# Birational classification of pointless del Pezzo surfaces of degree $8$

Andrey Trepalin

###### Abstract.

Let $\Bbbk$ be an algebraically nonclosed field of characteristic $0$. We show
that two pointless quadric surfaces over $\Bbbk$ are birationally equivalent
if and only if they are isomorphic. Also we describe minimal surfaces
birationally equivalent to a given pointless quadric surface.

This work is supported by the Russian Science Foundation under grant
№18-11-00121.

##  1\. Introduction

Let $\Bbbk$ be a field of characteristic $0$. We want to study biregular and
birational classification of quadrics in $\mathbb{P}^{3}_{\Bbbk}$. If $\Bbbk$
is algebraically closed then all smooth quadrics are isomorphic, but if
$\Bbbk$ is not algebraically closed then the classification is non-trivial.
The biregular classification is well-known (see Section 2 for the details),
and the aim of this paper is to show that birational classification is almost
the same.

Note that if there is a $\Bbbk$-point on a quadric surface $Q$ in
$\mathbb{P}^{3}_{\Bbbk}$, then one can consider a projection from this point
to $\mathbb{P}^{2}_{\Bbbk}$. This projection is a birational map
$Q\dashrightarrow\mathbb{P}^{2}_{\Bbbk}$. Therefore any quadric with a
$\Bbbk$-point is $\Bbbk$-rational. In particular, any two such quadrics are
birationally equivalent over $\Bbbk$. But if there are no $\Bbbk$-points on a
quadric then it is obviously not $\Bbbk$-rational. Thus in this paper we study
pointless quadrics in $\mathbb{P}^{3}_{\Bbbk}$. The main result of this paper
is the following theorem.

######  Theorem 1.1.

Let $\Bbbk$ be an algebraically nonclosed field of characteristic $0$, and let
$Q_{1}$ and $Q_{2}$ be two pointless quadrics in $\mathbb{P}^{3}_{\Bbbk}$.
Then $Q_{1}$ is birationally equivalent to $Q_{2}$ if and only if $Q_{1}\cong
Q_{2}$.

To prove this theorem we study pointless del Pezzo surfaces of degree $8$. Let
us recall the definition of a del Pezzo surface.

######  Definition 1.2.

A del Pezzo surface is a smooth projective surface $X$ such that the
anticanonical class $-K_{X}$ is ample. The number $d=K_{X}^{2}$ is called the
degree of a del Pezzo surface $X$.

Any quadric in $\mathbb{P}^{3}_{\Bbbk}$ is a del Pezzo surface of degree $8$.
Moreover, over algebraically closed field any del Pezzo surface of degree $8$
is either isomorphic to $\mathbb{P}^{1}_{\Bbbk}\times\mathbb{P}^{1}_{\Bbbk}$,
or isomorphic to the blowup of $\mathbb{P}^{2}_{\Bbbk}$ at a point. Therefore
for an arbitrary field $\Bbbk$ of characteristic $0$ and a del Pezzo surface
$X$ of degree $8$ there are two possibilities: either
$\overline{X}=X\otimes_{\Bbbk}\overline{\Bbbk}$ is isomorphic to
$\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$, or
$\overline{X}$ is isomorphic to the blowup of
$\mathbb{P}^{2}_{\overline{\Bbbk}}$ at a point. In the latter case there is a
unique $(-1)$-curve on $\overline{X}$. This curve is defined over $\Bbbk$,
therefore we can contract this curve and get a morphism
$X\rightarrow\mathbb{P}^{2}_{\Bbbk}$. In particular,
$X(\Bbbk)\neq\varnothing$. Thus for any pointless del Pezzo surface $X$ of
degree $8$ the surface $\overline{X}$ is isomorphic to a quadric in
$\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$.

If $X$ is a pointless del Pezzo surface of degree $8$ then
$\operatorname{Pic}\left(\overline{X}\right)=\operatorname{Pic}\left(\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}\right)\cong\mathbb{Z}^{2}$.
Therefore the Picard number $\rho(X)=\operatorname{rk}\operatorname{Pic}(X)$
is equal to $1$ or $2$, since
$\operatorname{Pic}\left(X\right)\subset\operatorname{Pic}\left(\overline{X}\right)$.

To establish more results about birational classification of pointless del
Pezzo surfaces of degree $8$ we need the following definition.

######  Definition 1.3.

A smooth surface $S$ is called minimal if any birational morphism of smooth
surfaces $S\rightarrow S^{\prime}$ is an isomorphism.

The classification of minimal geometrically rational surfaces is well-known.

######  Theorem 1.4 (cf. [Isk79, Theorems 1, 4, 5]).

Let $S$ be a minimal geometrically rational surface. Then either $S$ admits a
conic bundle structure over a conic with $\rho(S)=2$, or $S$ is a del Pezzo
surface with $\rho(S)=1$. In the former case $K_{S}^{2}\notin\\{3,5,6,7\\}$,
and for $K_{S}^{2}=8$ the surface $S$ is not isomorphic to the blowup of
$\mathbb{P}^{2}_{\Bbbk}$ at a point.

Note that any pointless del Pezzo surface $X$ of degree $8$ is minimal. If
$\rho(X)=1$ then it is obvious, and if $\rho(X)=2$ then $X$ is isomorphic to
$C_{1}\times C_{2}$, where $C_{1}$ and $C_{2}$ are smooth conics, and the
projections $X\rightarrow C_{1}$ and $X\rightarrow C_{2}$ define two
structures of conic bundles on $X$.

We want to study which surfaces are birationally equivalent to pointless del
Pezzo surfaces of degree $8$. Obviously, for any variety one can equivariantly
blow up a collection of points defined over $\Bbbk$ and get a birationally
equivalent variety. Therefore for a given pointless del Pezzo surface $X$ of
degree $8$ the natural question is to find minimal surfaces birationally
equivalent to $X$.

For the case $\rho(X)=2$ the answer seems to be well-known for specialists.
For example, many results for this case can be found in [Kol05]. We give the
following theorem for the completeness.

######  Theorem 1.5.

Let $\Bbbk$ be an algebraically nonclosed field of characteristic $0$, and let
$X$ be a pointless del Pezzo surface of degree $8$ such that $\rho(X)=2$. Then
$X$ is a product of two smooth conics and there are two following
possibilities.

  1. (1)

The surface $X$ is isomorphic to $C_{1}\times C_{2}$, where $C_{1}$ and
$C_{2}$ are two non-trivial smooth conics not isomorphic to each other. If the
Brauer product $C_{3}=C_{1}*C_{2}$ (for the definition see Definition 3.1 or
[Kol05, 7]) is defined, then any minimal surface birationally equivalent to
$X$ is isomorphic to $C_{1}\times C_{2}$, $C_{1}\times C_{3}$ or $C_{2}\times
C_{3}$. Otherwise, any minimal surface birationally equivalent to $X$ is
isomorphic to $X$.

  2. (2)

The surface $X$ is isomorphic to $C\times C$ or
$C\times\mathbb{P}^{1}_{\Bbbk}$, where $C$ is a non-trivial smooth conic. Then
any minimal surface birationally equivalent to $X$ is isomorphic to $C\times
C$, $C\times\mathbb{P}^{1}_{\Bbbk}$ or to a $\Bbbk$-form of Hirzebruch surface
$\mathbb{F}_{2k}$ admitting a conic bundle structure over $C$. Such
$\Bbbk$-form is unique up to isomorphism for a given $k$.

For the case $\rho(X)=1$ we prove the following theorem.

######  Theorem 1.6.

Let $\Bbbk$ be an algebraically nonclosed field of characteristic $0$, and let
$X$ be a pointless del Pezzo surface of degree $8$ such that $\rho(X)=1$. Then
any minimal surface birationally equivalent to $X$ is isomorphic to $X$.

Note that a minimal surface $X$ is called birationally rigid, if for any
birational map $X\dashrightarrow X^{\prime}$ to a minimal surface $X^{\prime}$
one has $X^{\prime}\cong X$ (for the precise definition of birational rigidity
in general case and some properties see e.g. [Ch05]). But in this definition
relatively minimal conic bundles (i. e. the conic bundles with the Picard
number $2$) are usually considered as minimal surfaces. Actually in Section
$4$ we show that a pointless del Pezzo surface $X$ of degree $8$ with
$\rho(X)=1$ is birationally equivalent to (non-minimal) del Pezzo surfaces of
degree $6$ admitting a structure of relatively minimal conic bundle. For a
given surface $X$ these surfaces are parametrised by quadratic extensions
$F/\Bbbk$ such that $X_{F}(F)\neq\varnothing$. Therefore $X$ is not
birationally rigid in terms of [Ch05].

Actually for any pointless del Pezzo surface $X$ of degree $8$ we describe all
possible minimal del Pezzo surfaces and relatively minimal conic bundles
birationally equivalent to $X$. Note that if $X$ is a minimal del Pezzo
surface $X$ with $\rho(X)=1$ of degree $1$, $2$ or $3$, or a pointless del
Pezzo surface of degree $4$, then $X$ is birationally rigid (see [Isk96,
Theorems 4.4 and 4.5]). On the other hand by [Isk96, Chapter 4] if $X$ is a
del Pezzo surface of degree $5$ or greater with $\rho(X)=1$ and
$X(\Bbbk)\neq\varnothing$ or $X\rightarrow\mathbb{P}^{1}_{\Bbbk}$ is a conic
bundle with $\rho(X)=2$, $X(\Bbbk)\neq\varnothing$ and $K_{X}^{2}\geqslant 5$
then $X$ is $\Bbbk$-rational. In particular, all these surfaces are
birationally equivalent to each other.

A non-trivial Severi–Brauer surface $X$ (i.e. a pointless del Pezzo surface of
degree $9$) is not birationally equivalent to any conic bundle, and if a
minimal del Pezzo surface $X^{\prime}$ is birationally equivalent to $X$ then
either $X^{\prime}\cong X$, or $X^{\prime}\cong X^{\mathrm{op}}$, where
$X^{\mathrm{op}}$ is the Severi–Brauer surface such that the central simple
algebras corresponding to $X$ and $X^{\mathrm{op}}$ are opposite (see [Sh20,
Corollary 2.4 and Theorem 2.10]).

Note that a geometrically rational surface $X$ with $K_{X}^{2}=7$ is never
minimal by [Isk79, Theorem 4], and a del Pezzo surface of degree $5$ always
has a $\Bbbk$-point (see [SD72]).

An interesting question is to describe all minimal surfaces birationally
equivalent to a given del Pezzo surface $X$ with $\rho(X)=1$, or a conic
bundle $X\rightarrow C$ with $\rho(X)=2$. The remaining cases are pointless
del Pezzo surfaces of degree $6$ and del Pezzo surfaces of degree $4$ with
$\Bbbk$-points, and conic bundles with at least $4$ degenerate fibres, since
the cases of pointless del Pezzo surfaces of degree $8$ and pointless conic
bundles with $0$ or $2$ degenerate fibres are considered in this paper.

