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The Brauer-Manin Obstruction of del Pezzo surfaces of degree 4

Manar Riman

General Exam University of Washington

August 9, 2016

Manar Riman Brauer-Manin Obstruction August 9, 2016

Overview

1

Motivation

2

The Brauer-Manin Obstruction The Brauer Group The Brauer-Manin Obstruction BSD Example

3

The Main Theorem

Manar Riman Brauer-Manin Obstruction August 9, 2016

Motivation

Let X be a nice variety, i.e, a smooth, projective, geometrically integral variety over a global field k with set of places Ωk. We are interested in existence of k-rational points on X. Xpkq :“{k-rational points on X} XpAkq :“{adelic points on X}. If X is nice then XpAkq “ ś

vPΩ Xpkvq.

If Xpkq ‰ H then Xpkvq ‰ H for every v P Ωk. The converse when it holds for a class of varieties is called the Hasse principle. The Hasse principle is interesting because finding kv points is an easier question. Question Does the Hasse principle always hold?

Manar Riman Brauer-Manin Obstruction August 9, 2016

Motivation

Let X be a nice variety, i.e, a smooth, projective, geometrically integral variety over a global field k with set of places Ωk. We are interested in existence of k-rational points on X. Xpkq :“{k-rational points on X} XpAkq :“{adelic points on X}. If X is nice then XpAkq “ ś

vPΩ Xpkvq.

If Xpkq ‰ H then Xpkvq ‰ H for every v P Ωk. The converse when it holds for a class of varieties is called the Hasse principle. The Hasse principle is interesting because finding kv points is an easier question. Question Does the Hasse principle always hold? Answer It holds for some classes of varieties, but not in general.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Example of the failure of the Hasse principle

Let X be the intersection of two quadrics in P4 given by the equations X : s2 “ xy ` 5z2 s2 ´ 5t2 “ x2 ` 3xy ` 2y2. X is an example of the failure of the Hasse principle, i.e, XpAQq ‰ H and XpQq “ H.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Manin constructed a smaller set XpAkqBr depending on the Brauer group such that: Xpkq Ă XpAkqBr Ă XpAkq. We say that there is a Brauer-Manin obstruction to the Hasse principle for X if XpAkq ‰ H and XpAkqBr “ H. Question Is the Brauer-Manin obstruction the only obstruction to the Hasse principle?

Manar Riman Brauer-Manin Obstruction August 9, 2016

Manin constructed a smaller set XpAkqBr depending on the Brauer group such that: Xpkq Ă XpAkqBr Ă XpAkq. We say that there is a Brauer-Manin obstruction to the Hasse principle for X if XpAkq ‰ H and XpAkqBr “ H. Question Is the Brauer-Manin obstruction the only obstruction to the Hasse principle? Answer No, the first counter example was constructed by Skorobogatov in 1999 [Sko99].

Manar Riman Brauer-Manin Obstruction August 9, 2016

However Colliot-Th´ el` ene and Sansuc conjectured that:

Colliot-Th´ el` ene and Sansuc [CTS81]

For a geometrically rational variety X over a number field, the Brauer-Manin obstruction is the only obstruction to the Hasse principle. Assuming Schinzel’s hypothesis and the finiteness of the Tate-Shafarevich groups for elliptic curves, Wittenberg proved this conjecture for some cases

• f X which assumed Br X “ Br k [Wit07].

We will study a consequence of the conjecture for a nice intersection of two quadrics in P4 also referred to as a del Pezzo surface of degree 4.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Springer’s Theorem [Lam05]

Let Q be a quadric over k and L{k an odd degree extension. Let X :“ V pQq. Then XpLq ‰ H ñ Xpkq ‰ H.

Amer-Brumer Theorem [Lam05]

Let Q1 and Q2 be two quadrics over k. Let X “ V pQ1, Q2q be the intersection of Q1 and Q2 over k, and Xλ “ V pQ1 ` λQ2q a quadric over kpλq where λ is an indeterminant. Then Xpkq ‰ H ð ñ Xλpkpλqq ‰ H. If X is the intersection of two quadrics, and L{k is an odd degree extension then we deduce that: XpLq ‰ H

AB

ù ñ XλpLpλqq ‰ H

Spr

ù ñ Xλpkpλqq ‰ H

AB

ù ñ Xpkq ‰ H.

