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changed title (finite-->finitely generated)
Sam Lichtenstein
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Is a torsion free abelian group finitely generated, if all of its localizations at primes p are finitely generated over Zp?

Background: When proving that the group of k http://latex.mathoverflow.net/png?k-isogenies M=\mathrm{Hom}\sb k(A,B) http://latex.mathoverflow.net/png?M%3D%5Cmathrm%7BHom%7D%5Fk%28A%2CB%29 between two abelian varieties is finitely generated, one first shows that the Tate map \mathbb{Z}\sb \ell\otimes\sb {\mathbb{Z}} M \to \mathrm{Hom}\sb {\mathbb{Z}\sb \ell}(T\sb \ell A,T\sb \ell B) http://latex.mathoverflow.net/png?%5Cmathbb%7BZ%7D%5F%5Cell%5Cotimes%5F%7B%5Cmathbb%7BZ%7D%7D%20M%20%5Cto%20%5Cmathrm%7BHom%7D%5F%7B%5Cmathbb%7BZ%7D%5F%5Cell%7D%28T%5F%5Cell%20A%2CT%5F%5Cell%20B%29 is injective. Since each Tate module is free of finite rank over \mathbb{Z}\sb \ell http://latex.mathoverflow.net/png?%5Cmathbb%7BZ%7D%5F%5Cell, it follows that the localization M\sb \ell http://latex.mathoverflow.net/png?M%5F%5Cell is \mathbb{Z}\sb \ell http://latex.mathoverflow.net/png?%5Cmathbb%7BZ%7D%5F%5Cell-finite. One then uses a little trick to deduce the \mathbb{Z} http://latex.mathoverflow.net/png?%5Cmathbb%7BZ%7D-finiteness of M http://latex.mathoverflow.net/png?M itself. (See Silverman I, for example.)

The above proof needs only a single prime \ell http://latex.mathoverflow.net/png?%5Cell. But disregarding issues of the characteristic of the field (which are apparently surmountable) we actually have an injective Tate map at every prime. Thus...

Question: Can the \mathbb{Z} http://latex.mathoverflow.net/png?%5Cmathbb%7BZ%7D-finiteness of M http://latex.mathoverflow.net/png?M be deduced directly from the \mathbb{Z}\sb \ell http://latex.mathoverflow.net/png?%5Cmathbb%7BZ%7D%5F%5Cell-finiteness of M\sb \ell http://latex.mathoverflow.net/png?M%5F%5Cell for all primes \ell http://latex.mathoverflow.net/png?%5Cell?

One can consider this a question about general torsion-free abelian groups M http://latex.mathoverflow.net/png?M. A non-counterexample to keep in mind is M=\mathbb{Z}[1/p] http://latex.mathoverflow.net/png?M%3D%5Cmathbb%7BZ%7D%5B1%2Fp%5D, for which M\sb \ell http://latex.mathoverflow.net/png?M%5F%5Cell is \mathbb{Z}\sb \ell http://latex.mathoverflow.net/png?%5Cmathbb%7BZ%7D%5F%5Cell-finite for all \ell\neq p http://latex.mathoverflow.net/png?%5Cell%5Cneq%20p.

(A google search shows that there is actually quite a body of literature on torsion-free abelian groups, so perhaps the answer to this question is well-known, but I'm not sure where to look...)

Sam Lichtenstein
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