# The most symmetric smooth cubic surface

Anastasia V. Vikulova  Steklov Mathematical Institute of Russian Academy of
Sciences, 8 Gubkin str., Moscow, 119991, Russia
<EMAIL_ADDRESS>
###### Abstract.

In this paper for any field of characteristic different from 2 we find the
largest automorphism group of a smooth cubic surface over this field.
Moreover, we prove that for a given field a smooth cubic surface with the
largest automorphism group is unique up to isomorphism.

This work was supported by the Russian Science Foundation under grant no.
23-11-00033, https://rscf.ru/en/project/23-11-00033/

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Cubic surfaces
  4. 4 The group $\mathfrak{S}_{5}$ and the Clebsch cubic surface
  5. 5 Characteristic $3$
  6. 6 The Fermat cubic surface
  7. 7 Fields without cube root of unity
  8. 8 Proof of the main results

##  1\. Introduction

The protagonist of this paper is a smooth cubic surface. Let us recall two
famous cubic surfaces which have their own names. The cubic surface given by
the equation

$x^{3}+y^{3}+z^{3}+t^{3}=0$

in $\mathbb{P}^{3}$ is called the Fermat cubic surface, and the cubic surface
given by the equations

$x+y+z+t+w=\frac{1}{3}\left(x^{3}+y^{3}+z^{3}+t^{3}+w^{3}-(x+y+z+t+w)^{3}\right)=0$

in $\mathbb{P}^{4}$ is called the Clebsch cubic surface. It is well known that
the Fermat cubic surface over an algebraically closed field of characteristic
different from $3$ is the smooth cubic surface with the largest automorphism
group (see, for instance, [6, Theorem 9.4.1] and [7, Theorem 1.1]). The
automorphism group of the Clebsch cubic surface is isomorphic to
$\mathfrak{S}_{5}$ over any field of characteristic different from $2$ and $5$
(see, for example, [7, Theorem 6.1]). Over a field of characteristic different
from $3$ the Clebsch cubic surface can be given by more simple equations

$x+y+z+t+w=x^{3}+y^{3}+z^{3}+t^{3}+w^{3}=0.$

Note that over a field of characteristic $5$ the Clebsch cubic surface is
singular at the point $[1:1:1:1:1]\in\mathbb{P}^{4}.$

In this paper we are going to study the largest automorphism groups of smooth
cubic surfaces over arbitrary fields. To start with we recall the results
about the largest automorphism groups of smooth cubic surfaces over an
algebraically closed field, which were established in [7] for positive
characteristic and in [9] for zero characteristic.

######  Theorem 1.1 ([7, Theorem 1.1] and [9, Theorem 5.3]).

Let $S$ be a smooth cubic surface over an algebraically closed field
$\mathbf{F}.$ Then

  1. (i)′

If $\mathbf{F}$ is of characteristic $2,$ then $|\mathrm{Aut}(S)|\leqslant
25\,920.$ The equality holds if and only if
$\mathrm{Aut}(S)\simeq\mathrm{PSU}_{4}(\mathbb{F}_{2})$ and $S$ is isomorphic
to the Fermat cubic surface.

  2. (i)′′

If $\mathbf{F}$ is of characteristic $2$ and $|\mathrm{Aut}(S)|<25\,920,$ then
$|\mathrm{Aut}(S)|\leqslant 192.$ The equality holds if and only if
$\mathrm{Aut}(S)\simeq(\mathbb{Z}/2\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$

  3. (ii)′

If $\mathbf{F}$ is of characteristic $3,$ then $|\mathrm{Aut}(S)|\leqslant
216.$ The equality holds if and only if
$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$
and $S$ is isomorphic to a cubic surface

(1.1) $t^{3}+tz^{2}-xy^{2}+x^{2}z=0.$

  4. (ii)′′

If $\mathbf{F}$ is of characteristic $3$ and $|\mathrm{Aut}(S)|<216,$ then
$|\mathrm{Aut}(S)|\leqslant 120.$ The equality holds if and only if
$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$ and $S$ is isomorphic to the Clebsch
cubic surface.

  5. (iii)′

If $\mathbf{F}$ is of characteristic $5,$ then $|\mathrm{Aut}(S)|\leqslant
648.$ The equality holds if and only if
$\mathrm{Aut}(S)\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ and
$S$ is isomorphic to the Fermat cubic surface.

  6. (iii)′′

If $\mathbf{F}$ is of characteristic $5$ and $|\mathrm{Aut}(S)|<648,$ then
$|\mathrm{Aut}(S)|\leqslant 108.$ The equality holds if and only if
$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/4\mathbb{Z}.$
Moreover, if $|\mathrm{Aut}(S)|<108,$ then $|\mathrm{Aut}(S)|\leqslant 54.$
The equality holds if and only if

$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/2\mathbb{Z}.$

  7. (iv)′

If $\mathbf{F}$ is of characteristic different from $2,$ $3$ and $5,$ then
$|\mathrm{Aut}(S)|\leqslant 648.$ The equality holds if and only if
$\mathrm{Aut}(S)\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ and
$S$ is isomorphic to the Fermat cubic surface.

  8. (iv)′′

If $\mathbf{F}$ is of characteristic different from $2,$ $3$ and $5$ and
$|\mathrm{Aut}(S)|<648,$ then we have $|\mathrm{Aut}(S)|\leqslant 120.$ The
equality holds if and only if $\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$ and $S$
is isomorphic to the Clebsch cubic surface.

Let us return to the examples of cubic surfaces. The other smooth cubic
surface which we are interested in is given by the equation

(1.2) $x^{2}t+y^{2}z+z^{2}y+t^{2}x=0$

in $\mathbb{P}^{3}.$ The interesting feature of cubic surface (1.2) is that up
to isomorphism it is the unique smooth cubic surface over
$\mathbb{F}_{2^{2k+1}}$ for any $k\in\mathbb{Z}_{\geqslant 0}$ with
automorphism group of maximal order (see [15, Theorem 1.4]). This automorphism
group is isomorphic to $\mathfrak{S}_{6}.$ Moreover, by [15, Corollary 1.7]
over $\mathbb{F}_{4^{k}}$ for any $k\in\mathbb{Z}_{\geqslant 0}$ cubic surface
(1.2) is isomorphic to the Fermat cubic surface. So as a result, over a finite
field of characteristic $2$ it is the smooth cubic surface with the largest
automorphism group. To sum up, we have the following result.