The plan of this paper is as follows. In Section $2$ we recall some notions
and properties of pointless del Pezzo surfaces of degree $8$ and give a
defininion of Amitsur subgroup
$\operatorname{Am}(X)\subset\operatorname{Br}\left(\Bbbk\right)$ (see [Lied17,
Definition 2.8]), that is a birational invariant. Also we give a description
of all geometrically rational surfaces with non-trivial
$\operatorname{Am}(X)$, and show how one can restore a del Pezzo surface $X$
of degree $8$ with $\rho(X)=1$ by $\operatorname{Am}(X_{\mathbb{L}})$, where
$\mathbb{L}$ is a splitting field of $X$: unique quadratic extension of
$\Bbbk$ such that $\rho\left(X_{\mathbb{L}}\right)=2$.

In Section $3$ we consider pointless del Pezzo surfaces of degree $8$ with the
Picard number $2$. Any such surface $X$ is isomorphic to a product of two
conics. We consider Sarkisov links for these surfaces, describe possibilities
for minimal surfaces birationally equivalent to $X$, and show that any other
surface is not birationally equivalent to $X$. As a result of this section we
prove Theorem 1.5.

In Section $4$ we consider pointless del Pezzo surfaces of degree $8$ with the
Picard number $1$. We show that for any such surface $X$ any Sarkisov link or
sequence of Sarkisov links leads to an isomorphic surface or a certain non-
minimal del Pezzo surface of degree $6$ obtaining a relatively minimal conic
bundle structure. Finally, we prove Theorems 1.6 and 1.1.

Acknowledgements. The author is grateful to Sergey Gorchinskiy for advice to
study the group $\operatorname{Am}(X)$. Also the author is grateful to Costya
Shramov for numerous discussions during the preparation and writing of this
paper.

######  Notation 1.7.

Throughout this paper $\Bbbk$ is any field of characteristic zero,
$\overline{\Bbbk}$ is its algebraic closure, and the Galois group
$\operatorname{Gal}\left(\overline{\Bbbk}/\Bbbk\right)$ is denoted by
$G_{\Bbbk}$. For a surface $X$ we denote $X\otimes\overline{\Bbbk}$ by
$\overline{X}$. For a surface $X$ we denote the Picard group by
$\operatorname{Pic}(X)$. The number
$\rho(X)=\operatorname{rk}\operatorname{Pic}(X)$ is the Picard number of $X$.
If two surfaces $X$ and $Y$ are birationally equivalent then we write
$X\approx Y$. If two divisors $A$ and $B$ are linearly equivalent then we
write $A\sim B$. The rational ruled (Hirzebruch) surface
$\mathbb{P}_{\mathbb{P}^{1}_{\overline{\Bbbk}}}\left(\mathcal{O}\oplus\mathcal{O}(n)\right)$
is denoted by $\mathbb{F}_{n}$. For a given quadratic extension
$\mathbb{L}/\Bbbk$ and a conic $C$ over $\mathbb{L}$ we denote by
$R_{\mathbb{L}/\Bbbk}C$ its Weil restriction of scalars.

##  2\. Preliminaries

In this section we review some results about biregular classification of del
Pezzo surfaces of degree $8$, and give an alternative description of this
classification in terms of the Brauer group.

Throughout this section we assume that $X$ is a del Pezzo surface of degree
$8$ over algebraically closed field $\Bbbk$ of characteristic $0$, such that
$\overline{X}\cong\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$.
The following lemma gives biregular classification of such surfaces.

######  Lemma 2.1 (cf. [ShV18, Lemma 3.4(i and ii)]).

Let $\Bbbk$ be a field of characteristic $0$, and let $X$ be a del Pezzo
surface of degree $8$, such that
$\overline{X}\cong\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$.
The following assertion hold.

  1. (i)

The surface $X$ is isomorphic to the product $C_{1}\times C_{2}$ of two conics
over $\Bbbk$, or to $R_{\mathbb{L}/\Bbbk}C$, where $\mathbb{L}$ is a quadratic
extension of $\Bbbk$ and $C$ is a conic over $\mathbb{L}$. In the former case
one has $\rho(X)=2$, while in the latter $\rho(X)=1$. Furthemore, in the
former case the (non-ordered) pair of conics $\\{C_{1},C_{2}\\}$ is uniquely
determined by $X$; in the latter case the extension $\mathbb{L}/\Bbbk$ and the
conic $C$ are uniquely determined by $X$ up to conjugation by the Galois group
$\operatorname{Gal}\left(\mathbb{L}/\Bbbk\right)$.

  2. (ii)

The surface $X$ is isomorphic to a quadric in $\mathbb{P}^{3}_{\Bbbk}$ if and
only if $X\cong C\times C$ or $X\cong R_{\mathbb{L}/\Bbbk}N_{\mathbb{L}}$, for
some conic $N$ over $\Bbbk$.

######  Corollary 2.2.

Let $\Bbbk$ be a field of characteristic $0$, and let $X$ be a del Pezzo
surface of degree $8$ with $\rho(X)=1$. In this case $X\cong
R_{\mathbb{L}/\Bbbk}C$. Let $C^{\prime}$ be a
$\operatorname{Gal}\left(\mathbb{L}/\Bbbk\right)$-conjugate conic of $C$. Then
$X_{\mathbb{L}}\cong C\times C^{\prime}$. In particular, $X$ is pointless if
and only if $X_{\mathbb{L}}$ is pointless.

For the both cases $\rho(X)=2$ and $\rho(X)=1$ we want to give an
interpretation of Lemma 2.1 in terms of the Brauer group
$\operatorname{Br}\left(\Bbbk\right)$ (for the definition of the Brauer group
and properties see, for example [GS18, Chapter 3]). Let us recall some facts
about this group.

######  Proposition 2.3 (cf. [GS18, Exercise 3.3.5]).

Let $V$ be a smooth projective geometrically irreducible variety over $\Bbbk$.
Then there exists an exact sequence

$0\rightarrow\operatorname{Pic}(V)\rightarrow\operatorname{Pic}(\overline{V})^{G_{\Bbbk}}\rightarrow\operatorname{Br}\left(\Bbbk\right)\rightarrow\operatorname{Br}\left(\Bbbk(V)\right).$

According to [Lied17, Definition 2.8] we denote the group
$\operatorname{Pic}(\overline{V})^{G_{\Bbbk}}/\operatorname{Pic}(V)\subset\operatorname{Br}\left(\Bbbk\right)$
by $\operatorname{Am}(V)$. The group $\operatorname{Am}(V)$ is called Amitsur
subgroup of $V$ in $\operatorname{Br}\left(\Bbbk\right)$. Note that
$\operatorname{Am}(V)$ is a birational invariant since it is the kernel of the
map
$\operatorname{Br}\left(\Bbbk\right)\rightarrow\operatorname{Br}\left(\Bbbk(V)\right)$,
and $\Bbbk(V)$ is a birational invariant (see [Lied17, Proposition 2.10]).

We want to describe minimal geometrically rational surfaces $S$ with non-
trivial $\operatorname{Am}(S)$. These surfaces are described in [CTKM08,
Proposition 5.3], but we obtain a more detailed description. By Theorem 1.4
the surface $S$ is either a del Pezzo surface with $\rho(S)=1$, or admits a
conic bundle structure over a smooth conic with $\rho(S)=2$.

Let us start from the case of conic bundle. We say that a conic is non-trivial
if it is not isomorphic to $\mathbb{P}^{1}_{\Bbbk}$. Note that for each conic
$C$ we can define a class $b(C)$, that is an element in $2$-torsion subgroup
of the Brauer group $\operatorname{Br}\left(\Bbbk\right)$ (see, for example,
[GS18, Section 3.3]). If two conics are not isomorphic then they have
different classes, and $b(C)$ is trivial if and only if $C$ is trivial.
Moreover, for a conic $C$ the class $b(C)$ is a generator of
$\operatorname{Am}(C)$.

If for a smooth projective geometrically irreducible variety $V$ there exists
a map $V\dashrightarrow C$, where $C$ is a smooth conic, then this map induces
embeddings $\operatorname{Pic}(C)\hookrightarrow\operatorname{Pic}(V)$ and
$\operatorname{Pic}(\overline{C})^{G_{\Bbbk}}\hookrightarrow\operatorname{Pic}(\overline{V})^{G_{\Bbbk}}$,
since we can consider the preimage of a general geometric point on $C$.
Therefore $\operatorname{Am}(C)\subset\operatorname{Am}(V)$. In particular, if
$C$ is non-trivial then $\operatorname{Am}(V)$ is non-trivial, since it
contains $b(C)$.

Now we can describe conic bundles $S\rightarrow C$ with non-trivial
$\operatorname{Am}(S)$.

######  Proposition 2.4.

Let $S$ be a surface admitting a conic bundle structure $S\rightarrow C$ over
a smooth conic $C$ over algebraically nonclosed field $\Bbbk$, such that
$\rho(S)=2$ and $\operatorname{Am}(S)$ is non-trivial. Then there are five
possibilities.

  1. (1)

The surface $S$ is isomorphic to $C_{1}\times C_{2}$, where $C_{1}$ and
$C_{2}$ are two non-trivial smooth conics not isomorphic to each other. The
group $\operatorname{Am}(S)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$ is
generated by $b(C_{1})$ and $b(C_{2})$.

  2. (2)

The surface $S$ is isomorphic to $C\times C$, where $C$ is a non-trivial
smooth conic. The group $\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ is
generated by $b(C)$.

  3. (3)

The surface $S$ is isomorphic to $C\times\mathbb{P}^{1}_{\Bbbk}$, where $C$ is
a non-trivial smooth conic. The group
$\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ is generated by $b(C)$.

  4. (4)

The surface $S$ is a $\Bbbk$-form of Hirzebruch surface $\mathbb{F}_{2k}$
admitting a conic bundle structure over a smooth non-trivial conic $C$, where
$k$ is a positive integer. The group
$\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ is generated by $b(C)$.

  5. (5)

The surface $S$ admits a conic bundle structure over a smooth non-trivial
conic $C$ with $2k$ degenerate geometric fibres, where $k$ is a positive
integer. The group $\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ is
generated by $b(C)$.

###### Proof.

Note that $\rho\left(\overline{S}\right)^{G_{\Bbbk}}=\rho\left(S\right)=2$.
Therefore $\operatorname{Pic}\left(S\right)$ is a sublattice of
$\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}$ of finite index.

We start from the case, when the conic bundle $S\rightarrow C$ has a
degenerate geometric fibre. Let us show that
$\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}$ is generated by
$-K_{\overline{S}}$ and the class $F$ of a geometric fibre. Otherwise there
exists a class $D$ in
$\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}$, such that $mD\sim-
K_{\overline{S}}+nF$. Let $E$ be an irreducible component of a geometric
degenerate fibre. Then $E$ is a $(-1)$-curve, and one has
$-K_{\overline{S}}\cdot E=1$ and $F\cdot E=0$. Therefore

$m(D\cdot E)=mD\cdot E=\left(-K_{\overline{S}}+nF\right)\cdot E=1.$

It is possible if and only if $m=\pm 1$, but in this case $D$ lies in the
lattice generated by $-K_{\overline{S}}$ and $F$.