Manar Riman Brauer-Manin Obstruction August 9, 2016

XpLq ‰ H ñ Xpkq ‰ H

Theorem

Assume CTS conjecture. Let X be an intersection of two quadrics over k, and L{k an odd degree extension. Then XpAkqBr “ H ù ñ XpALqBr “ H. Reason: XpAkqBr “ H ñ Xpkq “ H ñ XpLq “ H

CTS

ù ù ñ XpALqBr “ H Our goal is to prove the theorem unconditionaly. Proving the theorem serves as evidence for Colliot-Th´ el` ene and Sansuc’s conjecture.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Outline for section 2

1

Motivation

2

The Brauer-Manin Obstruction The Brauer Group The Brauer-Manin Obstruction BSD Example

3

The Main Theorem

Manar Riman Brauer-Manin Obstruction August 9, 2016

The Brauer Group of a field

Definition 1 (algebraic)

We define the Brauer group of a field Br k to be: Br k :“ tCSA{ku Brauer Equivalence. We say A „ A

1 iff A bk Mnpkq » A 1 bk Mmpkq for some n, m.

Definition 2 (cohomological)

Let ks be the seperable closure of k. Then Br k “ H2pGalpks{kq, kˆ

s q “ H2 ´ etpk, Gmq.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Example

Let L{k be a cyclic extension of degree n. Fix a generator σ of GalpL{kq and a P kˆ. We define the cyclic algebra pσ, aq to be pσ, aq :“ Lrxsσ xn ´ a where Lrxsσ is the twisted polynomial ring with multiplication defined as xℓ “ ℓσx.

Example

Let a, b P k˚. The quaternion algebra pa, bq2 is generated by i, j as a k-algebra such that i2 “ a, j2 “ b and ij “ ´ji. It is a cyclic algebra corresponding to kp?aq{k.

Theorem [GS06]

The algebra pσ, aq is trivial in Br k if and only if a P NormL{kpLˆq.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Brauer group of a local field

The Brauer group of a nonarchimedean local field kv is completely determined by the invariant isomorphism from class field theory: invv : Br kv

„

Ý Ñ Q{Z. If Lw is a finite extension of kv then the following diagram commutes Br kv

invkv

Ý Ý Ý Ý Ñ Q{Z § § đ § § đrLw:kvs Br Lw

invLw

Ý Ý Ý Ý Ñ Q{Z.

Example [Mil]

Let Lw{kv be an unramified cyclic extension and σ P GalpLw{kvq. Then invkpσ, aq “ vpaq rLw : kvs P Q{Z.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Brauer group of a global field

For k a global field, the fundamental exact sequence of global class field theory completely characterizes Br k: 0 Ñ Br k Ñ ‘vPΩk Br kv

ř

v invv

Ý Ý Ý Ý Ñ Q{Z.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Brauer group of a scheme

Let X be a scheme. As a generalization of the Brauer group of a field we define the cohomological Brauer group of X as follows.

Definition

Br X :“ H2

´ etpX, Gmq.

In practice we often use the following exact sequence for a regular integral noetherian scheme X to test whether an algebra A P Br KpXq is in Br X . 0 Ñ Br X Ñ Br kpXq ‘Bx Ý Ý Ñ ‘xPX p1qH1pkpxq, Q{Zq where X p1q is the set of codimension 1 points of X and kpxq is the residue field corresponding to x.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Brauer group of a scheme

Let X be a scheme. As a generalization of the Brauer group of a field we define the cohomological Brauer group of X as follows.

Definition

Br X :“ H2

´ etpX, Gmq.

In practice we often use the following exact sequence for a regular integral noetherian scheme X to test whether an algebra A P Br KpXq is in Br X . 0 Ñ Br X Ñ Br kpXq ‘Bx Ý Ý Ñ ‘xPX p1qH1pkpxq, Q{Zq where X p1q is the set of codimension 1 points of X and kpxq is the residue field corresponding to x. Fact: Let L{KpXq be cyclic and of prime degree p that is unframified at x. Then pL{KpXq, aq P ker Bx iff x splits completely in L or vxpaq ” 0 ppq.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Let A P Br X. We define XpAkqA :“ tpPvq P XpAkq : ÿ

vPΩk

invvpA pPvqq “ 0u.

Definition

The Brauer-Manin set is XpAkqBr :“ č

A PBr X

XpAkqA .

Manar Riman Brauer-Manin Obstruction August 9, 2016

The Brauer-Manin Obstruction

XpAkqA :“ tpPvq P XpAkq : ÿ

vPΩk

invvpA pPvqq “ 0u It follows from the fundamental sequence of class field theory Xpkq XpAkq Br k ‘v Br kv Q{Z

evA evA ř

v invv

that Xpkq Ă XpAkqBr Ă XpAkq.