######  Theorem 1.2 ([15, Theorem 1.4]).

Let $S$ be a smooth cubic surface over a finite field $\mathbf{F}$ of
characteristic $2.$

  1. (i)

If $\mathbf{F}=\mathbb{F}_{4^{k}},$ then $|\mathrm{Aut}(S)|\leqslant 25\,920.$
Moreover, $|\mathrm{Aut}(S)|=25\,920$ if and only if

$\mathrm{Aut}(S)\simeq\mathrm{PSU}_{4}(\mathbb{F}_{2})$

and $S$ is isomorphic to the Fermat cubic surface.

  2. (ii)

If $\mathbf{F}=\mathbb{F}_{2^{2k+1}},$ then $|\mathrm{Aut}(S)|\leqslant 720.$
Moreover, $|\mathrm{Aut}(S)|=720$ if and only if

$\mathrm{Aut}(S)\simeq\mathfrak{S}_{6}$

and $S$ is isomorphic to cubic surface (1.2).

The goal of this paper is to find the largest automorphism group of a smooth
cubic surface over an arbitrary field of characteristic different from $2.$ So
we are going to prove the following theorem.

######  Theorem 1.3.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different from $2.$

  * •

Assume that the characteristic of $\mathbf{F}$ is equal to $3.$

    1. (i)3

If $\mathbf{F}$ contains a non-trivial fourth root of unity, then
$|\mathrm{Aut}(S)|\leqslant 216.$ Moreover, $|\mathrm{Aut}(S)|=216$ if and
only if

$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$

and $S$ is isomorphic to smooth cubic surface (1.1).

    2. (ii)3

If $\mathbf{F}$ does not contain a non-trivial fourth root of unity, then

$|\mathrm{Aut}(S)|\leqslant 120.$

Moreover, $|\mathrm{Aut}(S)|=120$ if and only if

$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$

and $S$ is isomorphic to the Clebsch cubic surface.

  * •

Assume that the characteristic of $\mathbf{F}$ is equal to $5.$

    1. (i)5

If $\mathbf{F}$ contains a non-trivial cube root of unity, then
$|\mathrm{Aut}(S)|\leqslant 648.$ Moreover, $|\mathrm{Aut}(S)|=648$ if and
only if

$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$

and $S$ is isomorphic to the Fermat cubic surface.

    2. (ii)5

If $\mathbf{F}$ does not contain a non-trivial cube root of unity, then

$|\mathrm{Aut}(S)|\leqslant 72.$

Moreover, $|\mathrm{Aut}(S)|=72$ if and only if

$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{2}\rtimes\mathfrak{D}_{4}$

and $S$ is isomorphic to smooth cubic surface (1.2).

  * •

Assume that the characteristic of $\mathbf{F}$ is different from $2,$ $3$ and
$5.$

    1. (i)≠2,3,5

If $\mathbf{F}$ contains a non-trivial cube root of unity, then
$|\mathrm{Aut}(S)|\leqslant 648.$ Moreover, $|\mathrm{Aut}(S)|=648$ if and
only if

$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$

and $S$ is isomorphic to the Fermat cubic surface.

    2. (ii)≠2,3,5

If $\mathbf{F}$ does not contain a non-trivial cube root of unity, then

$|\mathrm{Aut}(S)|\leqslant 120.$

Moreover, $|\mathrm{Aut}(S)|=120$ if and only if

$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$

and $S$ is isomorphic to the Clebsch cubic surface.

Note that in the paper [13] the classification of automorphism groups of
smooth cubic surfaces over all fields of characteristic zero was given. One
can deduce Theorem 1.3 for fields of characteristic zero from these results
(see [13, Theorem 1.1]).

From Theorem 1.3 we immediately get the following corollary, which, together
with Theorem 1.2, completes the classification of the largest automorphism
groups of smooth cubic surfaces over finite fields.

######  Corollary 1.4.

Let $S$ be a smooth cubic surface over a finite field $\mathbf{F}$ of odd
characteristic.

  * •

Assume that the characteristic of $\mathbf{F}$ is equal to $3.$

    1. (i)3

If $\mathbf{F}=\mathbb{F}_{9^{k}},$ then $|\mathrm{Aut}(S)|\leqslant 216.$
Moreover, $|\mathrm{Aut}(S)|=216$ if and only if

$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$

and $S$ is isomorphic to smooth cubic surface (1.1).

    2. (ii)3

If $\mathbf{F}=\mathbb{F}_{3^{2k+1}},$ then $|\mathrm{Aut}(S)|\leqslant 120.$
Moreover, $|\mathrm{Aut}(S)|=120$ if and only if

$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$

and $S$ is isomorphic to the Clebsch cubic surface.

  * •

Assume that the characteristic of $\mathbf{F}$ is equal to $5.$

    1. (i)5

If $\mathbf{F}=\mathbb{F}_{25^{k}},$ then $|\mathrm{Aut}(S)|\leqslant 648.$
Moreover, $|\mathrm{Aut}(S)|=648$ if and only if

$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$

and $S$ is isomorphic to the Fermat cubic surface.

    2. (ii)5

If $\mathbf{F}=\mathbb{F}_{5^{2k+1}},$ then $|\mathrm{Aut}(S)|\leqslant 72.$
Moreover, $|\mathrm{Aut}(S)|=72$ if and only if

$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{2}\rtimes\mathfrak{D}_{4}$

and $S$ is isomorphic to smooth cubic surface (1.2).

