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

I have some questions about this.


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?


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
up vote 2 down vote accepted

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.

In the scenario of (1), we are dealing with a similar situation, only the mapping is between 2 complex manifolds. But the same result holds: For a holomorphic map between complex manifolds, the preimage of a point is discrete, unless the map is identically constant. That is the result your book is alluding to.

For (2), if a set is discrete, then each point is open, so we have an open cover in which each set in the cover consists of only one point. Now if the set is compact also, then ...

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I'm confused with the term 'discrete' here. Does it just mean a set of isolated points or involve discrete topology? So 'the preimage is discrete' means the topology of preimage equipped with relative topology is discrete topology? – hxhxhx88 Jan 6 '13 at 14:09
A subset is a set of isolated points if and only if it has the discrete topology as a subspace. – Zhen Lin Jan 6 '13 at 14:21
@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
@Ted, OK, I see. – hxhxhx88 Jan 7 '13 at 1:23
@ZhenLin, yeah, I realized.. – hxhxhx88 Jan 7 '13 at 1:24

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