# Why the kernel of isogeny is finite?

It is said that the kernel of a isogeny is finite because it is discrete and complex tori are compact.

1.

Following is my reason for the kernel is discrete.

Suppose $$\varphi:\mathbb{C}/\Lambda\rightarrow\mathbb{C}/\Lambda'$$

is an isogeny. Then there exists $m\in \mathbb{C}$ such that $m\Lambda=\Lambda'$. So the kernel of $\varphi$ is $\left(\frac{1}{m}\Lambda'\right)/\Lambda$. Intuitively, it is discrete, I think. But I don't know how to reason it.

There is a hint in the book I'm reading saying that if the kernel is not discrete, complex analysis shows that the map is zero.

Can anyone tell me why?

2.

Why can we deduce finiteness from discreteness and compactness?

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Regarding (2), a discrete compact set is finite. This is easy point-set topology. – Zhen Lin Jan 6 '13 at 9:41
@ZhenLin, Oh I see, I misunderstand the term "discrete" here.. – hxhxhx88 Jan 6 '13 at 14:03

You may know the result from complex analysis that for a nonconstant holomorphic function defined on an open set $\Omega \to \mathbb{C}$, the zero set must be discrete. More generally, the preimage of any point of such a function is discrete.
@hxhxhx88 I don't see what's the difference. If you're concerned about whether we could have a set of points like $\{1/n : n \in \mathbb{Z\}}$, remember the set of zeroes must be a closed subset of the domain. So if 0 is in the domain, then the set of zeroes could not be exactly this set. – Ted Jan 6 '13 at 19:24