  * •

Assume that the characteristic of $\mathbf{F}$ is at least $7.$

    1. (i)⩾7

If the cardinality of $\mathbf{F}$ is congruent to $1$ modulo $3,$ then
$|\mathrm{Aut}(S)|\leqslant 648.$ Moreover, $|\mathrm{Aut}(S)|=648$ if and
only if

$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$

and $S$ is isomorphic to the Fermat cubic surface.

    2. (ii)⩾7

If the cardinality of $\mathbf{F}$ is congruent to $2$ modulo $3,$ then
$|\mathrm{Aut}(S)|\leqslant 120.$ Moreover, $|\mathrm{Aut}(S)|=120$ if and
only if

$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$

and $S$ is isomorphic to the Clebsch cubic surface.

In the following proposition we study isomorphism relations among the Fermat
cubic surface, the Clebsch cubic surface and cubic surfaces (1.1) and (1.2)
over fields of arbitrary characteristic.

######  Proposition 1.5.

Let $\mathbf{F}$ be a field.

  1. (i)

The Fermat cubic surface is singular (in fact, non-reduced) if and only if
$\mathbf{F}$ is of characteristic $3.$ The Clebsch cubic surface is singular
if and only if $\mathbf{F}$ is of characteristic $5.$ Cubic surface (1.1) is
singular if and only if $\mathbf{F}$ is of characteristic $2.$ Cubic surface
(1.2) is singular if and only if $\mathbf{F}$ is of characteristic $3.$

  2. (ii)

Cubic surface (1.1) is isomorphic neither to the Fermat cubic surface, nor to
the Clebsch cubic surface, nor to cubic surface (1.2).

  3. (iii)

If $\mathbf{F}$ is of characteristic $2,$ then cubic surface (1.2) and the
Clebsch cubic surface are isomorphic.

  4. (iv)

If $\mathbf{F}$ is of characteristic different from $3,$ then cubic surface
(1.2) is isomorphic to the Fermat cubic surface if and only if $\mathbf{F}$
contains a non-trivial cube root of unity.

  5. (v)

If $\mathbf{F}$ is of characteristic different from $2,$ the Clebsch cubic
surface is isomorphic neither to the Fermat cubic surface, nor to cubic
surface (1.2).

We are truly excited about the following question.

######  Question 1.6.

Does Theorem 1.2 hold without assuming finiteness of the field $\mathbf{F}?$

Now let us describe the structure of the paper. In Section 2 we prove some
auxiliary lemmas and recall some facts about Weyl group $W(\mathrm{E}_{6}),$
its subgroups and elements. In Section 3 we recall some basic facts about
smooth cubic surfaces and discuss their automorphisms. In Section 4 we study
smooth cubic surfaces with automorphism group $\mathfrak{S}_{5}$ and prove
that such cubic surfaces are isomorphic to the Clebsch cubic surface. In
Section 5 we study smooth cubic surfaces and their automorphisms over fields
of characteristic $3.$ In Section 6 we study smooth cubic surfaces with
automorphism group
$\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$ and prove
that such cubic surfaces are isomorphic to the Fermat cubic surface. In
Section 7 we study smooth cubic surfaces and their automorphism groups over
fields that do not contain non-trivial cube root of unity. In Section 8 we
prove Theorem 1.3 and Proposition 1.5.

Notation. Let $X$ be a variety defined over a field $\mathbf{F}.$ If
$\mathbf{F}\subset\mathbf{L}$ is an extension of $\mathbf{F},$ then we will
denote by $X_{\mathbf{L}}$ the variety

$X_{\mathbf{L}}=X\times_{\mathrm{Spec}(\mathbf{F})}\mathrm{Spec}(\mathbf{L})$

over $\mathbf{L}.$ By $\overline{\mathbf{F}}$ we denote the algebraic closure
of $\mathbf{F}.$ By $\mathbf{F}^{sep}$ we denote the separable closure of
$\mathbf{F}.$ By $\mathfrak{S}_{n}$ we denote the symmetric group of degree
$n,$ by $\mathfrak{A}_{n}$ we denote the alternating group of degree $n$ and
by $\mathfrak{D}_{n}$ we denote the dihedral group of order $2n.$ By
$\mathcal{H}_{3}(\mathbb{F}_{3})$ we denote the Heisenberg group of upper
triangular $3\times 3$ matrices over the field $\mathbb{F}_{3}$ with diagonal
entries equal to $1.$

Acknowledgment. The author would like to extend her sincere gratitude to C. A.
Shramov for proposing this problem, constant support and for teaching the
author to write readable mathematical papers. Also the author would like to
thank A. S. Trepalin for deep reading of the paper and for useful advice.

##  2\. Preliminaries

In this section we prove some auxiliary lemmas and recall some facts
concerning the Weyl group $W(\mathrm{E}_{6}).$ Let us start with a general
observation on del Pezzo surfaces of small degree; in the sequel almost
everywhere we use it without explicit reference.

######  Theorem 2.1 (cf. [10, Theorem 4.2] and [12, Lemma 7.1]).

Let $X$ be a del Pezzo surface of degree $d\leqslant 5$ over a field
$\mathbf{F}.$ Then the following holds.

  1. (i)

We have
$\mathrm{Aut}(X_{\overline{\mathbf{F}}})\simeq\mathrm{Aut}(X_{\mathbf{F}^{sep}}).$

  2. (ii)

The lines on $X_{\overline{\mathbf{F}}}$ are defined over $\mathbf{F}^{sep}.$

  3. (iii)

We have
$\mathrm{Pic}(X_{\overline{\mathbf{F}}})\simeq\mathrm{Pic}(X_{\mathbf{F}^{sep}}).$

###### Proof.

The first two assertions can be proved as follow. Let $\mathcal{S}$ be either
$\mathrm{Aut}(X_{\overline{\mathbf{F}}}),$ or the Hilbert scheme
$\mathdutchcal{Hilb}_{l,X_{\overline{\mathbf{F}}}}$ of lines on
$X_{\overline{\mathbf{F}}}.$ The scheme $\mathcal{S}$ is zero-dimensional.
First of all, we show that $\mathcal{S}$ is a smooth scheme. Then we apply [2,
Corollary 2.2.13] to obtain that the set of points which is defined over
$\mathbf{F}^{sep}$ on $\mathcal{S}$ is dense. Since $\mathcal{S}$ is finite we
get that all points are defined over $\mathbf{F}^{sep}.$

