TITLE: Arithmetic geometry examples
QUESTION [33 upvotes]: (This is inspired by Algebraic geometry examples.)
I want to collect here (counter)examples in arithmetic geometry.

Curves violating the Hasse principle: The Selmer curve $3X^3 + 4Y^3 + 5Z^3 = 0$. It is a nontrivial element of the Tate–Shafarevich group of the elliptic curve $3\cdot4\cdot5\cdot X^3 + Y^3 + Z^3 = 0$. It is also an example of an abelian variety for which finiteness of Sha is known. In fact, $|\mathrm{III}(E/\mathbf{Q})| = 3^2$.
Non-isogeneous elliptic curves having the same Hasse–Weil $L$-series: $y^2 = x^3 \pm ix + 3$ over $K = \mathbf{Q}(i)$ (cf. Cornell–Silverman–Stevens, p. 32).
Counterexample to the Hasse norm theorem for non-cyclic extensions: $L = \mathbf{Q}(\sqrt{13},\sqrt{17})$ Galois with $G =\mathbf{Z}/2 \times \mathbf{Z}/2$, see Cassels–Fröhlich, p. 360, Exercise 5.3.

tbc

REPLY [11 votes]: It is a widely-used fact that a complex torus with "complex multiplication" is algebraic (i.e., an abelian variety) in the sense of the GAGA theorem; the proof goes via Riemann forms.  But the analogue over $p$-adic fields is false: there exists 
a non-algebraizable formal CM abelian scheme over a $p$-adic discrete valuation ring, so (using Neron-Ogg-Shafarevich) its generic fiber in the sense of Raynaud is a rigid-analytic smooth connected proper group with "complex multiplication" yet is not algebraic (in the sense of $p$-adic GAGA). 
More specifically, for $p \equiv \pm 2 \bmod 5$, consider the simple abelian surface over $\kappa = {\mathbf{F}}_{p^2}$ with Weil number $\pm p \zeta_5$ and endomorphism ring $\mathbf{Z}[\zeta_5]$ (this exists by Honda-Tate theory, and for each sign it is unique up to $\mathbf{Z}[\zeta_5]$-linear isogeny). This lifts to a formal abelian scheme $A$ with action by $\mathbf{Z}[\zeta_5]$ over $W(\kappa)$. But for $K := W(\kappa)[1/p]$ the induced $K$-linear action of $\mathbf{Q}(\zeta_5)$ on the 2-dimensional $K$-vector space ${\rm{Lie}}(A)[1/p]$ is given by a pair of embeddings $\mathbf{Q}(\zeta_5) \rightrightarrows K$ related through complex conjugation, so it is not a CM type and hence $A$ is not algebraic. 
For details, see 4.1.2 (up through 4.1.2.3) in this link.