Introduction to work of Hassett-Pirutka-Tschinkel and Schreieder

Abstract

Hassett, Pirutka and Tschinkel gave the first examples of families X → B of smooth, projective, connected, complex varieties having some rational fibres and some other fibres which are not even stably rational. This used the specialisation method of Voisin, as extended by Pirutka and myself. Under specific circumstances, a simplified version of the specialisation method was produced by Schreieder, leading to a simpler proof of the HPT example. I describe the method in its simplest form.

Appendices

Appendix 1: Conics over a Discrete Valuation Ring

Let R be a dvr with residue field k of characteristic not 2. Let K be the fraction field. A smooth conic over K admits a regular model $\mathcal {X}$ given in ${\mathbb P}^2_{R}$ either by an equation

$$\displaystyle \begin{aligned}x^2-ay^2-bz^2=0\end{aligned} $$

with a, b ∈ R ^∗ (case (I)) or a regular model $\mathcal {X}$ given by an equation

$$\displaystyle \begin{aligned}x^2-ay^2-\pi z^2=0\end{aligned}$$

with a ∈ R ^∗ and π a uniformizing parameter (case (II)). Moreover, in the second case one may assume that a is not a square in the residue field κ.

Proposition 1

Let R be a dvr with residue field k of characteristic not 2. Let K be the fraction field. Let W→Spec(R) be a proper flat morphism with W regular and connected. Assume that the generic fibre is a smooth conic over K. Then:

• (a) The natural map Br(R)→Br(W) is onto.

• (b) For Y ⊂ W an integral divisor contained in the special fibre of W→Spec(R), and β ∈Br(W), the image of β under restriction Br(W)→Br(Y ) belongs to the image of Br(κ)→Br(Y ).

Proof

By purity for the Brauer group of a two-dimensional regular scheme, to prove (a), one may assume that $W=\mathcal {X}$ as above. Let $X=\mathcal {X}\times _{R}K$. It is well known that the map Br(K)→Br(X) is onto, with kernel spanned by the quaternion symbol (a, b)_K in case (I) and by (a, π)_K in case (II).

Let $\beta \in \mathrm {{Br}}(\mathcal {X}) \subset \mathrm {{Br}}(X)$. Let α ∈Br(K) be some element with image β _K. We have the exact sequence

$$\displaystyle \begin{aligned}0 {\rightarrow} \mathrm{{Br}}(R)\{2\} {\rightarrow} \mathrm{{Br}}(K)\{2\} {\rightarrow} H^1(\kappa,{\mathbb Q}_{2}/{\mathbb Z}_{2})\end{aligned}$$

Comparison of residues on Spec(R) and on $\mathcal {X}$ shows that the residue δ _R(α) is either 0 or is equal to the nontrivial class in $H^1(k(\sqrt {\overline a})/k,{\mathbb Z}/2)$, and this last case may happen only in case (II). In the first case, we have α ∈Br(R), hence $\beta -\alpha _{\mathcal {X}}=0$ in Br(X) hence also in $\mathrm {{Br}}(\mathcal {X})$ since $\mathcal {X}$ is regular. In the second case, we have

$$\displaystyle \begin{aligned}\delta_{R}(\alpha)=\delta_{R}((a,\pi))\end{aligned}$$

hence α = (a, π) + γ with γ ∈Br(R). We then get

$$\displaystyle \begin{aligned}\beta= (a,\pi)_{K(X)} + \gamma_{K(X)} \in \mathrm{{Br}}(K(X)).\end{aligned}$$

But (a, π)_K(X) = 0. Thus $\beta -\gamma _{\mathcal {X}} \in \mathrm {{Br}}(\mathcal {X}) \subset \mathrm {{Br}}(K(X))$ vanishes, hence $\beta = \gamma _{\mathcal {X}} \in \mathrm {{Br}}(\mathcal {X})$. The map $\mathrm {{Br}}(R) {\rightarrow } \mathrm {{Br}}(\mathcal {X})$ is thus surjective. This gives (a) for $\mathcal {X}$ hence for W, and (b) immediately follows. □

Exercise

Artin-Mumford type examples are specific singular conic bundles X in the total space of a rank 3 projective bundle over ${\mathbb P}^2_{{\mathbb C}}$ whose unramified Brauer group is non trivial. Using Proposition 1 and Theorem 5.1, deform such examples into conic bundles of the same type with smooth ramification locus and whose total space is not stably rational. As in Sect. 4, there is no need to compute an explicit resolution of singularities of X.

Appendix 2: Quadric Surfaces over a Discrete Valuation Ring

The following section was written up to give details on some tools and results used in [19, Thm. 4]. As demonstrated above, this section turns out not to be necessary to vindicate the HPT example. But it is useful for more general examples.