Let us prove smoothness of $\mathcal{S}.$ It is enough to prove that the
Zariski tangent space $T_{p}\mathcal{S}$ to $\mathcal{S}$ at any point
$p\in\mathcal{S}$ satisfies the equality

(2.1) $\dim T_{p}\mathcal{S}=0.$

Assume that $\mathcal{S}=\mathrm{Aut}(X_{\overline{\mathbf{F}}}).$ It is well-
known that

$T_{p}\mathrm{Aut}(X_{\overline{\mathbf{F}}})\simeq
H^{0}(X_{\overline{\mathbf{F}}},TX_{\overline{\mathbf{F}}}),$

where $TX_{\overline{\mathbf{F}}}$ is the tangent bundle of
$X_{\overline{\mathbf{F}}}.$ Since $X_{\overline{\mathbf{F}}}$ is a del Pezzo
surface of degree $d$ at most $5$ over an algebraically closed field, it is a
blowup of $9-d>3$ points. The second Chern class of $T{\mathbb{P}^{2}}$ is
equal to $3.$ Therefore, there are no global sections of tangent bundle of
$\mathbb{P}^{2}$ vanishing at $9-d$ points. Since the global sections of
$TX_{\overline{\mathbf{F}}}$ correspond to global sections on
$T\mathbb{P}^{2}$ vanishing in $9-d$ points we get that
$H^{0}(X_{\overline{\mathbf{F}}},TX_{\overline{\mathbf{F}}})=0.$ Therefore, we
get equality (2.1) and, as a result, the smoothness of the scheme
$\mathrm{Aut}(X_{\overline{\mathbf{F}}}).$

Assume that $\mathcal{S}=\mathdutchcal{Hilb}_{l,X_{\overline{\mathbf{F}}}}.$
It is well-known that

$T_{l}\mathdutchcal{Hilb}_{l,X_{\overline{\mathbf{F}}}}\simeq
H^{0}(l,\mathcal{N}_{l/X_{\overline{\mathbf{F}}}}),$

where $\mathcal{N}_{l/X_{\overline{\mathbf{F}}}}$ is the normal bundle to the
line $l$ in $X_{\overline{\mathbf{F}}}.$ The degree of the bundle
$\mathcal{N}_{l/X_{\overline{\mathbf{F}}}}$ is equal to $-1.$ Therefore,
$H^{0}(X_{\overline{\mathbf{F}}},TX_{\overline{\mathbf{F}}})=0.$ So we get
equality (2.1) and, as a result, the smoothness of the scheme
$\mathdutchcal{Hilb}_{l,X_{\overline{\mathbf{F}}}}.$

Assertion (iii) follows from assertion (ii).

∎

######  Remark 2.2.

From Theorem 2.1(i) we immediately get that Theorem 1.1 holds under the
assumption that the field $\mathbf{F}$ is separably closed instead of
algebraically closed.

Later we will study conjugate elements in
$\mathrm{PGL}_{n}(\overline{\mathbf{F}})$ over $\mathbf{F}$ and
$\mathbf{F}^{sep}.$ First of all, we need the following auxiliary lemma about
conjugate matrices over $\mathbf{F}$ and over an algebraic extension
$\mathbf{F}\subset\mathbf{L}.$

######  Lemma 2.3.

Let $\mathbf{F}\subset\mathbf{L}$ be an algebraic extension of a field
$\mathbf{F}.$ Let $A$ and $B$ be two matrices in $\mathrm{Mat}_{n\times
n}(\mathbf{F}).$ Assume that they are conjugate in $\mathrm{Mat}_{n\times
n}(\mathbf{L}).$ Then they are conjugate in $\mathrm{Mat}_{n\times
n}(\mathbf{F}).$

###### Proof.

Since $A$ and $B$ are conjugate in $\mathrm{Mat}_{n\times n}(\mathbf{L}),$ we
obtain

$CAC^{-1}=B$

for some $C\in\mathrm{GL}_{n\times n}(\mathbf{L}).$ Let us consider the system
of linear equations

(2.2) $CA=BC$

on the entries of $C.$ Let $\Lambda$ be a subspace in $\mathrm{Mat}_{n\times
n}(\mathbf{L})$ of matrices $C$ satisfying (2.2). Then $\Lambda$ is defined
over $\mathbf{F},$ since $A$ and $B$ are matrices with entries in
$\mathbf{F}.$ Denote by $\Lambda(\mathbf{F})$ the set of $\mathbf{F}$-points
on $\Lambda.$ Since $A$ and $B$ are conjugate in
$\mathrm{Mat}_{n}(\mathbf{L})$ we get that
$\Lambda\cap\mathrm{GL}_{n}(\mathbf{L})$ is non-empty. Moreover, since
$\Lambda(\mathbf{F})$ is dense in $\Lambda$ we get that
$\Lambda(\mathbf{F})\cap\mathrm{GL}_{n}(\mathbf{L})$ is also non-empty. Thus,
we get that there is $C\in\mathrm{GL}_{n}(\mathbf{F})$ such that (2.2) holds.
This means that the elements $A$ and $B$ are conjugate in
$\mathrm{Mat}_{n\times n}(\mathbf{F}).$

∎

######  Remark 2.4.

Let us recall what conjugacy in the projective linear group means. Let
$\mathbf{F}$ be a field. Consider the natural homomorphism

$\mathrm{GL}_{n}(\mathbf{F})\stackrel{{\scriptstyle\rho}}{{\twoheadrightarrow}}\mathrm{PGL}_{n}(\mathbf{F}).$

