# Arithmetic geometry examples

(This is inspired by Algebraic geometry examples.)

I want to collect here (counter)examples in arithmetic geometry.

1. 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, $|\mathrm{III}(E/\mathbf{Q})| = 3^2$.

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).

3. 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

• It would be good, if someone is giving a "counterexample", that the correct result be mentioned and the missing hypothesis be pointed out. For instance, with the non-isogenous elliptic curves having the same $L$-series, it should be noted that over the rationals elliptic curves with the same $L$-series are isogenous. Another possibility is abelian varieties with a finite Tate-Shafarevich group of nonsquare size; for ell. curves the size must be a square, but in higher dimensions that is false. See mathoverflow.net/questions/9924/…. Mar 18 '12 at 17:39

The Diophantine equation $x^2 - 34y^2 = -1$ has no integer solutions, even though it has solutions in ${\bf Z}_p$ for all $p$ (including $p = \infty$ if we understand "${\bf Z}_\infty$" as $\bf R$). This is the first example of the failure of the Hasse principle for the minus case of the Fermat-Pell equation $x^2 - \Delta y^2 = \pm 1$ (with $\Delta$ a fixed positive integer that is not a square), or equivalently for the existence of units of norm $-1$ in ${\bf Z}[\sqrt{\Delta}]$. It can also be regarded as the first example of a nontrivial element of the "Tate-Šafarevič group" for the torus $x^2 - \Delta y^2 = +1$ (since $x^2 - \Delta y^2 = -1$ is a principal homogeneous space for that torus).

[NB the equation $x^2 - 34y^2 = -1$ does have rational solutions, such as $(x,y) = (5/3,1/3)$. Indeed Minkowski already showed that a quadratic equation in any number of variables has a rational solution iff it has a solution in each ${\bf Q}_p$ and in ${\bf R}$; Hasse generalized this from ${\bf Q}$ to an arbitrary number field.]

[Added later:] In general $x^2 - \Delta y^2 = -1$ has solutions in every ${\bf Z}_p$ iff $\Delta$ is either a product of primes congruent to $1 \bmod 4$ or twice such a product; equivalently, iff $\Delta$ is the sum of two coprime squares. If such $\Delta$ is of the form $n^2 \pm 2$ then $(n + \sqrt\Delta)^2 / 2$ is a unit of norm $+1$, and is fundamental unless $\Delta=2$. This accounts for infinitely many examples, including the first two, $\Delta = 34 = 5^2 + 3^2 = 6^2 - 2$ and $\Delta = 146 = 11^2 + 5^2 = 12^2 + 2$ (see OEIS sequence A031398). The infinitude may be shown with a polynomial identity such as $$(2t^2+2t+1)^2 + (2t+1)^2 = (2t^2+2t+2)^2 - 2$$ which recovers $\Delta = 34$ for $t=1$. It's then a natural question to ask: as $M \rightarrow \infty$, among those positive $\Delta < M$ that are sums of two coprime squares, for what fraction does $x^2 - \Delta y^2 = -1$ have solutions? I guess that it is conjectured, but not known, that there is a positive limit strictly smaller than $1$.

• There was talk of such examples on this site a while ago: mathoverflow.net/questions/47442/… Mar 20 '12 at 7:51
• @Kevin: I see that you gave the example of $x^2 - 37y^2 = 3$ in a comment to F.Voloch's gold-star answer to that question. Most of the answers have a different flavor, but Borovoi's example (from his paper with Rudnick) is impressive in that he gives an indefinite ternary quadratic form that does not integrally represent $1$ even though there is no local obstruction -- this is presumably the same kind of example as the cubic surfaces that have rational points over ever ${\bf Q}_p$ but not over ${\bf Q}$. Mar 21 '12 at 3:23

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.

• This is really cool! Dec 21 '17 at 16:59

The residual representation of $G_{\mathbb Q_{p}}$ attached to an eigencuspform is markedly different depending on whether $p$ divides the coefficient $a_{p}$, the non-ordinary case, or not, the ordinary case (the representation is reducible if and only if $p$ does not divide $a_{p}$; this translates into very different behaviors for $p$-adic families of cuspforms). But what does $p$ divides $a_{p}$ mean? It means more precisely that, after a choice of an embedding $i_{p}$ of $\bar{\mathbb Q}$ inside $\bar{\mathbb Q}_{p}$, the $p$-adic norm of $i_{p}(a_{p})$ is not 1.

The eigencuspform $f=q+\alpha q^{2}-\alpha q^{3}+(\alpha^{2}-2)q^{4}+(-\alpha^{2}+1)q^{5}+\cdots\in S_{2}(\Gamma_{0}(389))$ where $\alpha$ is a root of $x^{3}-4x-2$ is $5$-ordinary for two of the embeddings of $\mathbb Q[X]/(X^3-4X-2)$ into $\bar{\mathbb Q}_{5}$ but not for the third one (because 1 is a root of $x^{3}-4x-2$ modulo 5).

• I could be wrong, but, assuming the situation for the elliptic curve case is representative of the general situation, shouldn't the statement in parentheses have the word "irreducible" instead of "reducible?" Jun 3 '12 at 3:27
• You are correct. Edited Jun 5 '12 at 14:48