Note that $\operatorname{Pic}\left(S\right)$ contains $-K_{S}$ and $2F$, since
there is a point of degree $2$ on $C$, and a fibre over this point has class
$2F$. Therefore $\operatorname{Am}(S)$ is non-trivial if and only if
$\operatorname{Pic}\left(S\right)$ does not contain $F$, and for this case
$\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ is generated by $b(C)$. This
is possible if and only if $C$ is a non-trivial conic. Also note that if
$S\rightarrow C$ has degenerate fibres over odd number of geometric points,
then $C\cong\mathbb{P}^{1}_{\Bbbk}$, since there is a point of odd degree on
$C$.

Now assume that the conic bundle $S\rightarrow C$ does not have degenerate
geometric fibres. Then either
$\overline{S}\cong\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$,
or $S$ is a $\Bbbk$-form of a Hirzebruch surface $\mathbb{F}_{m}$, where
$m\geqslant 1$. In the latter case there is a unique section $H$ of
$S\rightarrow C$ such that $H^{2}=-m$. The group
$\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}=\operatorname{Pic}\left(\overline{S}\right)$
is generated by the class of $H$ and a class $F$ of a geometric fibre. The
group $\operatorname{Pic}\left(S\right)$ contains $2F$ and the class of $H$,
since $H$ is unique and therefore defined over $\Bbbk$. Therefore
$\operatorname{Am}(S)$ is non-trivial if and only if
$\operatorname{Pic}\left(S\right)$ does not contain $F$, and for this case one
has $\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$. This is possible if and
only if $C$ is a non-trivial conic. Also note that $-K_{S}\sim 2H+(m+2)F$,
therefore $mF\in\operatorname{Pic}\left(S\right)$, and if
$\operatorname{Am}(S)$ is non-trivial then $m$ is even.

Now assume that $S\cong C_{1}\times C_{2}$ and
$\overline{S}\cong\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$.
Note that
$\operatorname{Pic}\left(\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}\right)$
is generated by the classes $A$ and $B$ of fibres of the projections on the
first and the second factors of
$\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$. If
$\rho(S)=2$ then $\operatorname{Pic}(S)$ contains $2A$ and $2B$. Therefore if
$\operatorname{Am}(S)$ is non-trivial then $\operatorname{Pic}(S)$ is $\langle
2A,2B\rangle$, $\langle 2A,A+B\rangle$, $\langle 2A,B\rangle$ or $\langle
A,2B\rangle$.

If $\operatorname{Pic}(S)$ contains $A$ or $B$ then $C_{1}$ or $C_{2}$
respectively is isomorphic to $\mathbb{P}^{1}_{\Bbbk}$, and $S\cong
C\times\mathbb{P}^{1}_{\Bbbk}$, where $C$ is a non-trivial smooth conic. For
this case $\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ is generated by
$b(C)$.

Let us show that $\operatorname{Pic}(S)$ contains $A+B$ if and only if $S$ is
isomorphic to a quadric in $\mathbb{P}^{3}_{\Bbbk}$. If $A+B$ lies in
$\operatorname{Pic}(S)$, then the linear system $|A+B|$ defines an embedding
$S\hookrightarrow\mathbb{P}^{3}_{\Bbbk}$ and the image is a quadric surface.
Conversely, if $S$ is isomorphic to a quadric in $\mathbb{P}^{3}_{\Bbbk}$ then
any hyperplane section of this quadric has class $A+B$ in
$\operatorname{Pic}(S)$. Therefore for this case we can apply Lemma 2.1 and
get that $S$ is isomorphic to $C\times C$, where $C$ is a smooth conic over
$\Bbbk$. If $\operatorname{Am}(S)$ is non-trivial then $C$ is non-trivial, and
for this case $\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ is generated
by $b(C)$.

If $\operatorname{Pic}(S)$ does not contain $A$, $B$ and $A+B$, then $C_{1}$
and $C_{2}$ are non-trivial and not isomorphic to each other. For this case
$\operatorname{Am}(S)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$ is
generated by $b(C_{1})$ and $b(C_{2})$. ∎

The surfaces considered in cases $(1)$–$(4)$ of Proposition 2.4 widely appear
in this paper. In the following example we show how to construct a surface
considered in case $(5)$. For this aim we slightly modify the constuction of
an exceptional conic bundle (see [DI09, Subsection 5.2]).

######  Example 2.5.

Let $C$ be a non-trivial conic, $p_{1}$, …$p_{2k}$ be a collection of $2k$
geometric points defined over $\Bbbk$, and $H\rightarrow C$ be a double cover
of $C$ branched into this collection of points. Assume that $\mathbb{F}$ is a
minimal field, such that all points $p_{i}$ are defined over $\mathbb{F}$.
Consider two conjugate geometric points $q_{1}$ and $q_{2}$ on
$\mathbb{P}^{1}_{\Bbbk}$ defined over a quadratic extension $\mathbb{L}/\Bbbk$
and not defined over $\mathbb{F}$.

Let $\iota_{1}$ be the involution of the double cover $H\rightarrow C$, and
$\iota_{2}$ be an involution acting on $\mathbb{P}^{1}_{\Bbbk}$ such that
$\iota_{2}\left(q_{i}\right)=q_{i}$. Consider an involution $\iota$ on
$H\times\mathbb{P}^{1}_{\Bbbk}$ that acts on $H$ and $\mathbb{P}^{1}_{\Bbbk}$
as $\iota_{1}$ and $\iota_{2}$ respectively. The quotient
$Y=\left(H\times\mathbb{P}^{1}_{\Bbbk}\right)/\langle\iota\rangle$ admits a
structure of a bundle $Y\rightarrow C\cong H/\langle\iota_{1}\rangle$ such
that its general fibre is a smooth conic. The images of fixed points of
$\iota$ are $A_{1}$-singularities. Therefore there are $4k$ singular points on
$Y$ lying in $2k$ fibres of $Y\rightarrow C$ over the points $p_{i}$.

Let $\widetilde{Y}\rightarrow Y$ be the minimal resolution of singularities,
and let $\widetilde{Y}\rightarrow S$ be the contraction of the proper
transforms of the $2k$ fibres of $Y\rightarrow C$ over the points $p_{i}$. We
obtain a conic bundle $S\rightarrow C$ with $2k$ degenerate fibres over the
points $p_{i}$. The components of these fibres are permuted by the group
$\operatorname{Gal}\left(\mathbb{L}/\Bbbk\right)$, therefore $S$ is minimal.
The group $\operatorname{Am}(S)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$
since it is generated by $b(C)$.

The group $\operatorname{Am}(S)$ is a birational invariant, therefore it is
obvious that any surface described in case $(1)$ of Proposition 2.4 cannot be
birationally equivalent to any of surfaces listed in cases $(2)$–$(5)$. In
Section $3$ for the surfaces listed in cases $(2)$–$(4)$ we show that two of
them $S_{1}$ and $S_{2}$ are birationally equivalent to each other if and only
if $\operatorname{Am}(S_{1})=\operatorname{Am}(S_{2})$. Surfaces described in
case $(5)$ are not birationally equivalent to surfaces from the other cases.

Note that for a given smooth conic $C$ and even positive integer $2k$ the
surface $S$ from case $(4)$ is unique, since it is isomorphic to
$\mathbb{P}_{C}\left(\mathcal{O}_{C}\oplus\mathcal{O}_{C}(-2k)\right)$.

Now we describe minimal del Pezzo surfaces $S$ with $\rho(S)=1$ and non-
trivial $\operatorname{Am}(S)$.

######  Proposition 2.6.

Let $S$ be a del Pezzo surface over algebraically nonclosed field $\Bbbk$,
such that $\rho(S)=1$ and $\operatorname{Am}(S)$ is non-trivial. Then there
are two possibilities.

  1. (1)

The surface $S$ is a non-trivial Severi–Brauer surface, ${K_{S}^{2}=9}$, and
$\operatorname{Am}(S)\cong\mathbb{Z}/3\mathbb{Z}$.

  2. (2)

The surface $S$ is a pointless del Pezzo surface of degree $8$ isomorphic to
$R_{\mathbb{L}/\Bbbk}C$, where $\mathbb{L}$ is a quadratic extension of
$\Bbbk$ and $C$ is a conic over $\mathbb{L}$, such that for any conic $N$ over
$\Bbbk$ the conics $C$ and $N_{\mathbb{L}}$ are not isomorphic. One has
$\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$.

###### Proof.

Note that $\rho\left(\overline{S}\right)^{G_{\Bbbk}}=\rho\left(S\right)=1$.
Therefore $\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}$ is
generated by a class $D$ such that for a certain number $k$ one has
$kD=-K_{S}$, since $-K_{S}$ obviously lies in
$\operatorname{Pic}\left(S\right)\subset\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}$.

Assume that there is a $(-1)$-curve $E$ on $\overline{S}$, then one has

$1=-K_{\overline{S}}\cdot E=kD\cdot E=k(D\cdot E).$

Therefore $k=1$, since $D\cdot E$ is integer. Thus in this case
$\operatorname{Pic}\left(S\right)=\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}$,
and $\operatorname{Am}(S)$ is trivial.

If there are no $(-1)$-curves on $\overline{S}$ then $\overline{S}$ is either
$\mathbb{P}^{2}_{\overline{\Bbbk}}$, or
$\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$. In
the former case $S$ is a del Pezzo surface of degree $9$, i.e. Severi–Brauer
surface. A trivial Severi–Brauer surface is just $\mathbb{P}^{2}_{\Bbbk}$, and
$\operatorname{Am}\left(\mathbb{P}^{2}_{\Bbbk}\right)$ is trivial. For a non-
trivial Severi–Brauer surface $S$ it is well-known that
$\operatorname{Pic}(S)$ is generated by $-K_{S}$, and
$\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}=\operatorname{Pic}\left(\mathbb{P}^{2}_{\overline{\Bbbk}}\right)^{G_{\Bbbk}}=\operatorname{Pic}\left(\mathbb{P}^{2}_{\overline{\Bbbk}}\right)$
is generated by the class of a line on $\mathbb{P}^{2}_{\overline{\Bbbk}}$.
Therefore $\operatorname{Am}(S)\cong\mathbb{Z}/3\mathbb{Z}$.

If
$\overline{S}\cong\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$
then $K_{S}^{2}=8$. As in the case $\rho(S)=2$, the group
$\operatorname{Pic}\left(\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}\right)$
is generated by the classes $A$ and $B$ of fibres of the projections on the
first and the second factors of
$\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$,
and $\operatorname{Pic}(S)$ contains $-K_{S}\sim 2A+2B$. If $\rho(S)=1$, then
the Galois group $G_{\Bbbk}$ permutes $A$ and $B$, and
$\operatorname{Pic}\left(\overline{S}\right)^{G_{\Bbbk}}$ is generated by the
class $A+B$. Therefore if $\operatorname{Am}(S)$ is not trivial, then
$\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ and $\operatorname{Pic}(S)$
does not contain $A+B$.

