Reference: Jacob Lurie’s talk
Rigid analytic geometry is a non-archimedean analogue of complex geometry.
What can you say about the set of rational solutions to any curve, like
\[y^2 = x^3+107x+1055\]
We know that \(E(\mathbb{Q})=\{(x,y): E\}\cup \{\infty\}\) form an abelian group. Mordell’s theorem tells you that \(E(\mathbb{Q})\) is always finitely generated and thus of the form
\[E(\mathbb{Q})\cong \mathbb{Z}^r\times E(\mathbb{Q})_{tors}\]
What can we say about the finite group \(E(\mathbb{Q})_{tors}\)?
If you want to get information for questions like this, it can be very useful to think about solutions not just over the rational numbers, say \(\mathbb{C}\) or \(\mathbb{Q}_p\).
Now \(E(\mathbb{C})\) is a Riemann surface of genus \(1\), looks like a torus. As a group it is
\[E(\mathbb{C})\cong \mathbb{C}/\Lambda \cong (\mathbb{R}/\mathbb{Z})^2\]
where \(\Lambda\) is a lattice in \(\mathbb{C}\). And the second isomorphism forgets about the complex structure, only remembering the Lie group structure. From this we can say that
\[E(\mathbb{C})_{tors}\cong (\mathbb{Q}/\mathbb{Z})^2\]
This immediately gives you some information: The finite group \(E(\mathbb{Q})\) is generated by at most \(2\) elements.
The \(N\)-torsion points of \(E\) are defined by polynomials, thus they are algebraic points living in certain subfields. Therefore
\[E(\overline{\mathbb{Q}})_{tors} = E(\mathbb{C})_{tors} = (\mathbb{Q}/\mathbb{Z})^2\]
the left hand side has an action of the Galois group \(G=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\), which is interestingly acting on \((\mathbb{Q}/\mathbb{Z})^2\). The fixed points of the action is
\[E(\mathbb{Q})_{tors} = E(\overline{\mathbb{Q}})_{tors}^G.\]
The problem is, most automorphisms of \(\overline{\mathbb{Q}}\) does not extend continuously to \(\mathbb{C}\).