(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.


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    $\begingroup$ Community wiki, I suppose? $\endgroup$ – Noam D. Elkies Mar 18 '12 at 16:30
  • $\begingroup$ Yes. How do I do that? $\endgroup$ – TKe Mar 18 '12 at 16:31
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    $\begingroup$ 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/…. $\endgroup$ – KConrad Mar 18 '12 at 17:39
  • $\begingroup$ This should be CW. $\endgroup$ – Benjamin Steinberg Mar 19 '12 at 0:46
  • $\begingroup$ Why put a bounty on a big-list / CW question? . . . $\endgroup$ – Noam D. Elkies Dec 20 '17 at 5:16

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 $1$.

  • $\begingroup$ There was talk of such examples on this site a while ago: mathoverflow.net/questions/47442/… $\endgroup$ – Kevin Buzzard Mar 20 '12 at 7:51
  • $\begingroup$ @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 ${\bf Q}$. $\endgroup$ – Noam D. Elkies 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 in this link.

  • $\begingroup$ This is really cool! $\endgroup$ – Daniel Litt 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).

  • $\begingroup$ 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?" $\endgroup$ – Keenan Kidwell Jun 3 '12 at 3:27
  • $\begingroup$ You are correct. Edited $\endgroup$ – Olivier Jun 5 '12 at 14:48

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