As in the proof of Proposition 2.4 one can easily show that
$\operatorname{Pic}(S)$ contains $A+B$ if and only if $S$ is isomorphic to a
quadric in $\mathbb{P}^{3}_{\Bbbk}$. Therefore for this case we can apply
Lemma 2.1 and get that $S$ is isomorphic to
$R_{\mathbb{L}/\Bbbk}N_{\mathbb{L}}$, where $N$ is a conic over $\Bbbk$. Thus
for the case $\operatorname{Am}(S)\cong\mathbb{Z}/2\mathbb{Z}$ the surface $S$
is not isomorphic to a quadric in $\mathbb{P}^{3}_{\Bbbk}$. By Lemma 2.1 the
surface $S$ is isomorphic to $R_{\mathbb{L}/\Bbbk}C$, where $\mathbb{L}$ is a
quadratic extension of $\Bbbk$ and $C$ is a conic over $\mathbb{L}$, such that
for any conic $N$ over $\Bbbk$ the conics $C$ and $N_{\mathbb{L}}$ are not
isomorphic. ∎

######  Remark 2.7.

It is well known, that two non-trivial Severi–Brauer surfaces $S$ and
$S^{\prime}$ are birationally equivalent if and only if $S\cong S^{\prime}$ or
$S^{\prime}\cong S^{\mathrm{op}}$, where $S^{\mathrm{op}}$ is the
Severi–Brauer surface such that the central simple algebras corresponding to
$S$ and $S^{\mathrm{op}}$ are opposite (see [Ami55]). Any other minimal
surface is not birationally equivalent to $S$ and $S^{\prime}$ by [Sh20,
Corollary 2.4 and Theorem 2.10]. Applying Propositions 2.4 and 2.6 we can
obtain an alternative proof of this fact.

By Proposition 2.6 a non-trivial Severi–Brauer surface $S$ has
$\operatorname{Am}(S)\cong\mathbb{Z}/3\mathbb{Z}$ generated by the class
$b(S)$ of $S$. If a surface $S^{\prime}$ is birationally equivalent to $S$
then
$\operatorname{Am}(S^{\prime})=\operatorname{Am}(S)\cong\mathbb{Z}/3\mathbb{Z}$.
Therefore $S$ is not birationally equivalent to any relatively minimal conic
bundle by Proposition 2.4. If $S^{\prime}$ is a minimal del Pezzo surface then
$S^{\prime}$ is a Severi–Brauer surface by Proposition 2.6, and
$b(S^{\prime})$ generates
$\operatorname{Am}(S^{\prime})=\operatorname{Am}(S)\cong\mathbb{Z}/3\mathbb{Z}$.
There are two possibilities: either $b(S^{\prime})=b(S)$ and $S^{\prime}\cong
S$, or $b(S^{\prime})=-b(S)$ and $S^{\prime}\cong S^{\mathrm{op}}$.

Note that non-isomorphic del Pezzo surfaces of degree $8$ with $\rho(X)=1$ can
have the same group $\operatorname{Am}(X)$. For example, if $X$ is a pointless
quadric in $\mathbb{P}^{3}_{\Bbbk}$ then $\operatorname{Am}(X)$ is trivial.
Therefore we want to consider more invariants for such surfaces.

For a given del Pezzo surface $X$ of degree $8$ with $\rho(X)=1$ there exists
a quadratic extension $\mathbb{L}/\Bbbk$, such that $X_{\mathbb{L}}\cong
C_{1}\times C_{2}$, where $C_{1}$ and $C_{2}$ are smooth conics over
$\mathbb{L}$. The field $\mathbb{L}$ is uniquely determinated by $X$, since
$\mathbb{L}=\overline{\Bbbk}^{K}$, where $K$ is the kernel of the action of
$\operatorname{Gal}\left(\overline{\Bbbk}/\Bbbk\right)$ on
$\operatorname{Pic}(\overline{X})$. For simplicity of notation we will say
that $\mathbb{L}$ is the splitting field of $X$.

We can consider the group
$\operatorname{Am}\left(X_{\mathbb{L}}\right)\subset\operatorname{Br}\left(\mathbb{L}\right)$.
This group is obviously a birational invariant. By Lemma 2.1 one has $X\cong
R_{\mathbb{L}/\Bbbk}C$, where $C$ is a smooth conic over $\mathbb{L}$. Let
$C^{\prime}$ be the conjugate conic. Then by Corollary 2.2 one has
$X_{\mathbb{L}}\cong C\times C^{\prime}$.

Applying Proposition 2.4 we can see that there are three possibilities for
$\operatorname{Am}(X_{\mathbb{L}})$:

  * •

$\operatorname{Am}(X_{\mathbb{L}})$ is trivial, if and only if
$C\cong\mathbb{P}^{1}_{\mathbb{L}}$;

  * •

$\operatorname{Am}(X_{\mathbb{L}})\cong\mathbb{Z}/2\mathbb{Z}$, if and only if
$C\cong C^{\prime}$ and $C$ is non-trivial;

  * •

$\operatorname{Am}(X_{\mathbb{L}})\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$,
if and only if $C$ and $C^{\prime}$ are not isomorphic.

Therefore for a given del Pezzo surface $X$ of degree $8$ with $\rho(X)=1$ we
can consider a pair of invariants
$\left(\mathbb{L},\operatorname{Am}(X_{\mathbb{L}})\right)$. We show that this
pair is a biregular invariant.

######  Theorem 2.8.

Let $X$ and $X^{\prime}$ be two del Pezzo surfaces of degree $8$ with
$\rho(X)=\rho(X^{\prime})=1$. Then $X$ and $X^{\prime}$ have the same
splitting field $\mathbb{L}$ and
$\operatorname{Am}(X_{\mathbb{L}})=\operatorname{Am}(X^{\prime}_{\mathbb{L}})$
if and only if $X\cong X^{\prime}$.

###### Proof.

If $X\cong X^{\prime}$ then they obviously have the same splitting field
$\mathbb{L}$ and
$\operatorname{Am}(X_{\mathbb{L}})=\operatorname{Am}(X^{\prime}_{\mathbb{L}})$.

Now assume that $X$ and $X^{\prime}$ have the same splitting field
$\mathbb{L}$ and
$\operatorname{Am}(X_{\mathbb{L}})=\operatorname{Am}(X^{\prime}_{\mathbb{L}})$
and show that $X\cong X^{\prime}$.

Note that $X\cong R_{\mathbb{L}/\Bbbk}C$, where
$b(C)\in\operatorname{Am}(X_{\mathbb{L}})$. For a
$\operatorname{Gal}\left(\mathbb{L}/\Bbbk\right)$-conjugate conic $C^{\prime}$
one has $R_{\mathbb{L}/\Bbbk}C\cong R_{\mathbb{L}/\Bbbk}C^{\prime}$ by Lemma
2.1. Moreover, $X_{\mathbb{L}}\cong C\times C^{\prime}$ by Corollary 2.2.

If $\operatorname{Am}(X_{\mathbb{L}})$ is trivial then $C\cong
C^{\prime}\cong\mathbb{P}^{1}_{\Bbbk}$. If
$\operatorname{Am}(X_{\mathbb{L}})\cong\mathbb{Z}/2\mathbb{Z}$ then there is a
unique non-trivial element in $\operatorname{Am}(X_{\mathbb{L}})$
corresponding to $C\cong C^{\prime}$.

If
$\operatorname{Am}(X_{\mathbb{L}})\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$
then there are three non-trivial elements in
$\operatorname{Am}(X_{\mathbb{L}})$. Two of these elements are
$\operatorname{Gal}\left(\mathbb{L}/\Bbbk\right)$-conjugate since in this case
$X_{\mathbb{L}}\cong C\times C^{\prime}$, where $C$ and $C^{\prime}$ are not
isomorphic. Therefore even if the third non-trivial element in
$\operatorname{Am}(X_{\mathbb{L}})$ corresponds to a conic $\widetilde{C}$
then the $\operatorname{Gal}\left(\mathbb{L}/\Bbbk\right)$-conjugate conic is
isomorphic to $\widetilde{C}$. Then for
$\widetilde{X}=R_{\mathbb{L}/\Bbbk}\widetilde{C}$ one has
$\widetilde{X}_{\mathbb{L}}\cong\widetilde{C}\times\widetilde{C}$, and
$\operatorname{Am}(\widetilde{X}_{\mathbb{L}})\cong\mathbb{Z}/2\mathbb{Z}$.
Thus if
$\operatorname{Am}(X_{\mathbb{L}})\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$
then we can definitely restore the pair of conjugate conics $C$ and
$C^{\prime}$ such that $X_{\mathbb{L}}\cong C\times C^{\prime}$ and $X\cong
R_{\mathbb{L}/\Bbbk}C\cong R_{\mathbb{L}/\Bbbk}C^{\prime}$.

Therefore in any case $X^{\prime}\cong R_{\mathbb{L}/\Bbbk}C\cong X$. ∎

In Section $4$ we show that the pair
$\left(\mathbb{L},\operatorname{Am}(X_{\mathbb{L}})\right)$ is also a
birational invariant.

######  Remark 2.9.

Actually, the consideration of such pairs was inspired by [ShV18, Lemma 3.4],
that originally was obtained to describe the automorphism groups of pointless
del Pezzo surfaces of degree $8$. For the case $\rho(X)=1$ we show connection
between the groups $\operatorname{Am}(X)$ and
$\operatorname{Am}\left(X_{\mathbb{L}}\right)$ and the structure of the group
$\operatorname{Aut}(X)$ in the following table.

$\operatorname{Am}(X)$ | $\operatorname{Am}(X_{\mathbb{L}})$ | Quadric | $C$ | $\operatorname{Aut}(X)$
---|---|---|---|---
$\langle\mathrm{id}\rangle\rule[-5.0pt]{0.0pt}{18.0pt}$ | $\langle\mathrm{id}\rangle$ | yes | $C\cong\mathbb{P}^{1}_{\mathbb{L}}$ | $\operatorname{PGL}_{2}(\mathbb{L})\rtimes\left(\mathbb{Z}/2\mathbb{Z}\right)$
$\langle\mathrm{id}\rangle\rule[-5.0pt]{0.0pt}{18.0pt}$ | $\mathbb{Z}/2\mathbb{Z}$ | yes |  $C\cong N_{\mathbb{L}}$, where $N$ is a conic over $\Bbbk$ | $\operatorname{Aut}\left(C\right)\rtimes\left(\mathbb{Z}/2\mathbb{Z}\right)$
$\mathbb{Z}/2\mathbb{Z}\rule[-5.0pt]{0.0pt}{18.0pt}$ | $\mathbb{Z}/2\mathbb{Z}$ | no |  $C\cong C^{\prime}$ is not isomorphic to $N_{\mathbb{L}}$ | $\operatorname{Aut}\left(X\right)/\operatorname{Aut}\left(C\right)\cong\mathbb{Z}/2\mathbb{Z}$
$\mathbb{Z}/2\mathbb{Z}\rule[-5.0pt]{0.0pt}{18.0pt}$ | $\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$ | no |  $C$ is not isomorphic to $C^{\prime}$ | $\operatorname{Aut}\left(X\right)\cong\operatorname{Aut}\left(C\right)$

In the third column we give an answer to the question whether the surface $X$
is isomorphic to a quadric in $\mathbb{P}^{3}_{\Bbbk}$. In the fourth row the
group $\operatorname{Aut}\left(X\right)$ is a non-split extension of
$\operatorname{Aut}\left(C\right)$ by $\mathbb{Z}/2\mathbb{Z}$, that means
that $\operatorname{Aut}\left(X\right)$ contains a normal subgroup
$\operatorname{Aut}\left(C\right)$ of index $2$ but is not isomorphic to any
semi-direct product
$\operatorname{Aut}\left(C\right)\rtimes\left(\mathbb{Z}/2\mathbb{Z}\right)$.

