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If $A$ is an abelian variety over a number field $K$, then the set of $K$-rational points $A(K)$ is a finitely generated group by the Mordell-Weil Theorem. By the classification of finitely generated abelian groups, we know that there exists a free part of this group, whose rank we can call the rank of the abelian variety.

Consider now an abelian variety over lets say $\mathbb{Q}_p$. Then the groups of $\mathbb{Q}_p$-points is not finitely generated anymore, see for instance (Is there an analogue of Mordell-Weil theorem for other fields?). So there is no (naive) generalization of the notion of rank to such abelian varities.

However, maybe there is a non-naive generalization of the notion of rank (lets call it ``generalized rank"), which is backwards compatible, i.e. such that if $A$ is an abelian variety over a number field $K$ and $\nu$ is place of the number field, we have $$\text{Mordell-Weil rank}(A)=\text{generalized rank}(A\times_K K_\nu)?$$ Possibly one would want this $\nu$ be such that $A$ has good reduction at $\nu$. One idea I had was to hope that the associated formal group law had a sensible analogue of a rank, however I don't see why and didn't find any reference to this in the literature.

curious math guy
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  • Did you look at the $p$-adic analog of complex torus $ \Bbb{C}/(\Bbb{Z+\tau Z})\cong \Bbb{C}^/\langle e^{2i\pi \tau }\rangle$ which is $ \Bbb{C}_p^/\langle q\rangle$ with $q\in \Bbb{Q}_p^*, v(q)\ne 0$. It is an elliptic curve and its $\Bbb{Q}_p$-points don't seem to have a rank. Otherwise BSD conjecture says the rank is the vanishing order of $L(E/\Bbb{Q},s)$ at $s=1$, but its Euler product doesn't converge there so you can't relate that to $L(E/\Bbb{Q}_p,s)$. A last attempt would be to look at the $p$-adic modular form corresponding to $f, L(f,s)=L(E,s)$. – reuns Mar 12 '20 at 21:52
  • Surely not: there should be plenty of elliptic curves $E, E'$ with different ranks, such that $E, E'$ are isomorphic over $K_v$. – Mathmo123 Mar 12 '20 at 22:46
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    For example, take $E:y^2+y = x^3-x^2$ over $\mathbb Q$, which has rank $0$, and its quadratic twist $E':y^2 +y=x^3 +x^2 −16x−66$ which has rank $1$. These curves are non-isomorphic over $\mathbb Q$ but isomorphic over $\mathbb Q(\sqrt{-7})$. So they will be isomorphic over $\mathbb Q_p$ for any prime $p$ that splits in $\mathbb Q(\sqrt{-7})$. – Mathmo123 Mar 12 '20 at 23:01

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