# isogenies between tori

Let Hom$(\mathbb{C}/\Lambda_1,\mathbb{C}/\Lambda_2)$ be the set of isogenies between $\mathbb{C}/\Lambda_1$ and $\mathbb{C}/\Lambda_2$, where $\Lambda_1,\Lambda_2$ are lattices.

I am asked to prove what the structure of this group is if:

$\Lambda_1=\mathbb{Z}+\mathbb{Z}i$ and $\Lambda_2=\mathbb{Z}+\mathbb{Z}2i$.

Now I know that if $\psi$ is an isogeny, there exists $\alpha\in\mathbb{C}$ such that $\psi(z\mod\Lambda_1)=\alpha z\mod\Lambda_2$ and $\alpha\Lambda_1\subset\Lambda_2$, and conversely every such $\alpha$ has a corresponding isogeny.

I tried finding the solution as follows: Let $\alpha=a+bi$, let $z=c+di$, where $a,b,c,d\in\mathbb{R}$. Then $\alpha z=ac-bd+(bc+ad)i$. Thus $ac-bd$ has to be an integer multiple of $1$, that is, an integer and $bc+ad$ has to be an integer multiple of 2. If this is correct, how do I continue and if it is not correct, how should it be done?

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Note that in your notation, $z$ should be an arbitrary element of the lattice $\Lambda_{1}$ (i.e., $c$ and $d$ are integers), not an arbitrary complex number.

Said another way (eliminating mention of $z$), you should have no trouble showing that $\psi(\Lambda_1)\subseteq\Lambda_2$ if and only if $\alpha\in\Lambda_2$ and $i\alpha\in\Lambda_2$.

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I agree, $c,d\in\mathbb{R}$, so then from $bc+ad$ is even, it follows that $b$ and $d$ are even and from $ac-bd$ integer, it follows that $b,d$ are integers. But wouldn't then $\psi(\Lambda_1)\subseteq\Lambda_2$ iff $\alpha=2\mathbb{Z}+2\mathbb{Z}i$. This is equivalent to saying $\alpha\in\Lambda_2$ and $i\alpha\in\Lambda_2$. But is there an easier way to see your answer immediately? –  user100659 Oct 18 '13 at 13:28
And in case $\Lambda_1=\mathbb{Z}+\mathbb{Z}i$ and $\Lambda_2=\mathbb{Z}+\mathbb{Z}\sqrt{-2}$. Would we get $ac-bd$ integer and $bc+ad$ a multiple of $\sqrt{2}$, which seems impossible so the group Hom will be empty? –  user100659 Oct 18 '13 at 13:37
To expand my second observation, since $1$ and $i$ generate $\Lambda_1$ (and $\Lambda_2$ is a lattice), $\psi(\Lambda_1)\subseteq\Lambda_2$ if and only if $\alpha=\psi(1)\in\Lambda_2$ and $i\alpha=\psi(i)\in\Lambda_2)$. (And "yes" to your question about $\Lambda_2=\mathbf{Z}+\mathbf{Z}\sqrt{-2}$. :) –  Andrew D. Hwang Oct 18 '13 at 13:40
Awesome, thanks! –  user100659 Oct 18 '13 at 13:45