##  3\. Sarkisov links for the case $\rho(X)=2$

In this section we consider Sarkisov links for pointless del Pezzo surfaces of
degree $8$ with the Picard number $2$, and prove Theorem 1.5.

Let $X$ be a pointless del Pezzo surface of degree $8$ with the Picard number
$2$. By Lemma 2.1 in this case $X$ is isomorphic to a product of two smooth
conics, such that at least one of these conics is non-trivial. Therefore by
Proposition 2.4 either
$\operatorname{Am}(X)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$, or
$\operatorname{Am}(X)\cong\mathbb{Z}/2\mathbb{Z}$.

We start from the case
$\operatorname{Am}(X)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$. Assume
that there exists a minimal surface $X^{\prime}$, and a birational map
$X\dashrightarrow X^{\prime}$. Then
$\operatorname{Am}(X^{\prime})=\operatorname{Am}(X)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$.
By Propositions 2.4 and 2.6 it is possible only if $X^{\prime}\cong
C^{\prime}_{1}\times C^{\prime}_{2}$, where $C^{\prime}_{1}$ and
$C^{\prime}_{2}$ are two smooth non-trivial non-isomorphic conics. Moreover,
[Kol05, Theorem 2] implies that if for a minimal surface $X^{\prime}\cong
C^{\prime}_{1}\times C^{\prime}_{2}$ one has
$\operatorname{Am}(X^{\prime})=\operatorname{Am}(X)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$,
then $X^{\prime}$ is birationally equivalent to $X$.

Actually we want to remind some results and constructions from [Kol05] to
describe birational maps from $X$ more explicitly.

######  Definition 3.1.

Let $C_{1}$ and $C_{2}$ be two smooth conics, and $b(C_{1})$ and $b(C_{2})$ be
the classes of these conics in $\operatorname{Br}\left(\Bbbk\right)$. If there
exists a conic $C_{3}$, such that $b(C_{3})=b(C_{1})+b(C_{2})$, then $C_{3}$
is called the Brauer product of $C_{1}$ and $C_{2}$. We denote the Brauer
product of $C_{1}$ and $C_{2}$ by $C_{1}*C_{2}$.

We want to show that $C_{1}*C_{2}$ is defined if and only if there exists a
quadratic extension $F/\Bbbk$, such that $C_{1}\otimes F\cong C_{2}\otimes
F\cong\mathbb{P}^{1}_{F}$. In the other words there are points of degree $2$
on $C_{1}$ and $C_{2}$, such that the corresponding geometric points are
defined over $F$. Note that this condition holds if $C_{1}\cong C_{2}$, or if
$C_{1}$ or $C_{2}$ is trivial.

######  Lemma 3.2.

Let $C_{1}$ and $C_{2}$ be two smooth conics. Suppose that there exist points
of degree $2$ on $C_{1}$ and $C_{2}$, such that the corresponding geometric
points are defined over a quadratic extension $F/\Bbbk$. Then there exists a
conic $C_{3}=C_{1}*C_{2}$, and $X\cong C_{1}\times C_{2}$ is birationally
equivalent to $C_{1}\times C_{3}$ and $C_{2}\times C_{3}$.

###### Proof.

Let $\pi_{1}:X\rightarrow C_{1}$ and $\pi_{2}:X\rightarrow C_{2}$ be the
projections. Let $A$ and $B$ be the classes in
$\operatorname{Pic}(\overline{X})$ of geometric fibres of $\pi_{1}$ and
$\pi_{2}$ respectively. Let $A_{1}$ and $A_{2}$ be conjugate fibres of
$\pi_{1}$ defined over $F$, and $B_{1}$ and $B_{2}$ be conjugate fibres of
$\pi_{2}$ defined over $F$. Then the pair of points $A_{1}\cap B_{1}$ and
$A_{2}\cap B_{2}$ are defined over $F$. We can equivariantly blow up this pair
of points, contract the proper transforms of $A_{1}$ and $A_{2}$, and get a
surface $X^{\prime}$.

Note that the birational map $X\dashrightarrow X^{\prime}$ respects the
projection $\pi_{1}:X\rightarrow C_{1}$. Therefore $X^{\prime}$ admits a
structure of a conic bundle over $C_{1}$. Moreover,
$K_{X^{\prime}}^{2}=K_{X}^{2}=8$, therefore $X^{\prime}$ is a $\Bbbk$-form of
a Hirzebruch surface $\mathbb{F}_{n}$. Also note that the proper transforms of
$B_{1}$ and $B_{2}$ are sections of the conic bundle $X^{\prime}\rightarrow
C_{1}$ with selfintersection number $0$. For $n>0$ there are no sections of
$\mathbb{F}_{n}\rightarrow C_{1}$ with selfintersection number $0$. Thus
$X^{\prime}\cong\mathbb{F}_{0}$ that is a product of two conics.

One of these conics is $C_{1}$, since $X^{\prime}$ admits a structure of a
conic bundle over $C_{1}$. Denote the other conic by $C_{3}$. Note that
preimage of a geometric point on $C_{3}$ under the sequence of maps
$X\dashrightarrow X^{\prime}\rightarrow C_{3}$ has class $A+B$ in
$\operatorname{Pic}\left(\overline{X}\right)$. Therefore
$b(C_{3})\in\operatorname{Br}\left(\Bbbk\right)$ is $b(C_{1})+b(C_{2})$, since
the preimages of geometric points on $C_{1}$ and $C_{2}$ under the projections
$\pi_{1}$ and $\pi_{2}$ have classes $A$ and $B$ in
$\operatorname{Pic}\left(\overline{X}\right)$ respectively. Thus
$C_{3}=C_{1}*C_{2}$, and $X\cong C_{1}\times C_{2}$ is birationally equivalent
to $X^{\prime}\cong C_{1}\times C_{3}$. In the same way we can show that $X$
is birationally equivalent to $C_{2}\times C_{3}$. ∎

The following lemma directly follows from the proof of [Kol05, Lemma 8].

######  Lemma 3.3.

Let $C_{1}$ and $C_{2}$ be two smooth conics, such that for any pair of points
of degree $2$ on $C_{1}$ and $C_{2}$ the corresponding geometric points are
not defined over any quadratic extension $\mathbb{F}/\Bbbk$. Then a conic with
class $b(C_{1})+b(C_{2})$ does not exist.

Now consider the case $\operatorname{Am}(X)\cong\mathbb{Z}/2\mathbb{Z}$. By
Proposition 2.4 the surface $X$ is isomorphic either to $C\times C$, or to
$C\times\mathbb{P}^{1}_{\Bbbk}$ where $C$ is a smooth non-trivial conic.
Moreover, by Lemma 3.2 for a given smooth non-trivial conic $C$ the surfaces
$C\times C$ and $C\times\mathbb{P}^{1}_{\Bbbk}$ are birationally equivalent.
In the following lemma we show that these surfaces are birationally equivalent
to certain $\Bbbk$-forms of Hirzebruch surfaces.

######  Lemma 3.4.

Let $C$ be a smooth non-trivial conic. Then the surfaces $C\times C$,
$C\times\mathbb{P}^{1}_{\Bbbk}$, and $\Bbbk$-forms of Hirzebruch surfaces
$\mathbb{F}_{2k}$ admitting a structure of conic bundle over $C$ are
birationally equivalent.

###### Proof.

The surfaces $C\times C$ and $C\times\mathbb{P}^{1}_{\Bbbk}$ are birationally
equivalent by Lemma 3.2, since $C*C\cong\mathbb{P}^{1}_{\Bbbk}$.

Now consider $X\cong C\times\mathbb{P}^{1}_{\Bbbk}$. Let $\pi_{1}:X\rightarrow
C$ and $\pi_{2}:X\rightarrow\mathbb{P}^{1}_{\Bbbk}$ be the projections. Then
for a $\Bbbk$-point $p\in\mathbb{P}^{1}_{\Bbbk}$, the preimage
$B=\pi_{2}^{-1}(p)$ is a section of $\pi_{2}$. Let $A_{1}$ and $A_{2}$ be
conjugate fibres of $\pi_{1}$. Then the points $A_{1}\cap B$ and $A_{2}\cap B$
are conjugate. We can equivariantly blow up this pair of points, contract the
proper transforms of $A_{1}$ and $A_{2}$, and get a surface $X^{\prime}$.

The birational map $X\dashrightarrow X^{\prime}$ respects the projection
$\pi_{1}:X\rightarrow C$. Therefore $X^{\prime}$ admits a structure of a conic
bundle over $C$. Moreover, $K_{X^{\prime}}^{2}=K_{X}^{2}=8$, therefore
$X^{\prime}$ is a $\Bbbk$-form of a Hirzebruch surface $\mathbb{F}_{n}$. Note
that the proper transform of $B$ is a section of the conic bundle
$X^{\prime}\rightarrow C$ with selfintersection number $-2$. Therefore $n=2$.
Thus $C\times\mathbb{P}^{1}_{\Bbbk}$ is birationally equivalent to a
$\Bbbk$-form of a Hirzebruch surface $\mathbb{F}_{2}$ admitting a structure of
a conic bundle over $C$.

Now consider a $\Bbbk$-form $S$ of a Hirzebruch surface $\mathbb{F}_{2k}$
admitting a structure of a conic bundle $\pi:S\rightarrow C$. There is a
unique section $H$ of $\pi$ such that $H^{2}=-2k$. Let $A_{1}$ and $A_{2}$ be
conjugate fibres of $\pi$. Then the points $A_{1}\cap H$ and $A_{2}\cap H$ are
conjugate. We can equivariantly blow up this pair of points, contract the
proper transforms of $A_{1}$ and $A_{2}$, and get a surface $S^{\prime}$.

The birational map $S\dashrightarrow S^{\prime}$ respects the projection
$\pi:S\rightarrow C$. Therefore $S^{\prime}$ admits a structure of a conic
bundle over $C$. Moreover, $K_{S^{\prime}}^{2}=K_{S}^{2}=8$, therefore
$S^{\prime}$ is a $\Bbbk$-form of a Hirzebruch surface $\mathbb{F}_{n}$. Note
that the proper transform of $H$ is a section of the conic bundle
$S^{\prime}\rightarrow C$ with selfintersection number $-2k-2$. Therefore
$n=2k+2$. Thus $S$ is birationally equivalent to a $\Bbbk$-form of a
Hirzebruch surface $\mathbb{F}_{2k+2}$ admitting a structure of conic bundle
over $C$.

Thus we see that all surfaces listed in Lemma 3.4 are birationally equivalent.
∎

Now we prove Theorem 1.5.