References: [21], [8, §3], [9, Thm. 2.3.1], [17, Thm. 3.17].

Let R be a discrete valuation ring, K its fraction field, π a uniformizer, κ = R∕(π) the residue field. Assume char(κ) ≠ 2.

Let $X \subset {\mathbb P}^3_{K}$ be a smooth quadric, defined by a nondegenerate four-dimensional quadratic form q. Up to scaling and changing of variables, there are four possibilities.

• (I) q =< 1, −a, −b, abd >  with a, b, d ∈ R ^∗.

• (II) q =< 1, −a, −b, π >  with a, b ∈ R ^∗ and π a uniformizing parameter of R.

• (III) q =< 1, −a, π, −π.b >  with a, b ∈ R ^∗ and π a uniformizing parameter of R. The class of a.b ∈ R ^∗ represents the discriminant of the quadratic form. Its image $\overline {a}.\overline {b} \in \kappa ^*$ is a square if and only if the discriminant of q is a square in the completion of K for the valuation defined by R.

Let ${\mathcal X} \subset {\mathbb P}^3_{R}$ be the subscheme cut out by q. Let Y∕κ be the special fibre.

In case (I), ${\mathcal X} /R$ is smooth.

In case (II), X is regular, the special fibre Y  is a cone over a smooth conic.

In case (III), the special fibre is given by the equation $x^2-\overline {a}y^2=0$ in ${\mathbb P}^3_{\kappa }$. If $\overline {a}$ is a square, this is the union of two planes intersecting along the line x = y = 0. If $\overline {a}$ is not a square, this is an integral scheme which over $\kappa (\sqrt {a})$ breaks up as the union of two planes. In both cases, the scheme $\mathcal {X}$ is singular at the two points given by $x=y=0, z^2-\overline {d}t^2=0$. See [21, §2].

Proposition 1

Let us assume char(κ) = 0.

In case (III), let $W {\rightarrow } \mathcal {X}$ be a projective, birational desingularisation of $\mathcal X$.

In case (I), the map $\mathrm {{Br}}(R) {\rightarrow } \mathrm {{Br}}(\mathcal {X})$ is onto. If d ∈ R is not a square, it is an isomorphism. If d is a square, the kernel is spanned by the class (a, b) ∈Br(R).

In case (II), the map $\mathrm {{Br}}(R) {\rightarrow } \mathrm {{Br}}(\mathcal {X})$ is an isomorphism.

In case (III), assume $\overline {a}.\overline {b}$ is not a square in κ. Then Br(R)→Br(W) is onto.

In case (III), if either $\overline {a}$ or $\overline {b}$ is a square, or if $\overline {a}.\overline {b}$ is not a square, then Br(R)→Br(W) is onto. An element of Br(K) whose image in Br(X) lies in Br(W) belongs to Br(R).

In case (III), assume $\overline {a}.\overline {b}$ is a square in κ. Then the image of (a, π) ∈Br(K) in Br(X) belongs to Br(W). It spans the quotient of Br(W) by the image of Br(R). If moreover $\overline {a}$ is not a square in κ, then it does not belong to the image of Br(R).

Proof

Let x be a codimension 1 regular point on $\mathcal {X}$ or on W, lying above the closed point of Spec(R). Let e _x denote its multiplicity in the fibre. We have a commutative diagram

$$\displaystyle \begin{aligned}\begin{array}{ccccccccc} \mathrm{{Br}}(K) & {\rightarrow} & H^1(\kappa,{\mathbb Q}/{\mathbb Z})\\ \downarrow & & \downarrow \\ \mathrm{{Br}}(X) &{\rightarrow} &H^1(\kappa(x), {\mathbb Q}/{\mathbb Z})\\ \end{array} \end{aligned}$$

The kernel of ${\mathrm{Br}}(K) {\rightarrow } H^1(\kappa ,{\mathbb Q}/{\mathbb Z})$ is Br(R).

In case (I) and (III), the special fibre Y  is geometrically integral over κ, the multiplicity is 1, the map $H^1(\kappa ,{\mathbb Q}/{\mathbb Z}) {\rightarrow } H^1(\kappa (x), {\mathbb Q}/{\mathbb Z})$ is thus injective. This is enough to prove the claim.