Let $a$ and $b$ be two elements in $\mathrm{PGL}_{n}(\mathbf{F})$ and let
$A\in\rho^{-1}(a)$ and $B\in\rho^{-1}(b).$ Two elements
$a,b\in\mathrm{PGL}_{n}(\mathbf{F})$ are conjugate, if there are
$\lambda\in\mathbf{F}^{*}$ and $C\in\mathrm{GL}_{n}(\mathbf{F})$ such that

$CAC^{-1}=\lambda B.$

######  Lemma 2.5.

Let $\mathbf{F}$ be a field of characteristic $p\geqslant 0.$ Let us consider
$n\in\mathbb{N}$ such that $\gcd(p,n)=1.$ Let $a$ and $b$ be two elements in
$\mathrm{PGL}_{n}(\mathbf{F}^{sep})$ such that they are conjugate in
$\mathrm{PGL}_{n}(\overline{\mathbf{F}}).$ Then they are conjugate in
$\mathrm{PGL}_{n}(\mathbf{F}^{sep}).$

###### Proof.

Let us consider the natural homomorphism

$\mathrm{GL}_{n}(\mathbf{F}^{sep})\stackrel{{\scriptstyle\rho}}{{\twoheadrightarrow}}\mathrm{PGL}_{n}(\mathbf{F}^{sep}).$

Let $A\in\rho^{-1}(a)$ and $B\in\rho^{-1}(b).$ As
$a,b\in\mathrm{PGL}_{n}(\mathbf{F}^{sep})$ are conjugate in
$\mathrm{PGL}_{n}(\overline{\mathbf{F}})$ we get that there are
$\lambda\in\overline{\mathbf{F}}^{*}$ and
$C\in\mathrm{GL}_{n}(\overline{\mathbf{F}})$ such that

(2.3) $CAC^{-1}=\lambda B.$

First of all, let us prove that $\lambda\in(\mathbf{F}^{sep})^{*}.$ Taking the
determinant of (2.3) we obtain the equality $\lambda^{n}\det(B)=\det(A).$ We
get that $\lambda^{n}\in(\mathbf{F}^{sep})^{*}.$ Since $n$ and the
characteristic of $\mathbf{F}$ are coprime we get that
$\lambda\in(\mathbf{F}^{sep})^{*}.$

Let us fix some $\lambda$ which satisfies (2.3). Applying Lemma 2.3 for
matrices $A$ and $\lambda B$ we get that there is
$C\in\mathrm{GL}_{n}(\mathbf{F}^{sep})$ such that (2.3) holds. This means that
the elements $a$ and $b$ are conjugate in
$\mathrm{PGL}_{n}(\mathbf{F}^{sep}).$

∎

######  Lemma 2.6.

Let $\mathbf{F}$ be a field. Let $\mathbf{F}\subset\mathbf{L}$ be an algebraic
extension. Let $a$ and $b$ be two elements in $\mathrm{PGL}_{n}(\mathbf{F}),$
such that they are conjugate in $\mathrm{PGL}_{n}(\mathbf{L}).$ Assume that
the trace of some preimage $\rho^{-1}(a)$ of the natural homomorphism

$\mathrm{GL}_{n}(\mathbf{F})\stackrel{{\scriptstyle\rho}}{{\twoheadrightarrow}}\mathrm{PGL}_{n}(\mathbf{F})$

is non-zero. Then $a$ and $b$ are conjugate in $\mathrm{PGL}_{n}(\mathbf{F}).$

###### Proof.

Let $A\in\rho^{-1}(a)$ and $B\in\rho^{-1}(b).$ As
$a,b\in\mathrm{PGL}_{n}(\mathbf{F})$ are conjugate in
$\mathrm{PGL}_{n}(\mathbf{L})$ we get that there are
$\lambda\in\mathbf{L}^{*}$ and $C\in\mathrm{GL}_{n}(\mathbf{L})$ such that

(2.4) $CAC^{-1}=\lambda B.$

Let us prove that $\lambda\in\mathbf{F}^{*}.$ Since by assumption the trace of
$A$ is non-zero, taking the trace of (2.4) we obtain the equality
$\lambda\mathrm{tr}(B)=\mathrm{tr}(A).$ So we get that
$\lambda\in\mathbf{F}^{*}.$

Let us fix some $\lambda$ which satisfies (2.4). Applying Lemma 2.3 for
matrices $A$ and $\lambda B$ we get that there is
$C\in\mathrm{GL}_{n}(\mathbf{F})$ such that (2.4) holds. This means that the
elements $a$ and $b$ are conjugate in $\mathrm{PGL}_{n}(\mathbf{F}).$

∎

Let us recall the results from [3, Table 9] or [14, Theorem 1] about the
conjugacy classes of the elements in the Weyl group $W(\mathrm{E}_{6}).$

######  Lemma 2.7.

Conjugacy classes of elements in the Weyl group $W(\mathrm{E}_{6})$ of orders
$2$, $3$, $5$, $8,$ $9$ and $10,$ their cardinalities, orders of their
centralizers, collections of eigenvalues of the representation on the root
lattice $\mathrm{E}_{6}$ are those given in Table 1.

Order of | Conjugacy class | Cardinality | Order of | Eigenvalues
---|---|---|---|---
element |  |  | centralizer |
$2$ | $A_{1}$ | $36$ | $1440$ | $-1,1,1,1,1,1$
$2$ | $A_{1}^{2}$ | $270$ | $192$ | $-1,-1,1,1,1,1$
$2$ | $A_{1}^{3}$ | $540$ | $96$ | $-1,-1,-1,1,1,1$
$2$ | $A_{1}^{4}$ | $45$ | $1152$ | $-1,-1,-1,-1,1,1$
$3$ | $A_{2}$ | $240$ | $216$ | $\omega,\omega^{2},1,1,1,1$
$3$ | $A_{2}^{2}$ | $480$ | $108$ | $\omega,\omega,\omega^{2},\omega^{2},1,1$
$3$ | $A_{2}^{3}$ | $80$ | $648$ | $\omega,\omega,\omega,\omega^{2},\omega^{2},\omega^{2}$
$5$ | $A_{4}$ | $5184$ | $10$ | $\xi,\xi^{2},\xi^{3},\xi^{4},1,1$
$8$ | $D_{5}$ | $6480$ | $8$ | $\varepsilon,\varepsilon^{3},\varepsilon^{5},\varepsilon^{7},-1,1$
$9$ | $E_{6}(a_{1})$ | $5760$ | $9$ | $\zeta,\zeta^{2},\zeta^{4},\zeta^{5},\zeta^{7},\zeta^{8}$
$10$ | $A_{4}\times A_{1}$ | $5184$ | $10$ | $\xi,\xi^{2},\xi^{3},\xi^{4},-1,1$
Table 1. Table of conjugacy classes of elements of order $2$, $3$, $5$, $8$,
$9$ and $10$ in $W(\mathrm{E}_{6})$.