###### Proof of Theorem 1.5.

Let $X$ be a pointless del Pezzo surface of degree $8$ with $\rho(X)=2$. Then
$\overline{X}\cong\mathbb{P}^{1}_{\overline{\Bbbk}}\times\mathbb{P}^{1}_{\overline{\Bbbk}}$,
and $X$ is a product of two conics. If $\operatorname{Am}(X)$ is trivial then
$X\cong\mathbb{P}^{1}_{\Bbbk}\times\mathbb{P}^{1}_{\Bbbk}$ and
$X(\Bbbk)\neq\varnothing$. Therefore $\operatorname{Am}(X)$ is non-trivial,
and by Proposition 2.4 either
$\operatorname{Am}(X)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$, or
$\operatorname{Am}(X)\cong\mathbb{Z}/2\mathbb{Z}$.

If $\operatorname{Am}(X)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}$ then $X$
is a product of two non-trivial smooth conics $C_{1}$ and $C_{2}$, not
isomorphic to each other. If the Brauer product $C_{3}=C_{1}*C_{2}$ is defined
then $X\approx C_{1}\times C_{3}\approx C_{2}\times C_{3}$ by Lemma 3.2.

If a minimal surface $X^{\prime}$ is birationally equivalent to $X$ then

$\operatorname{Am}(X^{\prime})=\operatorname{Am}(X)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{2}.$

By Propositions 2.4 and 2.6 this is possible only if $X^{\prime}$ is a product
of two smooth conics $C^{\prime}_{1}$ and $C^{\prime}_{2}$. Moreover,
$b(C^{\prime}_{1})$ and $b(C^{\prime}_{2})$ are distinct non-trivial elements
of $\operatorname{Am}(X)$. Therefore if $C_{3}$ is defined then $X^{\prime}$
is isomorphic to $C_{1}\times C_{2}$, $C_{1}\times C_{2}$ or $C_{2}\times
C_{3}$, and otherwise $X^{\prime}\cong C_{1}\times C_{2}\cong X$. We have
proved Theorem 1.5(1).

If $\operatorname{Am}(X)\cong\mathbb{Z}/2\mathbb{Z}$ then either $X\cong
C\times C$, or $X\cong C\times\mathbb{P}^{1}_{\Bbbk}$, where $C$ is a non-
trivial smooth conic. By Lemma 3.4 the surface $X$ is birationally equivalent
to $C\times C$, $C\times\mathbb{P}^{1}_{\Bbbk}$, and $\Bbbk$-forms of
Hirzebruch surfaces $\mathbb{F}_{2k}$ admitting a structure of conic bundle
over $C$.

Assume that a minimal surface $X^{\prime}$ is birationally equivalent to $X$
and not listed before. Then the map $X\dashrightarrow X^{\prime}$ can be
decomposed into a sequence of Sarkisov links:

$X=X_{0}\dashrightarrow X_{1}\dashrightarrow\ldots\dashrightarrow
X_{n-1}\dashrightarrow X_{n}=X^{\prime}.$

In particular there exists a Sarkisov link $f:S\dashrightarrow S^{\prime}$,
such that $S$ is listed in Lemma 3.4, and $S^{\prime}$ is not listed. For
geometrically rational surfaces Sarkisov links are described in [Isk96,
Theorem 2.6]. All surfaces listed in Lemma 3.4 admit a structure of a conic
bundle over $C$. Therefore $f$ has type $\mathrm{II}$, $\mathrm{III}$ or
$\mathrm{IV}$. Links of types $\mathrm{II}$ and $\mathrm{IV}$ preserve
$K_{S}^{2}$ and $\operatorname{Am}(S)$. Therefore in this case $S^{\prime}$ is
one of the surfaces listed in Proposition 2.4(2)–(4), and also in Lemma 3.4.
For $S$ with $\rho(S)=2$ and $K_{S}^{2}=8$ the only possible link of type
$\mathrm{III}$ is a contraction of $(-1)$-section on a $\Bbbk$-form of a
Hirzebruch surface $\mathbb{F}_{1}$. But such surface is not listed in Lemma
3.4, therefore this link is impossible, and $X^{\prime}$ must be isomorphic to
a surface listed in Lemma 3.4. We have proved Theorem 1.5(2). ∎

##  4\. Sarkisov links for the case $\rho(X)=1$

In this section we consider Sarkisov links for pointless del Pezzo surfaces of
degree $8$ with the Picard number $1$, and prove Theorem 1.6.

Let $X$ be a pointless del Pezzo surface of degree $8$ with the Picard number
$1$. By Lemma 2.1 in this case $X$ is isomorphic to $R_{\mathbb{L}/\Bbbk}C$,
where $\mathbb{L}$ is a quadratic extension of $\Bbbk$ and $C$ is a conic over
$\mathbb{L}$. The field $\mathbb{L}$ is uniquely determined by $X$. The pair
$\left(\mathbb{L},\operatorname{Am}(X_{\mathbb{L}})\right)$ is a biregular
invariant of $X$ by Theorem 2.8. We want to find possibilities for minimal
surfaces $X^{\prime}$ birationally equivalent to $X$.

Note that by Corollary 2.2 one has $X_{\mathbb{L}}\cong C\times C^{\prime}$,
where $C^{\prime}$ is a conic conjugate to $C$ under the action of
$\operatorname{Gal}\left(\mathbb{L}/\Bbbk\right)$. Note that $X$ is pointless
if and only if $X_{\mathbb{L}}$ is pointless too. In particular, if $X$ is
pointless then $\operatorname{Am}(X_{\mathbb{L}})$ is non-trivial.

######  Lemma 4.1.

Let $X$ be a pointless del Pezzo surface of degree $8$ with $\rho(X)=1$. Then
any point on $X$ has even degree.

###### Proof.

Assume that there exists a point of odd degree on $X$. Then there exists a
point of odd degree on $X_{\mathbb{L}}$, and images of this point under the
projection on $C$ and $C^{\prime}$ are points of odd degree. There are no
points of odd degree on any non-trivial conic (see [Kol16, Theorem 53]).
Therefore $C\cong C^{\prime}\cong\mathbb{P}^{1}_{\mathbb{L}}$, and thus
$X\left(\Bbbk\right)\neq\varnothing$.

We obtain a contradiction. Therefore any point on $X$ has even degree. ∎

Now we want to consider Sarkisov links for the surface $X$. These links are
described in [Isk96, Theorem 2.6 (i and ii)]. In our case such a link must
have tipe $\mathrm{I}$ or $\mathrm{II}$, since $X$ is a del Pezzo surface with
$\rho(X)=1$. Any link of type $\mathrm{I}$ is a blowup $X_{1}\rightarrow X$ of
a point of degree $d$, where $X_{1}$ has a structure of a conic bundle over a
smooth conic and $\rho(X_{1})=2$. Any link of type $\mathrm{II}$ is a
composition $\sigma_{1}\circ\sigma^{-1}$, where $\sigma:Z\rightarrow X$ is a
blowup of a point of degree $d$, and $\sigma_{1}:Z\rightarrow X_{1}$ is a
contraction of a set of disjoint conjugate $(-1)$-curves, that differs from
the exceptional divisor of $\sigma$. In our case $d=2$, $d=4$ or $d=6$ by
Lemma 4.1. For $d=2$ the corresponding link has type $\mathrm{I}$, and for
$d=4$ and $d=6$ the corresponding link has type $\mathrm{II}$. For each of
these three links we want to describe $X_{1}$.

######  Lemma 4.2.

Let $X$ be a del Pezzo surface of degree $8$ with $\rho(X)=1$. Let
$X\dashrightarrow X_{1}$ be a Sarkisov link corresponding to the blowup of a
point of degree $6$. Then $X_{1}\cong X$.

###### Proof.

This case is considered in [Isk96, Theorem 2.6.(ii), $K_{X}^{2}=8$, $d=6$],
and there is explicitely written that $X_{1}\cong X$. ∎

######  Lemma 4.3.

Let $X$ be a del Pezzo surface of degree $8$ with $\rho(X)=1$. Let
$X\dashrightarrow X_{1}$ be a Sarkisov link corresponding to the blowup of a
point of degree $4$. Then $X_{1}\cong X$.

###### Proof.

This case is considered in [Isk96, Theorem 2.6.(ii), $K_{X}^{2}=8$, $d=4$],
and for this case $K_{X_{1}}^{2}=8$ and $\rho(X_{1})=1$. Therefore by Lemma
2.1 one has $R_{\mathbb{L}_{1}/\Bbbk}C_{1}$, where $\mathbb{L}_{1}$ is the
splitting field of $X_{1}$ and $C_{1}$ is a smooth conic over
$\mathbb{L}_{1}$. Let us show that $\mathbb{L}_{1}=\mathbb{L}$. It is
sufficient to prove that $\rho(\left(X_{1}\right)_{\mathbb{L}})=2$.

Let $\pi_{1}:X_{\mathbb{L}}\rightarrow C$ and
$\pi_{2}:X_{\mathbb{L}}\rightarrow C^{\prime}$ be the projections. Let $A$ and
$B$ be the classes in $\operatorname{Pic}(\overline{X})$ of geometric fibres
of $\pi_{1}$ and $\pi_{2}$ respectively. The map $X\dashrightarrow X_{1}$ is a
composition $\sigma_{1}\circ\sigma^{-1}$, where $\sigma:Z\rightarrow X$ is a
blowup of a point of degree $4$ and $\sigma_{1}:Z\rightarrow X_{1}$ is a
contraction of a set of disjoint conjugate $(-1)$-curves. Let $E_{1}$,
$E_{2}$, $E_{3}$ and $E_{4}$ be the classes in
$\operatorname{Pic}(\overline{Z})$ of geometrically irreducible components of
the exceptional divisor of $\sigma$.

The surface $Z$ is a del Pezzo surface of degree $4$, and thus there are $16$
$(-1)$-curves on $Z$, having classes $E_{i}$, $A-E_{i}$, $B-E_{i}$ and
$A+B+E_{i}-\sum\limits_{j=1}^{4}E_{j}$ in $\operatorname{Pic}(\overline{Z})$.
The four $(-1)$-curves with classes $A+B+E_{i}-\sum\limits_{j=1}^{4}E_{j}$ are
conjugate and disjoint, therefore $\sigma_{1}:Z\rightarrow X_{1}$ is the
contraction of these curves.

Let $k$ be the number of
$\operatorname{Gal}\left(\overline{\Bbbk}/\mathbb{L}\right)$-orbits on the set
$E_{i}$ (actually $k=1$ or $k=2$). Then the number of
$\operatorname{Gal}\left(\overline{\Bbbk}/\mathbb{L}\right)$-orbits on the set
$A+B+E_{i}-\sum\limits_{j=1}^{4}E_{j}$ is also $k$, since
$A+B-\sum\limits_{j=1}^{4}E_{j}$ is
$\operatorname{Gal}\left(\overline{\Bbbk}/\mathbb{L}\right)$-invariant.
Therefore

$\rho(\left(X_{1}\right)_{\mathbb{L}})=\rho(Z_{\mathbb{L}})-k=\rho(X_{\mathbb{L}})=2.$

The del Pezzo surfaces $X$ and $X_{1}$ of degree $8$ with
$\rho(X)=\rho(X_{1})=1$ have the same splitting field and
$\operatorname{Am}(X_{\mathbb{L}})=\operatorname{Am}((X_{1})_{\mathbb{L}})$,
since $X\approx X_{1}$. Therefore $X\cong X_{1}$ by Theorem 2.8.