Manar Riman Brauer-Manin Obstruction August 9, 2016

The Brauer-Manin Obstruction

XpAkqA :“ tpPvq P XpAkq : ÿ

vPΩk

invvpA pPvqq “ 0u It follows from the fundamental sequence of class field theory Xpkq XpAkq Br k ‘v Br kv Q{Z

evA evA ř

v invv

that Xpkq Ă XpAkqBr Ă XpAkq.

Definition

We say that there is a Brauer-Manin obstruction to the Hasse principle for X if XpAkq ‰ H and XpAkqBr “ H.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Birch and Swinnerton-Dyer Example

Recall the example given by the intersection of two quadrics in P4: X : s2 “ xy ` 5z2 s2 ´ 5t2 “ x2 ` 3xy ` 2y2. We will show that this example illustrates the Brauer-Manin obstruction to the Hasse principle, i.e, XpAQq ‰ H and XpAQqBr “ H. XpRq ‰ H Every p ‰ 2 splits in either Qp?´1q, Qp ? 5q or Qp?´5q. If p ‰ 2 splits in Qp?´1q then p1 : 1 : 1 : 0 : ?´1q P XpQpq. If p ‰ 2 splits in Qp ? 5q then p0 : 0 : ? 5 : 1; 1q P XpQpq. If p ‰ 2 splits in Qp?´5q then p1 : 0 : 0 : 0 :

1 ?´5q P XpQpq.

p´5 : 1 : 0 : 1 :

6 ?´15q P XpQ2q.

reason: one can show that 2 splits in Qp?´15q Hence XpAQq “ ś

v XpQvq ‰ H.

Manar Riman Brauer-Manin Obstruction August 9, 2016

X : s2 “ xy ` 5z2 s2 ´ 5t2 “ x2 ` 3xy ` 2y2 We want to prove that XpAQqBr “ H. Let A :“ p5, px ` yq{xq2. Other representations of A are A “ p5, px `2yq{xq2 “ p5, px `yq{yq2 “ p5, px `2yq{yq2 P Br KpXq. It is enough to show that XpAQqA “ H. A P Br X by the exact sequence 0 Ñ Br Xr2s Ñ BrpkpXqqr2s ‘Bc Ý Ý Ñ ‘cPX p1qH1 ˆ kpcq, 1 2Z{Z ˙ . First consider the open set U “ X ´ Z where Z “ V pxpx ` yq, xpx ` 2yq, ypx ` yq, ypx ` 2yqq. reason: Up1q “ X p1q and Br X » Br U

Manar Riman Brauer-Manin Obstruction August 9, 2016

s2 “ xy ` 5z2 s2 ´ 5t2 “ x2 ` 3xy ` 2y2 A “ p5, px `yq{xq2 “ p5, px `2yq{xq2 “ p5, px `yq{yq2 “ p5, px `2yq{yq2 If v P Ω splits in Qp ? 5q{Q then invvpA pPvqq “ 0 for all Pv P XpQvq. reason: 5 P Qˆ2

v

If v is inert in Qp ? 5q{Q then invvpA pPvqq “ 0 for all Pv P XpQvq. reason: invvppσ, bqq “ vpbq{rQvp ? 5q : Qvs for unramified valuations 0 “ vp1q “ v `x0 ` 2y0 y0 ´ x0 ` y0 y0 ˘ ě mintvpx0{y0 ` 2q, vpx0{y0 ` 1qu If v “ 5 then invvpA pPvqq “ 1{2 P Q{Z. reason: the equation 5x2 ` ppx0 ` y0q{x0qy2 “ z2 has no solution in Q5 and one can prove that this is equivalent to A pPvq is not trivial Finally ř

v invvA pPvq “ 1{2 for all pPvq P XpAQq. Hence XpAQqBr “ H.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Outline for section 3

1

Motivation

2

The Brauer-Manin Obstruction The Brauer Group The Brauer-Manin Obstruction BSD Example

3

The Main Theorem

Manar Riman Brauer-Manin Obstruction August 9, 2016

Goal

Let X be an intersection of two quadrics over k and L{k an odd degree

• extension. Prove that

XpAkqBr “ H ù ñ XpALqBr “ H. We specialize to the case Br X{ Br k “ xA y » Z{2Z. In this case the theorem reduces to: XpAkqA “ H ù ñ XpALqA “ H.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Statements 1 and 2 are sufficient to prove our goal. Statement 1 If invvA : Xpkvq Ñ 1

2Z{Z are constant for all valuations v

• n k then so are invwA for all valuations w on L where L{k

is a finite extension. Statement 2 Assume that invvA and invwA are constant and L{k is an