Here $\omega\in\mathbb{C}$ is a non-trivial cube root of unity,
$\xi\in\mathbb{C}$ is a non-trivial fifth root of unity,
$\varepsilon\in\mathbb{C}$ is a primitive eighth root of unity and
$\zeta\in\mathbb{C}$ is a primitive ninth root of unity.

There are $7$ conjugacy classes of elements of order $6$ in
$W(\mathrm{E}_{6}),$ however we will need only one conjugacy class of elements
of order $6$ which is denoted by $E_{6}(a_{2}).$

######  Lemma 2.8 ([14, Theorem 1]).

Let $g$ be an element in $W(\mathrm{E}_{6})$ of order $6$ lying in the
conjugacy class $E_{6}(a_{2}).$ Then its eigenvalues of the representation on
the root lattice $\mathrm{E}_{6}$ are

$-\omega,-\omega,-\omega^{2},-\omega^{2},\omega,\omega^{2},$

where $\omega\in\mathbb{C}$ is a non-trivial cube root of unity.

######  Lemma 2.9.

Let $g\in W(\mathrm{E}_{6})$ be an element of order $8.$ Then $g^{4}$ lies in
the conjugacy class $A_{1}^{4}.$

###### Proof.

By Lemma 2.7 there is a unique conjugacy class of elements of order $8$ in
$W(\mathrm{E}_{6})$ and the collection of eigenvalues of the representation of
$g$ on the root lattice $\mathrm{E}_{6}$ is

$\varepsilon,\varepsilon^{3},\varepsilon^{5},\varepsilon^{7},-1,1,$

where $\varepsilon\in\mathbb{C}$ is a primitive eighth root of unity.
Therefore, the collection of eigenvalues of the representation of $g^{4}$ on
the root lattice $\mathrm{E}_{6}$ is

$-1,-1,-1,-1,1,1.$

By Lemma 2.7 this means that $g^{4}$ lies in the conjugacy class $A_{1}^{4}.$

∎

######  Lemma 2.10.

Let $g,c\in W(\mathrm{E}_{6})$ be elements of order $2$ and $3,$ respectively,
such that they commute. Assume that $g$ lies in the conjugacy class
$A_{1}^{4}$ and $h=gc$ lies in the conjugacy class $E_{6}(a_{2}).$ Then $c$
lies in the conjugacy class $A_{2}^{3}.$

###### Proof.

Let us consider the representation of the elements $c,$ $g$ and $h$ on the
root lattice $\mathrm{E}_{6}.$ Since the elements $g,$ $c$ and $h$ commute,
they can be diagonalizable in the same basis. By Lemma 2.7 the collection of
eigenvalues of $g$ on the root lattice $\mathrm{E}_{6}$ is

$-1,-1,-1,-1,1,1.$

By Lemma 2.8 the collection of eigenvalues of $h$ is

$-\omega,-\omega,-\omega^{2},-\omega^{2},\omega,\omega^{2}.$

Therefore, by Lemma 2.7 the collection of eigenvalues of $gh=c$ can only
coincide with the collection of eigenvalues of an element lying in the
conjugacy class $A_{2}^{3}.$

∎

######  Lemma 2.11.

The centralizer of an element of order $5$ in $W(\mathrm{E}_{6})$ is
isomorphic to $\mathbb{Z}/10\mathbb{Z}.$

###### Proof.

By Lemma 2.7 the order of the centralizer of an element $g$ of order $5$ in
$W(\mathrm{E}_{6})$ is equal to $10.$ This means that the centralizer of $g$
in the Weyl group is isomorphic either to $\mathbb{Z}/10\mathbb{Z},$ or to a
dihedral group of order $10,$ since these two groups are the only groups of
order $10$ up to isomorphism. However, the second case is impossible because
$g$ lies in its centralizer.

∎

######  Lemma 2.12.

The centralizer of any subgroup in $W(\mathrm{E}_{6})$ isomorphic to
$\mathfrak{S}_{5}$ lies in a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}.$

###### Proof.

By Lemma 2.11 the centralizer
$\mathrm{C}_{W(\mathrm{E}_{6})}(\mathfrak{S}_{5})$ of the subgroup
$\mathfrak{S}_{5}$ in $W(\mathrm{E}_{6})$ lies in the subgroup
$\mathbb{Z}/10\mathbb{Z}\simeq G\subset W(\mathrm{E}_{6}),$ as $G$ is the
centralizer $\mathrm{C}_{W(\mathrm{E}_{6})}(\tau)$ of an element $\tau$ of
order $5$ in $\mathfrak{S}_{5}.$ Since
$\tau\in\mathrm{C}_{W(\mathrm{E}_{6})}(\tau)$ and since $\tau$ does not lie in
the centre of $\mathfrak{S}_{5},$ we get the desired.

∎

######  Lemma 2.13.

Let $\tau\in W(\mathrm{E}_{6})$ be an element of order $10.$ Then $\tau^{5}$
lies in the conjugacy class $A_{1}.$

###### Proof.

This directly follows from Lemma 2.7. The eigenvalues of an element lying in
$A_{4}\times A_{1}$ are

$\xi,\xi^{2},\xi^{3},\xi^{4},-1,1,$

where $\xi\in\mathbb{C}$ is a non-trivial fifth root of unity. We have
$g=\tau^{5}.$ So the eigenvalues of $\tau^{5}$ are

$-1,1,1,1,1,1$

which are the eigenvalues of elements in conjugacy class $A_{1}.$

∎

From [14, Theorem 1] we immediately get the following useful lemma.