Let us consider case (III). The map Br(K)→Br(X) is onto. Let α ∈Br(K). Let $\rho \in H^1(\kappa ,{\mathbb Q}/{\mathbb Z})$ be its residue. On the (singular) normal model given by q =< 1, −a, π, −π.b >  over R, if $\overline {a} \in \kappa$ is a square, the fibre Y  contains geometrically integral components of multiplicity 1 given by the components of $x^2-\overline {a}y^2=0$. By the above diagram, $\rho =0 \in H^1(\kappa ,{\mathbb Q}/{\mathbb Z})$. We can also use the model given by q =< 1, −b, π, −π.a > . If $\overline {b} \in \kappa$ is a square, we conclude that $\rho =0 \in H^1(\kappa ,{\mathbb Q}/{\mathbb Z})$. Let us assume that $\rho \neq 0 \in H^1(\kappa ,{\mathbb Q}/{\mathbb Z})$. Thus $\overline {a}$ and $\overline {b}$ are nonsquares. On the first model, the kernel of $H^1(\kappa , {\mathbb Q}/{\mathbb Z}) {\rightarrow } H^1(\kappa (Y),{\mathbb Q}/{\mathbb Z})$ coincides with the kernel of $H^1(\kappa , {\mathbb Q}/{\mathbb Z}) {\rightarrow } H^1(\kappa (\sqrt {\overline {a}}, {\mathbb Q}/{\mathbb Z})$, which is the ${\mathbb Z}/2$-module spanned by the class of $\overline {a}$ in $\kappa ^*/\kappa ^{*2}=H^1(\kappa ,{\mathbb Z}/2)$. On the second model, the kernel of $H^1(\kappa , {\mathbb Q}/{\mathbb Z}) {\rightarrow } H^1(\kappa (Y),{\mathbb Q}/{\mathbb Z})$ is the ${\mathbb Z}/2$-module spanned by the class of $\overline {b}$ in $\kappa ^*/\kappa ^{*2}=H^1(\kappa ,{\mathbb Z}/2)$. We thus conclude that $\overline {a}.\overline {b}$ is a square in κ, and that the residue of α coincides with $\overline {a}$, i.e. is equal to the residue of (a, π) ∈Br(K) (or to the residue of (b, π)).

It remains to show that if $\overline {a}.\overline {b}$ is a square in κ, then (a, π) has trivial residues on W and more generally with respect to any rank one discrete valuation v on the function field K(X) of X. One may restrict attention to those v which induce the R-valuation on K. Let S ⊂ K(X) be the valuation ring of v and let λ be its residue field. There is an inclusion κ ⊂ λ. In K(X) we have an equality

$$\displaystyle \begin{aligned}(x^2-ay^2)=\pi.(z^2-b),\end{aligned}$$

where both sides are nonzero. Thus in Br(K(X)), we have the equality

$$\displaystyle \begin{aligned}(a,\pi)= (a,x^2-ay^2) + (a, z^2-b) = (a,z^2-b),\end{aligned}$$

where the last equality comes from the classical (a, x ² − ay ²) = 0. To compute residues, we may go over to completions. In the completion of R, a.b is a square. It is thus a square in the completion of K(X) at v. But then in this completion (a, z ² − b) = (b, z ² − b) = 0 Hence the residue of (a, π) at v is zero. □

Proposition 2

Assume char(κ) = 0. Let ${\mathcal {X}} \subset {\mathbb P}^3_{R}$ be as above, and let $W {\rightarrow } \mathcal {X}$ be a proper birational map with W regular. Let β ∈Br(W) and let Y ⊂ W be an integral divisor contained in the special fibre of W→Spec(R). Then the image of β in Br(κ(Y )) belongs to the image of Br(κ)→Br(κ(Y )).

Proof

In case (I) and (II), and in case (III) when $\overline {a}.\overline {b}$ is a square in κ, this is clear since then the map Br(R)→Br(W) is onto.

Suppose we are in case (III). To prove the result, we may make a base change from R to its henselisation. Then ab is square in R. The group Br(W) is spanned by the image of Br(R) and the image of the class (a, π). The equation of the quadric may now be written

$$\displaystyle \begin{aligned}X^2-aY^2+\pi Z^2- a\pi T^2=0.\end{aligned}$$

This implies that (a, −π) vanishes in the Brauer group of the function field κ(W) of W. Since W is regular, the map Br(W)→Br(κ(W)) is injective. Since (a, −π)_κ(W) belongs to Br(W) and spans Br(W) modulo the image of Br(R), this completes the proof. □

One may rephrase the above results in a simpler fashion.

Proposition 3

Assume char(κ) = 0. Let ${\mathcal {X}} \subset {\mathbb P}^3_{R}$ be as above, and let $W {\rightarrow } \mathcal {X}$ be a proper birational map with W regular.

• (i) If R is henselian, then the map Br(R)→Br(W) is onto.

• (ii) For any element β ∈Br(W) and Y ⊂ W an integral divisor contained in the special fibre of W→Spec(R), the image of β under restriction Br(W)→Br(Y ) belongs to the image of Br(κ)→Br(Y ).

Upon use of Merkurjev’s geometric lemmas [16, §1], and use of Tsen’s theorem, one then gets [19, Prop. 7] of Schreieder.

