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In Chapter III, Theorem 7.4 of The Arithmetic of Elliptic Curves (first edition), Silverman gives the following lemma and proof:

Lemma: Let $M \subset Hom(E_1, E_2)$ be a finitely generated subgroup, and let $M^{div} = \{ \phi \in Hom(E_1, E_2) : [m] \circ \phi \in M$ for some integer $m \geq 1\}$. Then $M^{div}$ is also finitely generated.

Proof: Extend the degree mapping to the finite dimensional real vector space $M \otimes \mathbb{R}$, which we equip with the natural topology inherited from $\mathbb{R}$. Then the degree mapping is clearly continuous, so the set $U = \{\phi \in M \otimes \mathbb{R} : deg(\phi) < 1 \}$ is an open neighborhood of $0$. Further, since $Hom(E_1, E_2)$ is a torsion-free $\mathbb{Z}$-module, there is a natural inclusion $$M^{div} \subset M \otimes \mathbb{R};$$ and clearly $$M^{div} \cap U = \{ 0 \},$$ since every non-zero isogeny has degree at least 1. Hence $M^{div}$ is a discrete subgroup of the finite dimensional vector space $M \otimes \mathbb{R}$, so it is finitely generated.

When I first saw it, this proof felt like absolute voodoo to me. Tensor an endomorphism ring with the reals? What, so you can take $\pi$ or $e$ times an endomorphism?

The point is clearer to me now -- real vector spaces are a nice place to argue that things are finitely generated -- but I'm not sure I would have seen to do this if the proof were left as an exercise.

Does this idea have any natural context? Are proofs like this commonplace in some other area of math?

Thank you!

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I wondered the same thing when I first saw that proof in Silverman! –  David Corwin Aug 6 '12 at 1:19
    
Though my guess as to an answer is that this is, in some sense, the (a?) fundamental idea behind applications of analysis to number theory. –  David Corwin Aug 6 '12 at 1:24
    
I'm confused by the last sentence in Qiaochu's first comment. $M^{div}$ is not a submodule but a supermodule of $M$. So how does one apply the result about submodules? –  Andreas Blass Aug 6 '12 at 1:55
    
@Andreas: my apologies; I misread the problem. –  Qiaochu Yuan Aug 6 '12 at 2:00
    
I think the reason the idea is surprising, as with all of these statements, is that it involves proving a purely arithmetic (or "algebraic," if that's how you call it) by passing to the transcendental realm. It almost seems that in some "moral" sense, the simplest and most natural proof should stay in the same realm. (One could say that the fact that this might not be the case is a testament to how strange mathematics is.) –  David Corwin Aug 6 '12 at 17:54
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How do you prove that the ring of integers in a number field is finitely generated? One usually embeds them in $\mathbb{R}^{r_1}\times\mathbb{C}^{r_2}$. How do you prove that the units in a number field are finitely generated? One generally embeds them in a hyperplane in $\mathbb{R}^{r_1+r_2-1}$. Then one shows that the group sits as a discrete subgroup, hence is finitely generated. Further, by looking at the co-volume of the resulting lattice, one obtains important arithmetic invariants, namely the discriminant and the regulator. Anyway, one can call almost any technique a trick, but the idea of embedding a group into a real or complex vector space and using volume estimates to prove discreteness is quite a well-established technique. And if one simply has the group and a positive definnite quadratic form, as in the case of End$(E)$ or the canonical height on $E(\mathbb{Q})$, then it is very natural to tensor with $\mathbb{R}$ and extend the quadratic form to put a Euclidean structure on the resulting vector space.

Of course, as Will Sawin says, one can alternatively view the embedding as being into the dual. But for example, using the height pairing on $E(\mathbb{Q})/tors$, the values of $\hat h$ are real, so even though $E(\mathbb{Q})/tors$ is a torsion-free $\mathbb{Z}$-module, the dual space that one uses has to be Hom$(E(\mathbb{Q})/tors,\mathbb{R})$.

I don't know that this really answers the question, but I hope it gives some indication of why it's not so unusual to think of embedding $M$ into a real vector space as the first step in proving that it is finitely generated.

Finally, I should mention that I first saw the argument for proving that End$(E)$ is finitely generated in Mumford's Abelian Varieties, but I don't know the origins of the idea.

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Indeed, those are excellent examples. Thank you! –  Frank Thorne Aug 6 '12 at 20:37
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Here is an alternate proof that should be less problematic. The degree is a positive-definite quadratic form with image in $\mathbb Z$ (that's how he's extending it to $\mathbb R$, right?), so we can construct a bilinear form with image in $\mathbb Z$: $x\cdot y= deg(x,y)-deg(x)-deg(y)$. This gives a map $M^{div} \to M^{\vee}$. $M^{\vee}$ is a finitely-generated group. The map has no kernel since if an element in $M^{div}$ were in the kernel, its multiple in $M$ would be as well, but it can't be since the degree is positive-definite.

$M \otimes R$ is just being used as a bigger version of $M^{\vee}$ to embed $M^{div}$ in, I think.

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I think it's nothing more than a "trick" like in all other solutions you find throughout algebra and even Putnam-type problems.
A similar trick establishes $\mathbb{R}\cong\mathbb{C}$ as abelian groups. Consider $\mathbb{R}$, $\mathbb{R}^2$ naturally as $\mathbb{Q}$-vector spaces, and then they have same dimension and are isomorphic.
The closest thing to a "natural" context I can argue is that vector spaces behave well, they're modules over a field and have a nice structure (basis and dimension). So turning your problem into a vector space one and implementing its simplicity seems like a good idea -- whether you can actually do this is another question.

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I disagree. I think there are two points: first, torsion-free is equivalent to embedding into the tensor product with $\mathbb{Q}$. Second, tensoring all the way up to $\mathbb{R}$ gives you a nicer topology which interacts well with the degree map. I don't regard either of these as clever tricks. –  Qiaochu Yuan Aug 6 '12 at 2:24
    
Also, it's clear why for checking isomorphism as abelian groups, you want to forget the topological structure and view them just as groups, and why remembering that they are $\mathbb Q$-vector space lets you take advantage of the simplicity of vector space theory to prove isomorphism. It is very obvious why that trick works. It is less obvious in this case. –  Will Sawin Aug 6 '12 at 2:29
    
I don't claim all tricks to be of the same standard or transparency. I just claim that they are tricks, and stem from the simplicity of vector spaces. In both examples (the one I gave and the OP's), vector space theory was implemented... in both examples, you had to get to that vector space (and you guys demonstrated how in your comments). But yes, I can see ourselves still asking for a context on how to turn our problem into a vector space one. –  Chris Gerig Aug 6 '12 at 3:05
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