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Suppose $A$ is a complex abelian variety. Then $A$ is a complex torus $\mathbb C^g/\Lambda$ where $\Lambda$ is a lattice. On the other hand abelian varieties over $\mathbb C_p$ can have good reduction. But $\mathbb C$ and $\mathbb C_p$ are isomorphic... so, shouldn't abelian varieties over $\mathbb C_p$ be of the form $\mathbb C_p^g/\Lambda$ where $\Lambda$ is now a lattice in $\mathbb C_p$? But then abelian varieties over $\mathbb C_p$ with good reduction wouldn't exist...

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$\mathbb C$ and $\mathbb C_p$ are isomorphic... as what? –  Brenin Feb 8 at 14:44
    
@Brenin They're isomorphic as fields. They're both algebraically closed fields of characteristic $0$ and cardinality $2^{\aleph_0}$. –  Alex Youcis Feb 8 at 21:35
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In $\mathbb C_p$ there is no lattice of positive rank ! The $p$-adic uniformization uses $({\mathbb C}_p^*)^g$ instead. –  Cantlog Feb 9 at 16:14

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Probably the reason is that a lattice in $\mathbb C$ (i.e. a discrete subgroup of $\mathbb C$) is not a discrete subgroup of $\mathbb C_p$ (due to the fact that the norms of the two fields are different!)

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