Appendix 3: A Remark on the Vanishing of Unramified Elements on Components of the Special Fibre

The following proposition, found in June 2017, gives some partial explanation for the vanishing on components of the special fibre which occurs in [18, Prop. 6, Prop. 7] and [19, Prop. 7] or in Proposition 4.1 above. Unfortunately the proof requires that the component be of multiplicity one in the fibre. Since this was written, in the case of quadric bundles, Schreieder [20, §9.2] has managed to use arguments as in [8, §3] to get information on what happens with the other components.

Proposition 1

Let A↪B be a local homomorphism of discrete valuation rings and let K ⊂ L be the inclusion of their fraction fields. Let κ ⊂ λ be the induced inclusion on their residue fields.

Let ℓ be a prime invertible in A.

Let i ≥ 2 be an integer and let $\alpha \in H^{i}(K,\mu _{\ell }^{\otimes i})$.

Assume:

• (i) B is unramified over A.

• (ii) The image of α in $H^{i}(L,\mu _{\ell }^{\otimes i})$ is unramified, and in particular is the image of a (well defined) element $\beta \in H^{i}(B,\mu _{\ell }^{\otimes i})$.

Then $\beta (\lambda ) \in H^{i}(\lambda ,\mu _{\ell }^{\otimes i})$ is in the image of $H^{i}(\kappa , \mu _{\ell }^{\otimes i}) {\rightarrow } H^{i}(\lambda ,\mu _{\ell }^{\otimes i})$.

Proof

We may assume that A and B are henselian. Then the residue map $\partial _{A} : H^{i}(K, \mu _{\ell }^{\otimes i}) {\rightarrow } H^{i-1}(\kappa , \mu _{\ell }^{\otimes (i-1)})$ is part of a split exact sequence [10, Appendix B] and [12, Cor. 6.8.8]

$$\displaystyle \begin{aligned}0 {\rightarrow} H^{i}(A, \mu_{\ell}^{\otimes i}) {\rightarrow} H^{i}(K, \mu_{\ell}^{\otimes i}) {\rightarrow} H^{i-1}(\kappa, \mu_{\ell}^{\otimes (i-1)}) {\rightarrow} 0,\end{aligned}$$

and all reduction maps $H^{j}(A, \mu _{\ell }^{\otimes i}) {\rightarrow } H^{j}(\kappa , \mu _{\ell }^{\otimes i})$, denoted ρ↦ρ(κ), are isomorphisms. We have the analogous split exact sequence

$$\displaystyle \begin{aligned}0 {\rightarrow} H^{i}(B, \mu_{\ell}^{\otimes i}) {\rightarrow} H^{i}(L, \mu_{\ell}^{\otimes i}) {\rightarrow} H^{i-1}(\lambda, \mu_{\ell}^{\otimes (i-1)}) {\rightarrow} 0,\end{aligned}$$

Let π ∈ A be a uniformizer. Given $\alpha \in H^{i}(K,\mu _{\ell }^{\otimes i})$, the residue $\partial _A({\alpha }) \in H^{i-1}(\kappa ,\mu _{\ell }^{\otimes {(i-1)}})$ is the image of some unique $\gamma \in H^{i-1}(A,\mu _{\ell }^{\otimes {(i-1)}})$. Let us denote by (π) the class of π in K ^∗∕K ^∗ℓ = H ¹(K, μ _ℓ). Then the difference $\alpha - (\pi ) \cup \gamma \in H^{i}(K,\mu _{\ell }^{\otimes i})$ has trivial residue. Thus there exists $\zeta \in H^{i}(A, \mu _{\ell }^{\otimes i})$ such that

$$\displaystyle \begin{aligned}\alpha = \zeta + (\pi) \cup \gamma \in H^{i}(K,\mu_{\ell}^{\otimes i}).\end{aligned}$$

By hypothesis, the restriction β of α to L is unramified. Thus $(\pi ) \cup \gamma \in H^{i}(L,\mu _{\ell }^{\otimes i})$ is unramified. Since B is unramified over A, the uniformizer π is also a uniformizer of B. Thus

$$\displaystyle \begin{aligned}0= \partial_{B}((\pi) \cup \gamma)= \gamma_{\lambda} \in H^{i-1}(\lambda,\mu_{\ell}^{\otimes {(i-1)}})\end{aligned}$$

hence $\gamma =0 \in H^{i-1}(B,\mu _{\ell }^{\otimes {(i-1)}}),$ from which follows $\beta =\zeta \in H^{i}(B,\mu _{\ell }^{\otimes i})$ and $\beta (\lambda ) \in H^{i}(\lambda ,\mu _{\ell }^{\otimes i})$ is the image of $\zeta (\kappa ) \in H^{i}(\kappa ,\mu _{\ell }^{\otimes i})$. □