• dd extension. If there is an odd number of places v on k so

that invvA pXpkvqq “ t1{2u then the same is true for L. claim 1: If XpAkqA “ H then invvA : Xpkvq Ñ 1

2Z{Z are constant for all

valuations v on k otherwise we can find Qv0 so that invv0A pQv0q “ ´ ř

v‰v0 invvA pPvq for a fixed pPvq P Ak.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Statement 1 If invvA : Xpkvq Ñ 1

2Z{Z are constant for all valuations v

• n k then so are invwA for all valuations w on L where L{k

is a finite extension. Statement 2 Assume that invvA and invwA are constant and L{k is an

• dd extension. If there is an odd number of places v on k so

that invvA pXpkvqq “ t1{2u then the same is true for L. claim 2: If invvA : Xpkvq Ñ 1

2Z{Z are constant for all places v on k and

XpAkqA “ H then there is an odd number of places v so that invvA pXpkvqq “ t1{2u.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Statement 2: Assume that invvA and invwA are constant and L{k is an

• dd extension. If there is an odd number of places v on k so that

invvA pXpkvqq “ t1{2u then the same is true for L. Xpkvq

evA

Ý Ý Ý Ý Ñ Br kv

invv

Ý Ý Ý Ý Ñ Q{Z § § đ § § đ § § đrLw:kvs XpLwq

evA

Ý Ý Ý Ý Ñ Br Lw

invw

Ý Ý Ý Ý Ñ Q{Z.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Statement 2: Assume that invvA and invwA are constant and L{k is an

• dd extension. If there is an odd number of places v on k so that

invvA pXpkvqq “ t1{2u then the same is true for L. Xpkvq

evA

Ý Ý Ý Ý Ñ Br kv

invv

Ý Ý Ý Ý Ñ Q{Z § § đ § § đ § § đrLw:kvs XpLwq

evA

Ý Ý Ý Ý Ñ Br Lw

invw

Ý Ý Ý Ý Ñ Q{Z. If invvA pXpkvqq “ t0u or rLw : kvs is even then invwA pXpLwqq “ t0u.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Statement 2: Assume that invvA and invwA are constant and L{k is an

• dd extension. If there is an odd number of places v on k so that

invvA pXpkvqq “ t1{2u then the same is true for L. Xpkvq

evA

Ý Ý Ý Ý Ñ Br kv

invv

Ý Ý Ý Ý Ñ Q{Z § § đ § § đ § § đrLw:kvs XpLwq

evA

Ý Ý Ý Ý Ñ Br Lw

invw

Ý Ý Ý Ý Ñ Q{Z. If invvA pXpkvqq “ t0u or rLw : kvs is even then invwA pXpLwqq “ t0u. If invvA pXpkvqq “ t1

2u and rLw : kvs is odd then

invwA pXpLwqq “ t1

2u.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Statement 2: Assume that invvA and invwA are constant and L{k is an

• dd extension. If there is an odd number of places v on k so that

invvA pXpkvqq “ t1{2u then the same is true for L. Xpkvq

evA

Ý Ý Ý Ý Ñ Br kv

invv

Ý Ý Ý Ý Ñ Q{Z § § đ § § đ § § đrLw:kvs XpLwq

evA

Ý Ý Ý Ý Ñ Br Lw

invw

Ý Ý Ý Ý Ñ Q{Z. If invvA pXpkvqq “ t0u or rLw : kvs is even then invwA pXpLwqq “ t0u. If invvA pXpkvqq “ t1

2u and rLw : kvs is odd then

invwA pXpLwqq “ t1

2u.

There is an odd number of places such that invvA pXpkvqq “ t 1

2u and

rL : ks “ ř

w{vrLw : kvs is odd.

Hence there is an odd number of places w|v such that invwA pXpLwqq “ t1

2u.

Manar Riman Brauer-Manin Obstruction August 9, 2016

It remains to prove Statement 1: If invvA : Xpkvq Ñ 1

2Z{Z are constant for all valuations v on k then so

are invwA for all valuations w on L where L{k is a finite extension.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Let kv be a finite extension of Qp, O be its ring of integers, and F its residue field. Let X be a regular nice scheme over O X :“ X ˆO Spec kv be its generic fiber X0 :“ X ˆO Spec F be its special fiber X ˝

0 be the non-singular locus of X0.