######  Lemma 2.14.

There are no elements of order $24$ in the Weyl group $W(\mathrm{E}_{6}).$

Now we recall the classification of maximal proper subgroups in
$\mathrm{PSU}_{4}(\mathbb{F}_{2}).$

######  Lemma 2.15 ([5, p. 26]).

The maximal proper subgroups in $\mathrm{PSU}_{4}(\mathbb{F}_{2})$ are those
given in Table 2.

Subgroup | Order
---|---
$\left(\mathbb{Z}/2\mathbb{Z}\right)^{4}\rtimes\mathfrak{A}_{5}$ | $960$
$\mathfrak{S}_{6}$ | $720$
$\left(\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/2\mathbb{Z}\right)_{\mathbin{\vbox{\hbox{\scalebox{0.4}{$\bullet$}}}}}\mathfrak{A}_{4}$ | $648$
$\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$ | $648$
$\left(\mathbb{Z}/2\mathbb{Z}^{\,\mathbin{\vbox{\hbox{\scalebox{0.4}{$\bullet$}}}}}\left(\mathfrak{A}_{4}\times\mathfrak{A}_{4}\right)\right)_{\mathbin{\vbox{\hbox{\scalebox{0.4}{$\bullet$}}}}}\mathbb{Z}/2\mathbb{Z}$ | $576$
Table 2. Table of maximal subgroups in $\mathrm{PSU}_{4}(\mathbb{F}_{2})$.

Here by $G_{\mathbin{\vbox{\hbox{\scalebox{0.4}{$\bullet$}}}}}H$ we denote a
group with normal subgroup isomorphic to $G$ and quotient is isomorphic to $H$
and by $G^{\,\mathbin{\vbox{\hbox{\scalebox{0.4}{$\bullet$}}}}}H$ we denote
the group $G_{\mathbin{\vbox{\hbox{\scalebox{0.4}{$\bullet$}}}}}H$ which is
not a split extension. All subgroups in Table 2 are unique up to conjugation.

######  Lemma 2.16.

Let $G$ be a subgroup in $\mathrm{PSU}_{4}(\mathbb{F}_{2})$ isomorphic to
$\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}.$ Then $G$ is
the automorphism group of the Fermat cubic surface over a separabely closed
field of characteristic different from $2$ and $3.$

###### Proof.

By Lemma 2.15 the subgroup
$\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$ in the group
$\mathrm{PSU}_{4}(\mathbb{F}_{2})$ is unique up to conjugation. By Theorem
1.1(iv)′ and Remark 2.2 it is the automorphism group of the Fermat cubic
surface over a separably closed field of characteristic different from $2$ and
$3.$

∎

##  3\. Cubic surfaces

In this section we introduce the notation related to smooth cubic surfaces and
recall the description of their geometry. More details on smooth cubic
surfaces can be found in [6, Chapter 9] or [11, Chapter iv@].

Let $S\subset\mathbb{P}^{3}$ be a smooth cubic surface over a field
$\mathbf{F}.$ Recall that by Theorem 2.1 the lines on
$S_{\overline{\mathbf{F}}}$ are defined over $\mathbf{F}^{sep}$ and we have
$\mathrm{Pic}(S_{\overline{\mathbf{F}}})\simeq\mathrm{Pic}(S_{\mathbf{F}^{sep}}).$
So $S_{\mathbf{F}^{sep}}$ is a blowup of $6$ points

$P_{1},\;P_{2},\;P_{3},\;P_{4},\;P_{5},\;P_{6}$

in general position on $\mathbb{P}^{2}_{\mathbf{F}^{sep}}$ (this means that no
$3$ points of this set lie on a line and not all points of this set lie on a
conic). In other words, there is a birational morphism

$\pi\colon S_{\mathbf{F}^{sep}}\to\mathbb{P}^{2}_{\mathbf{F}^{sep}}$

which blows down the preimages of points $P_{1},\ldots,P_{6}.$ Denote by

$E_{1},\;E_{2},\;E_{3},\;E_{4},\;E_{5},\;E_{6}$

the preimages of $P_{1},$ $P_{2},$ $P_{3},$ $P_{4},$ $P_{5}$ and $P_{6},$
respectively. The exceptional curves are lines on
$S_{\mathbf{F}^{sep}}\subset\mathbb{P}^{3}_{\mathbf{F}^{sep}}.$ The self-
intersection of any line on a smooth cubic surface is equal to $-1.$ It is
well known that there are exactly $27$ lines on a smooth cubic surface. We
denote the other $21$ lines by $Q_{i}$ and $L_{ij}=L_{ji}$ for
$i,j\in\\{1,\ldots,6\\}$ and $i\neq j.$ Denote by $H$ the pullback of the
class of a line on $\mathbb{P}^{2}_{\mathbf{F}^{sep}}.$ Then the $27$ lines on
the cubic surface $S_{\mathbf{F}^{sep}}$ are the following effective divisors

$\displaystyle E_{i}\quad\text{for}\quad i\in\\{1,\ldots,6\\};$ $\displaystyle
L_{ij}\sim H-E_{i}-E_{j}\quad\text{for}\quad i\neq
j;\;i,j\in\\{1,\ldots,6\\};$ $\displaystyle Q_{i}\sim
2H+E_{i}-\sum_{j=1}^{6}E_{j}\quad\text{for}\quad i\in\\{1,\ldots,6\\}.$

Their intersections are:

