The Brauer-Manin Obstruction and III[2]

Abstract

We discuss the Brauer-Manin obstruction on del Pezzo surfaces of degree 4. We outline a detailed algorithm for computing the obstruction and provide associated programs in MAGMA. This is illustrated with the computation of an example with an irreducible cubic factor in the singular locus of the defining pencil of quadrics (in contrast to previous examples, which had at worst quadratic irreducible factors). We exploit the relationship with the Tate-Shafarevich group to give new types of examples of III [2], for families of curves of genus 2 of the form y2 = f(x), where f(x) is a quintic containing an irreducible cubic factor.

Homogeneous spaces and degree 4 del Pezzo surfaces

“…We shall instead use a initial change of basis over K so that one of the matrices is diagonalised, and then use only the splitting field of f (x) to perform the simultaneous diagonalisation. It will then be clear how to perform a further change of variable to obtain C, δ which are defined over K. We first see, given C, δ, how V δ appears when diagonalised (see also the proof of Lemma 17 in [3]). …”

“…Note also that this can be extended to any δ ∈ A * ; that is, we can define V δ , as above, derived from C, δ, for any δ ∈ A * . We recall the following result (Lemma 17 in [3]). …”

“…The following example is the first using this method to have only two Weierstrass points (whereas the examples in [3], [4] have three Weierstrass points; see also the examples in [6], [12], [14]). …”

Rational points on curves

“…In some cases, it is possible to show that there are non-trivial such elements, thereby improving the upper bound on the rank. Two techniques that have been suggested and also used are visualization [6] and the Brauer-Manin obstruction on certain related varieties [1,25,32].…”

Del Pezzo surfaces of degree four violating the Hasse principle are Zariski dense in the moduli scheme

“…3]. It is given in P 4 Q by the equations Meanwhile, more counterexamples to the Hasse principle have been constructed, see, e.g., [5,Ex. 15 and 16] is a counterexample to the Hasse principle if k is an integer such that 64k 2 + 40k + 5 is a prime number.…”