Lemma

Let Z1, Z2, . . . , Zn be the connected components of X ˝

0 . Consider P a

k-point of X lying in the image of X pOq and reducing to a point P0 of X ˝

0 . Then the following diagram commutes:

Br Xpp

1q

ś

i Bi

Ý Ý Ý Ý Ñ ś

i H1pZi, Q{Zqpp

1q

§ § đP˚ § § đP˚ Br kvpp

1q

»

Ý Ý Ý Ý Ñ H1pF, Q{Zqpp

1q. Manar Riman Brauer-Manin Obstruction August 9, 2016

Theorem

Assume that X ˝

0 has geometrically irreducible components Z1, Z2, . . . , Zn,

and that the residue field F is big enough. Let A P Br X be a normalized

• algebra. If the evaluation map evA : X pOkv q Ñ Br kv is constant then so

is evA : X pOLw q Ñ Br Lw where Lw{kv is a finite extension. Sketch of the proof: (see also [Bri15]) Given evA : X pOkv q Ñ Br kv is constant, we want to show that pBiqkv A “ 0 for all i. Let Y Ñ Zi be the torsor representing pBiqkv A in H1pZi, Z{2q Suppose pBiqkv A ‰ 0 for some i. Then pBiqkv A has order 2 in H1pZi, Z{2q. There exists BpY q such that when |F| ą BpY q, Y pFq ‰ H and so is Y BαpFq where α P Br kr2s. reason: Weil conjectures We can pick B independent of the torsor Y Ñ Zi and independent of

• Zi. reason: H1pZi, Z{2q is finite and the components are finite.

Let P0 be in the image of Y BαpFq Ñ ZipFq. Y Bα

P0 pFq ‰ H ñ rY Bα P0 s “ 0 ñ rYP0s “ Bα.

Manar Riman Brauer-Manin Obstruction August 9, 2016

rYP0s “ Bα A pPq “ P˚pA q “ ´α, a contradiction. Br Xpp

1q

Bi

Ý Ý Ý Ý Ñ H1pZi, Q{Zqpp

1q

§ § đP˚ § § đP˚ Br kvpp

1q

»

Ý Ý Ý Ý Ñ H1pF, Q{Zqpp

1q

So pBiqkv A “ 0 for every i. Hence pBiqLw A “ 0 by the diagram Br Xpp

1q

Bk

Ý Ý Ý Ý Ñ H1ppZiqkv , Q{Zqpp

1q

§ § đ § § đ Br XLw pp

1q

BL

Ý Ý Ý Ý Ñ H1ppZiqLw , Q{Zqpp

1q.

To prove that evA : X pOLw q Ñ Br Lw is constant, let P, Q P X pOLw q. We have that P˚pA q “ Q˚pA q by the map H1pFLw , Q{Zqpp

1q Ñ

ź

i

H1ppZiqLw , Q{Zqpp

1q.

Hence evA : X pOLw q Ñ Br Lw is constant.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Theorem

Assume that X ˝

0 has geometrically irreducible components Z1, Z2, . . . , Zn,

and that the residue field F is big enough. Let A P Br X be a normalized

• algebra. If the evaluation map evA : X pOkv q Ñ Br kv is constant then so

is evA : X pOLw q Ñ Br Lw where Lw{kv is a finite extension.

Corollary

Let X be an intersection of two quadrics over k{Q such that Br X{ Br k “ xA y. Consider a place v P Ωk with sufficiently large residue

• field. Let L{k be a finite extension. Assume that X admits a proper

regular model X over the ring of integers Ov. Suppose that the connected components of the special fiber X ˝

0 are geometrically irreducible.

If invvA : Xpkvq Ñ 1

2Z{Z is constant then so is invwA : XpLwq Ñ 1 2Z{Z

for every place w of L lying over v.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Thank you!

Colliot-Th´ el` ene, Jean-Louis and Sansuc, Jean-Jacques On the Chow groups of certain rational surfaces: a sequel to a paper of S. Bloch Duke Mathematical Journal, 1981. Skorobogatov, Alexei N. Beyond the Manin obstruction Inventiones Mathematicae, 1999. Bright, Martin Bad reduction of the Brauer-Manin obstruction Journal of the London Mathematical Society. Second Series, 2015.

Manar Riman Brauer-Manin Obstruction August 9, 2016

Lam, T. Y. Introduction to quadratic forms over field American Mathematical Society, Providence, RI, 2005. Milne, James Class Field Theory unpublished, available at http://www.jmilne.org/math/CourseNotes/CFT.pdf. Gille, Philippe and Szamuely, Tam´ as Central simple algebras and Galois cohomology Cambridge University Press, Cambridge, 2006. Wittenberg, Olivier Intersections de deux quadriques et pinceaux de courbes de genre 1/Intersections of two quadrics and pencils of curves of genus 1 Springer, Berlin, 2007.

Manar Riman Brauer-Manin Obstruction August 9, 2016