∎

######  Remark 4.4.

Note that in Lemmas 4.2 and 4.3 the surface $X$ is not necessary pointless.

The link of type $\mathrm{I}$ corresponding to the blowup $Z\rightarrow X$ of
two conjugate geometric points is considered in [Isk96, Theorem 2.6.(i),
$K_{X}^{2}=8$]. In this case $Z$ admits a structure of a conic bundle
$\pi:Z\rightarrow C$ with two degenerate geometric fibres, where $C$ is a
conic such that $b(C)$ is the generator of $\operatorname{Am}(X)$ (in
particular, $C\cong\mathbb{P}^{1}_{\Bbbk}$ if $\operatorname{Am}(X)$ is
trivial). In the following proposition we collect some facts about such conic
bundles following [Isk79] and [Isk96].

######  Proposition 4.5.

Let $\pi:S\rightarrow C$ be a conic bundle over a smooth conic $C$ with two
degenerate geometric fibres, such that $\rho(S)=2$. Then $K_{S}^{2}=6$, and
$S$ has the following properties.

  1. (i)

The surface $S$ is a del Pezzo surface of degree $6$ (see [Isk79, Theorem 5]).

  2. (ii)

There are six $(-1)$-curves on $S$: four components of the singular fibres,
and two disjoint conjugate sections $E_{1}$ and $E_{2}$ of $\pi$. These curves
form a hexagon.

  3. (iii)

The surface $S$ is not minimal (see [Isk79, Theorem 4]).

  4. (iv)

There exists a unique link $S\rightarrow Y$ of type $\mathrm{III}$ that is the
contraction of the pair $E_{1}$ and $E_{2}$. The surface $Y$ is a del Pezzo
surface of degree $8$ with $\rho(Y)=1$ (see [Isk96, Theorem 2.6(iii)]).

  5. (v)

Any other link $S\dashrightarrow S_{1}$ has type $\mathrm{II}$, and preserves
a structure of a conic bundle over $C$ (see [Isk96, Theorem 2.6]). Such link
is a composition $\sigma_{1}\circ\sigma^{-1}$, where $\sigma$ is a blowup of a
point of degree $d$ such that the corresponding geometric points do not lie on
the degenerate fibres, and $\sigma_{1}$ is the contraction of the proper
transforms of $d$ fibres containing these points. In particular, the conic
bundles $S\rightarrow C$ and $S_{1}\rightarrow C$ have degenerate fibres over
the same points on $C$, $\rho(S_{1})=2$ and $K_{S_{1}}^{2}=6$. Therefore all
described above properties hold for $S_{1}$.

Now we can consider the link of type $\mathrm{I}$.

######  Lemma 4.6.

Let $X$ be a pointless del Pezzo surface of degree $8$ with $\rho(X)=1$.
Consider a sequence of Sarkisov links

$X\leftarrow Z=Z_{0}\dashrightarrow Z_{1}\dashrightarrow\ldots\dashrightarrow
Z_{n-1}\dashrightarrow Z_{n}=Z^{\prime}\rightarrow X^{\prime},$

where the first link has type $\mathrm{I}$, the last link has type
$\mathrm{III}$, and all other links have type $\mathrm{II}$. Then
$X^{\prime}\cong X$.

###### Proof.

The surface $Z$ admits a structure of a conic bundle over a smooth conic
$\pi:Z\rightarrow C$ with two degenerate fibres. Denote the components of
these fibres by $A_{1}$, $A_{2}$, $B_{1}$ and $B_{2}$. These components are
$(-1)$-curves on $Z$. By Proposition 4.5(ii) there are two disjoint conjugate
sections $E_{1}$ and $E_{2}$ of $\pi$ that are $(-1)$-curves. Without loss of
generality we can assume that

$A_{1}\cdot E_{1}=E_{1}\cdot B_{1}=B_{1}\cdot A_{2}=A_{2}\cdot
E_{2}=E_{2}\cdot B_{2}=B_{2}\cdot A_{1}=1.$

Let $\Gamma$ be the kernel of the action of
$\operatorname{Gal}\left(\overline{\Bbbk}/\Bbbk\right)$ on the set of
$(-1)$-curves on $Z$, and $\mathbb{F}=\overline{\Bbbk}^{\Gamma}$. Then
$\rho\left(Z_{\mathbb{F}}\right)=\rho\left(\overline{Z}\right)=4$, and
$\rho(X_{\mathbb{F}})=2$. Therefore $\mathbb{L}$ is a subfield of
$\mathbb{F}$. Note that the pairs $A_{1}$ and $A_{2}$, $B_{1}$ and $B_{2}$,
and $E_{1}$ and $E_{2}$ are defined over $\mathbb{L}$, but no one of these
curves is defined over $\mathbb{L}$, since $X_{\mathbb{L}}$ is pointless.
Therefore the group $\operatorname{Gal}\left(\mathbb{F}/\mathbb{L}\right)$ has
order $2$ and pairwisely permutes $E_{1}$ and $E_{2}$, $A_{1}$ and $A_{2}$,
and $B_{1}$ and $B_{2}$.

By Proposition 4.5(v) all surfaces $Z_{i}$ are del Pezzo surfaces of degree
$6$ admitting a structure of a conic bundle $Z_{i}\rightarrow C$ with two
degenerate fibres. Let $f:Z\dashrightarrow Z^{\prime}$ be the composition of
the Sarkisov links $Z_{i}\dashrightarrow Z_{i+1}$. Then
$A^{\prime}_{1}=f(A_{1})$, $A^{\prime}_{2}=f(A_{2})$,
$B^{\prime}_{1}=f(B_{1})$ and $B^{\prime}_{2}=f(B_{2})$ are components of the
degenerate fibres of $Z^{\prime}\rightarrow C$. By Proposition 4.5(ii) there
are two disjoint conjugate sections $E^{\prime}_{1}$ and $E^{\prime}_{2}$ of
$\pi^{\prime}:Z^{\prime}\rightarrow C$ that are $(-1)$-curves. We can assume
that $A^{\prime}_{1}\cdot E^{\prime}_{1}=1$, and want to show that
$E^{\prime}_{1}\cdot B^{\prime}_{1}=1$.

Let $R=f^{-1}_{*}\left(E^{\prime}_{1}\right)$ be the proper transform of
$E^{\prime}_{1}$ on $Z$. If we blow up a point $p$ on $\overline{Z}$ and
contract a proper transform of the fibre containing $p$, then for the proper
transform $\widetilde{R}$ of $R$ one has $\widetilde{R}^{2}-R^{2}=\pm 1$ ($-1$
if $p\in R$, and $1$ otherwise). By Proposition 4.5(v) any link
$Z_{i}\dashrightarrow Z_{i+1}$ is a composition $\sigma_{1}\circ\sigma^{-1}$,
where $\sigma$ is a blowup of a point of degree $d$ such that the
corresponding geometric points do not lie on degenerate fibres, and
$\sigma_{1}$ is the contraction of the proper transforms of $d$ fibres
containing these points. Moreover, $d$ is even, since otherwise there is a
point of odd degree on $X$ that is impossible by Lemma 4.1. Thus
$R^{2}-\left(E^{\prime}_{1}\right)^{2}$ is even, and $R^{2}$ is odd.

Note that the group $\operatorname{Pic}(\overline{Z})\cong\mathbb{Z}^{4}$ is
generated by $A_{1}$, $E_{1}$, $B_{1}$ and the class of a fibre $F\sim
A_{1}+B_{2}\sim A_{2}+B_{1}$. Therefore $R\sim aA_{1}+eE_{1}+bB_{1}+fF$. The
curve $R$ is a section of $\pi:Z\rightarrow C$, thus $e=1$, since
$F^{2}=F\cdot A_{1}=F\cdot B_{1}=0$ and $F\cdot E_{1}=1$. One has $A_{1}\cdot
R=A^{\prime}_{1}\cdot E^{\prime}_{1}=1$, thus $a=0$, since $A_{1}\cdot
B_{1}=A_{1}\cdot F=0$, $A_{1}\cdot E_{1}=1$ and $A_{1}^{2}=-1$. Therefore
$R\sim E_{1}+bB_{1}+fF$, and

$R^{2}=E_{1}^{2}+b^{2}B_{1}^{2}+f^{2}F^{2}+2bE_{1}\cdot B_{1}+2fE_{1}\cdot
F+2bfB_{1}\cdot F=-1-b^{2}+2b+2f.$

The number $b$ is even, since the intersection number $R^{2}$ is odd. One has

$B_{1}\cdot R=B_{1}\cdot\left(E_{1}+bB_{1}+fF\right)=1-b.$

Therefore $b=0$, since $b$ is even and $B_{1}\cdot R\geqslant 0$. Thus
$E^{\prime}_{1}\cdot B^{\prime}_{1}=R\cdot B_{1}=1$.

Now we see that for the set of $(-1)$-curves on $Z^{\prime}$ we have

$A^{\prime}_{1}\cdot E^{\prime}_{1}=E^{\prime}_{1}\cdot
B^{\prime}_{1}=B^{\prime}_{1}\cdot A^{\prime}_{2}=A^{\prime}_{2}\cdot
E^{\prime}_{2}=E^{\prime}_{2}\cdot B^{\prime}_{2}=B^{\prime}_{2}\cdot
A^{\prime}_{1}=1.$

The group $\operatorname{Gal}\left(\mathbb{F}/\mathbb{L}\right)$ has order $2$
and pairwisely permutes $A^{\prime}_{1}$ and $A^{\prime}_{2}$,
$B^{\prime}_{1}$ and $B^{\prime}_{2}$, and also $E^{\prime}_{1}$ and
$E^{\prime}_{2}$. Therefore $\rho\left(Z^{\prime}_{\mathbb{L}}\right)=3$. By
Proposition 4.5(iv) the map $Z^{\prime}\rightarrow X^{\prime}$ is the
contraction of the disjoint conjugate curves $E^{\prime}_{1}$ and
$E^{\prime}_{2}$. Each of those curves is not defined over $\mathbb{L}$,
therefore $\rho\left(X^{\prime}_{\mathbb{L}}\right)=2$. It means that
$\mathbb{L}$ is the splitting field of $X^{\prime}$. One has
$\operatorname{Am}(X_{\mathbb{L}})=\operatorname{Am}(X^{\prime}_{\mathbb{L}})$,
since $X\approx X^{\prime}$. Therefore $X\cong X^{\prime}$ by Theorem 2.8. ∎

######  Remark 4.7.

Note that Lemma 4.6 does not work if $X(\Bbbk)\neq\varnothing$. Consider a
sequence of Sarkisov links

$X\leftarrow Z\dashrightarrow Z^{\prime}\rightarrow X^{\prime},$

where the first link has type $\mathrm{I}$ and the last link has type
$\mathrm{III}$. The second link corresponds to the blowup of a $\Bbbk$-point
$p$ on a smooth fibre of $Z\rightarrow C$ and the contraction of the proper
transform of the fibre passing through $p$, and has type $\mathrm{II}$. Let us
show that $X$ and $X^{\prime}$ are not isomorphic to each other. We use the
notation established in the proof of Lemma 4.6 for the $(-1)$-curves on $Z$
and $Z^{\prime}$. We want to show that $E^{\prime}_{1}\cdot B^{\prime}_{1}=0$.

