the exists an $\alpha \in \mathbb C$ such that $\alpha \Lambda_1 \subseteq \Lambda_2$.
Therefore $\mathrm{Hom}(E_1, E_2) \cong \{\alpha \in \mathbb C: \alpha \Lambda_1 \subseteq \Lambda_2\}$ and $\mathrm{End}(E_1)\cong\{\alpha \in \mathbb C: \alpha\Lambda_1 \subseteq \Lambda_1\}$.
My question regards applying this theory to (basic) examples.
For example, define
$\Lambda_1 = \langle 1, i\rangle$,
$\Lambda_2 = \langle 1, 2i\rangle$,
$\Lambda_3 = \langle 1, \sqrt{2}i\rangle$,
$\Lambda_4 = \langle 1, 1+\sqrt{2}i\rangle$ and correspondingly the elliptic curves $E_i(\mathbb C) \cong \mathbb C/\Lambda_i$ for $i=1,2,3,4$.
What is $\mathrm{End}(E_i)$ for $i \in \{1,2,3,4\}$?
Or, what is (the structure of) $\mathrm{Hom}(E_i, E_j)$ for $j>i$ and $i,j \in \{1,2,3,4\}$?
Which elliptic curves $E_i$ are isogenous/isomorphic/nothing to $E_j$ for $i,j \in \{1,2,3,4\}$ with $j>i$?
These are a lot of specific questions, I do not expect that all of them are answered, any thought/hint into solving any of them would be very much appreciated.
My attempt for some of them:
$\Lambda_3$ and $\Lambda_4$ are homothetic as
$\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}(\sqrt{2}i) = \sqrt{2}i+1$. Thus $E_3$ and $E_4$ are isomorphic.
Also, for any $k\in\mathbb Z[i]$ we have $k\Lambda_2\subseteq\Lambda_2\subset\Lambda_1$ thus $E_1$ and $E_2$ are isogenous, $\mathrm{Hom}(E_1,E_2) \cong \mathbb Z[i]$, and $\mathrm{End}(E_2) \cong \mathbb Z[i]$. But how to show that $E_1$ and $E_2$ are not isomorphic?
In addition $\sqrt{2}i\Lambda_1 = \langle \sqrt{2}i,-\sqrt{2} \rangle \not\subset \Lambda_3$ as $-\sqrt{2} \notin \Lambda_3$.
But how can I then show that $E_1$ and $E_3$ are neither isomorphic nor isogenous and $\mathrm{Hom}(E_1,E_3) \cong \varnothing$?
Best Answer
To show that $E_1$ and $E_2$ are not isomorphic, you can use the result at the bottom of page 10 of this book by Milne, and argue that if $\tau'$ is purely irrational and $\tau$ is rational, there is no possible combination of $a,b,c,d\in \mathbb{Z}$ such that $\tau'=\frac{a+b\tau}{c+d\tau}$. If you don't like this, you can argue that you have a pair $\{1,2i\}$ and another pair $\{1,\sqrt{2}i\}$, and one contains exactly one purely irrational number and the other none. So you cannot choose a number that when multiplied by one of the pairs gives you the other.
Also, note that $\Lambda_3$ and $\Lambda_4$ are the same lattice (draw some points and you'll se why).
To see that $\Lambda_1$ and $\Lambda_3$ are not isogenous (and thus also not isomorphic), define $\alpha=x+iy\in\mathbb{C}^\times$ and suppose $\alpha\Lambda_1\subset \Lambda_3$. An element of $\Lambda_1$ can be written as $a+ib$ with $a,b\in \mathbb{Z}$. Multiplying $\alpha$ with an element of this form gives you the following: $$(x+iy)(a+ib)=(xa-yb)+i(xb+ya)$$ We want this to be in $\Lambda_3=\mathbb{Z}+\sqrt{2}i\mathbb{Z}$. For that, first we need $xa-yb\in \mathbb{Z}$. Setting $a=1,b=0$ gives $x\in \mathbb{Z}$. Setting $a=0,b=1$ gives $y\in \mathbb{Z}$. I'll leave it to you to check that under these conditions, it is impossible that $xb+ya\in \sqrt{2}i\mathbb{Z}$. Thus, such $\alpha$ cannot exist and they are not isogenous.
I hope this answers your questions :)