Let $R$ be the proper transform of $E^{\prime}_{1}$ on $Z$. Then $R^{2}$ is
even, and $R\sim E_{1}+bB_{1}+fF$. Thus

$R^{2}=E_{1}^{2}+b^{2}B_{1}^{2}+f^{2}F^{2}+2bE_{1}\cdot B_{1}+2fE_{1}\cdot
F+2bfB_{1}\cdot F=-1-b^{2}+2b+2f.$

The number $b$ is odd, since the intersection number $R^{2}$ is even. One has

$B_{1}\cdot R=B_{1}\cdot\left(E_{1}+bB_{1}+fF\right)=1-b.$

Therefore $b=1$, since $b$ is odd and $B_{1}\cdot R\geqslant 0$. Thus
$E^{\prime}_{1}\cdot B^{\prime}_{1}=R\cdot B_{1}=0$.

Now we see that for the set of $(-1)$-curves on $Z^{\prime}$ we have

$A^{\prime}_{1}\cdot E^{\prime}_{1}=E^{\prime}_{1}\cdot
A^{\prime}_{2}=A^{\prime}_{2}\cdot B^{\prime}_{1}=B^{\prime}_{1}\cdot
E^{\prime}_{2}=E^{\prime}_{2}\cdot B^{\prime}_{2}=B^{\prime}_{2}\cdot
A^{\prime}_{1}=1.$

The group $\operatorname{Gal}\left(\mathbb{F}/\mathbb{L}\right)$ has order $2$
and pairwisely permutes $A^{\prime}_{1}$ and $A^{\prime}_{2}$,
$B^{\prime}_{1}$ and $B^{\prime}_{2}$, and the $(-1)$-curves $E^{\prime}_{1}$
and $E^{\prime}_{2}$ are defined over $\mathbb{L}$. Therefore
$\rho\left(Z^{\prime}_{\mathbb{L}}\right)=3$. By Proposition 4.5(iv) the map
$Z^{\prime}\rightarrow X^{\prime}$ is the contraction of the disjoint
conjugate curves $E^{\prime}_{1}$ and $E^{\prime}_{2}$. Each of those curves
is defined over $\mathbb{L}$, therefore
$\rho\left(X^{\prime}_{\mathbb{L}}\right)=1$, that means that $\mathbb{L}$ is
not the splitting field of $X^{\prime}$. Hence $X$ and $X^{\prime}$ are not
isomorphic to each other by Theorem 2.8.

As a by-product of Lemma 4.6 we obtain the following proposition.

######  Proposition 4.8.

Let $Z$ be a pointless del Pezzo surface of degree $6$ admitting a structure
of a relatively minimal conic bundle $Z\rightarrow C$, and $Z\dashrightarrow
Z_{1}$ be a Sarkisov link of type $\mathrm{II}$. Then $Z\cong Z_{1}$.

We decompose the proof into several lemmas that are interesting themselves.

######  Lemma 4.9.

Let $C$ be a conic over algebraically nonclosed field $\mathbb{L}$ of
characteristic $0$. Then the group $\operatorname{Aut}(C)$ transitively acts
on the set of points of degree $2$, such that the corresponding geometric
points are defined over $\mathbb{L}\left(\sqrt{a}\right)$ for a given
$a\in\mathbb{L}\setminus\mathbb{L}^{2}$.

###### Proof.

Let $\left(p_{1},p_{2}\right)$ and $\left(q_{1},q_{2}\right)$ be two pairs of
conjugate points on $C$ such that the points $p_{1}$, $p_{2}$, $q_{1}$ and
$q_{2}$ are defined over $\mathbb{L}\left(\sqrt{a}\right)$. Consider an
embedding $C\hookrightarrow\mathbb{P}^{2}_{\mathbb{L}}$. The two lines passing
through $\left(p_{1},p_{2}\right)$ and $\left(q_{1},q_{2}\right)$ are defined
over $\mathbb{L}$. Therefore we can choose coordinates in
$\mathbb{P}^{2}_{\mathbb{L}}$ such that these lines are given by $x=0$ and
$y=0$, and the homogeneous coordinates of $p_{1}$, $p_{2}$, $q_{1}$ and
$q_{2}$ are

$(0:\sqrt{a}:1),\qquad(0:-\sqrt{a}:1),\qquad(\sqrt{a}:0:1),\qquad(-\sqrt{a}:0:1)$

respectively. Then $C$ is given by the equation

$x^{2}+y^{2}-az^{2}=\theta xy,$

and the automorphism $x\leftrightarrow y$ acts on $C$ and permutes the pairs
$\left(p_{1},p_{2}\right)$ and $\left(q_{1},q_{2}\right)$. ∎

######  Lemma 4.10.

Let $S$ be a pointless del Pezzo surface of degree $8$ with $\rho(S)=1$ over
algebraically nonclosed field $\Bbbk$ of characteristic $0$. Then the group
$\operatorname{Aut}(S)$ transitively acts on the set of points of degree $2$,
such that the corresponding geometric points are defined over
$\Bbbk\left(\sqrt{a}\right)$ for a given $a\in\Bbbk\setminus\Bbbk^{2}$.

###### Proof.

By Lemma 2.1 one has $S\cong R_{\mathbb{L}/\Bbbk}C$, where $\mathbb{L}$ is the
splitting field of $S$ and $C$ is a conic over $\mathbb{L}$. Let
$\left(p_{1},p_{2}\right)$ and $\left(q_{1},q_{2}\right)$ be two pairs of
conjugate points on $C$ such that the points $p_{1}$, $p_{2}$, $q_{1}$ and
$q_{2}$ are defined over $\Bbbk\left(\sqrt{a}\right)$. Note that
$\mathbb{L}\neq\Bbbk\left(\sqrt{a}\right)$ since $S_{\mathbb{L}}$ is pointless
by Corollary 2.2.

By Corollary 2.2 one has $S_{\mathbb{L}}\cong C\times C^{\prime}$, where
$C^{\prime}$ is the
$\operatorname{Gal}\left(\mathbb{L}/\Bbbk\right)$-conjugate conic of $C$. Let
$\pi_{1}:S_{\mathbb{L}}\rightarrow C$ be the projection. One has
$\operatorname{Aut}^{0}\left(S\right)\cong\operatorname{Aut}(C)$ by Remark
2.9, and the group $\operatorname{Aut}^{0}\left(S\right)$ faithfully acts on
$C$, since $S\cong R_{\mathbb{L}/\Bbbk}C$. Consider an element
$g\in\operatorname{Aut}^{0}\left(S\right)$ such that the induced action of $g$
on $C$ maps the pair $\left(\pi_{1}(p_{1}),\pi_{1}(p_{2})\right)$ to the pair
$\left(\pi_{1}(q_{1}),\pi_{1}(q_{2})\right)$ (such $g$ exists by Lemma 4.9).

Note that the
$\operatorname{Gal}\left(\mathbb{L}\left(\sqrt{a}\right)/\Bbbk\right)$-orbit
of the fibre $F=\pi_{1}^{-1}(\pi_{1}(q_{1}))$ consists of the two fibres of
$S_{\mathbb{L}}\rightarrow C$ passing through $q_{1}$ and $q_{2}$, and the two
fibres of $S_{\mathbb{L}}\rightarrow C^{\prime}$ passing through $q_{1}$ and
$q_{2}$. These fibres have exactly four common geometric points: the conjugate
points $q_{1}$ and $q_{2}$, and the two other conjugate points $r_{1}$ and
$r_{2}$. Note that $r_{1}$ and $r_{2}$ are not defined over
$\Bbbk\left(\sqrt{a}\right)$, since otherwise $F$ is defined over
$\Bbbk\left(\sqrt{a}\right)$ and $S$ splits over quadratic extension
$\Bbbk\left(\sqrt{a}\right)\neq\mathbb{L}$. Therefore the pair
$\left(g(p_{1}),g(p_{2})\right)$ coincides with the pair
$\left(q_{1},q_{2}\right)$, and we are done. ∎

Now we can prove Proposition 4.8.

###### Proof of Proposition 4.8.

Consider the contractions of the negative sections $Z\rightarrow X$ and
${Z_{1}\rightarrow X_{1}}$ (see Proposition 4.5(iv)). The del Pezzo surfaces
$X$ and $X_{1}$ are isomorphic by Lemma 4.6. By Lemma 4.10 there exists an
automorphism of $X$ that maps the points of the blowup of $Z\rightarrow X$ to
the points of the blowup $Z_{1}\rightarrow X_{1}$. Therefore ${Z\cong Z_{1}}$.
∎

Now we prove Theorems 1.6 and 1.1.

###### Proof of Theorem 1.6.

Let $X$ be a pointless del Pezzo surface of degree $8$ with $\rho(X)=1$.
Assume that a minimal surface $X^{\prime\prime}$ is birationally equivalent to
$X$. Then $X\dashrightarrow X^{\prime\prime}$ can be decomposed into a
sequence of Sarkisov links:

$X=X_{0}\dashrightarrow X_{1}\dashrightarrow\ldots\dashrightarrow
X_{n-1}\dashrightarrow X_{n}=X^{\prime\prime}.$

By Lemma 4.1 any Sarkisov link corresponds to the blowup of $X$ at a point of
degree $6$, $4$ or $2$. For the cases of degree $6$ and $4$ this link
immediately gives a surface $X_{1}\cong X$ by Lemmas 4.2 and 4.3 respectively.
For the case of degree $2$ the sequence of links passes through some non-
minimal del Pezzo surfaces of degree $6$ (see Proposition 4.5(iii)) and
terminates at a del Pezzo surface $X^{\prime}\cong X$ by Lemma 4.6. Therefore
any minimal del Pezzo surface, obtained in a sequence of Sarkisov links
starting from $X$, is isomorphic to $X$. In particular, $X^{\prime\prime}\cong
X$. ∎

###### Proof of Theorem 1.1.

Consider two pointless birationally equivalent quadric surfaces $Q_{1}$ and
$Q_{2}$. These surfaces are minimal. If $\rho(Q_{1})=1$ then by Theorem 1.5
one has $Q_{1}\cong Q_{2}$.

If $\rho(Q_{1})=2$ then by Lemma 2.1 one has $Q_{1}\cong C_{1}\times C_{1}$.
By Theorem 1.5 any minimal surface birationally equivalent to $Q_{1}$ has
Picard number $2$, and therefore $Q_{2}\cong C_{2}\times C_{2}$ by Lemma 2.1.
Thus by Theorem 1.5(2) one has $Q_{1}\cong Q_{2}$. ∎

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