$\displaystyle E_{i}^{2}=Q_{i}^{2}=L_{ij}^{2}=-1\quad\text{for}\quad i\neq j;$
$\displaystyle E_{i}\cdot E_{j}=Q_{i}\cdot Q_{j}=0\quad\text{for}\quad i\neq
j;$ $\displaystyle E_{i}\cdot Q_{j}=L_{ij}\cdot E_{i}=L_{ij}\cdot
Q_{i}=1\quad\text{for}\quad i\neq j;$ $\displaystyle L_{ij}\cdot
L_{kl}=1,\quad\text{where}\;i,j,k,l\in\\{1,\ldots,6\\}\;\text{are pairwise
distinct;}$ $\displaystyle L_{ij}\cdot
L_{jk}=0,\quad\text{where}\;i,j,k\in\\{1,\ldots,6\\}\;\text{are pairwise
distinct}.$

Let us consider the set

$\Delta=\\{\alpha\in\mathrm{Pic}(S_{\mathbf{F}^{sep}})\mid
K_{S_{\mathbf{F}^{sep}}}\cdot\alpha=0\;\text{and}\;\alpha^{2}=-2\\}.$

Then this set is the root system $\mathrm{E}_{6}.$ The Weyl group
$W(\mathrm{E}_{6})$ acts on the Picard group
$\mathrm{Pic}(S_{\mathbf{F}^{sep}})$ by orthogonal transformations with
respect to the intersection form. This action restricts to the configuration
of lines on $S_{\mathbf{F}^{sep}}.$ It fixes $K_{S_{\mathbf{F}^{sep}}}$ and
preserves the cone of effective divisors.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}.$ By [11, Theorem
23.8(ii)] the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ acts on
$\mathrm{Pic}(S_{\mathbf{F}^{sep}})$ and preserves the canonical class of $S$
and the intersection form. This means that there is a homomorphism

$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})\to\mathrm{Aut}(\mathrm{Pic}(S_{\mathbf{F}^{sep}}))\subset
W(\mathrm{E}_{6}).$

Recall the following important theorem.

######  Theorem 3.1 (see, for instance, [6, Corollary 8.2.40]).

The automorphism group of a smooth cubic surface $S$ is isomorphic to a
subgroup in $W(\mathrm{E}_{6}).$

######  Remark 3.2.

Although Theorem 3.1 in [6] is proved over fields of characteristic zero,
exactly the same proof also works in the case of positive characteristic.

Let $\Gamma$ be the image of theabsolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6}).$ We remind
the reader that $\Gamma$ lies in the centralizer of $\mathrm{Aut}(S),$ since
any $\phi\in\mathrm{Aut}(S)$ is defined over $\mathbf{F}.$

Recall the definition of the graph of lines on a projective variety. This is
the graph such that its vertices correspond to lines on a projective variety
and there is an edge between two vertices if and only if their corresponding
lines intersect. Hereinafter, we need the following definition.

######  Definition 3.3.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}.$ We say that $5$
lines $l_{1},\ldots,l_{5}$ on $S$ are in star position if their graph is a
star (see Figure 1).

00000 Figure 1. Lines in star position

Let us go back to the conjugacy classes of elements in $W(\mathrm{E}_{6})$ in
the light of the action on the Picard group of a smooth cubic surface over an
algebraically closed field.

######  Lemma 3.4.

Let $S$ be a smooth cubic surface over a separably closed field. Consider an
element $g\in W(\mathrm{E}_{6})$ acting on the configuration of lines on $S.$
The following assertions hold.

  1. (i)

Assume that $g$ is of order $5.$ There are $2$ lines on $S$ which are
invariant under the action of $g.$ Under the action of $g$ there are $3$
invariant quintuples of skew lines on $S,$ and in any quintuple the lines are
transitively permuted by $g.$ Moreover, under the action of $g$ there are $2$
invariant quintuple of lines in star position on $S,$ and in any quintuple the
lines are transitively permuted by $g.$

  2. (ii)

Assume that $g$ is of order $10.$ There is a unique pair of skew lines on $S$
which is invariant under the action of $g,$ and the lines in this pair are
interchanged by $g.$ There is a unique quintuple of skew lines on $S$ which is
invariant under the action of $g,$ and the lines in this quintuple are
transitively permuted by $g.$ Under the action of $g$ there are two invariant
quintuples of lines in star position on $S,$ and in any quintuple the lines
are transitively permuted by $g.$ Moreover, $10$ remaining lines on $S$ form a
$g$-invariant decuple of lines on $S,$ and the lines in this decuple are
transitively permuted by $g.$

###### Proof.

Over an algebraically closed field the lemma follows from [1, Table 7.1,
column 8]. From Theorem 2.1(ii) we obtain the statement for a separably closed
field.

∎

######  Lemma 3.5.

Let $S$ be a smooth cubic surface over a separably closed field. Let $l_{1}$
and $l_{2}$ be two intersecting lines on $S.$ Then there is a unique line
$l_{3}\subset S$ which intersects both of $l_{1}$ and $l_{2}.$

###### Proof.

Consider the hyperplane $T$ in $\mathbb{P}^{3}$ such that $l_{1},l_{2}\subset
T.$ Then

(3.1) $T\cap S=l_{1}\cup l_{2}\cup l_{3}$

for some line $l_{3}\subset S.$ Assume that there is another line $l_{4}$
which does not lie in $T$ and which intersects both $l_{1}$ and $l_{2}.$ By
(3.1) we get that

(3.2) $-K_{S}=l_{1}+l_{2}+l_{3}.$

Therefore, on the one hand, we get $-K_{S}\cdot l_{4}=1,$ and on the other
hand, by (3.2) we obtain

$-K_{S}\cdot l_{4}=l_{1}\cdot l_{4}+l_{2}\cdot l_{4}+l_{3}\cdot
l_{4}=2+l_{3}\cdot l_{4}\neq 1.$

This contradiction gives us the uniqueness of the line which intersects both
$l_{1}$ and $l_{2}.$

∎

######  Lemma 3.6.

Let $S$ be a smooth cubic surface over a separably closed field. Consider an
element $g\in\mathrm{Aut}(S)$ of order $5$ acting on $S.$ Then the two lines
which are invariant under the action of $g$ are skew lines.

###### Proof.

Let $l_{1}$ and $l_{2}$ be two lines on $S$ which are invariant under the
action of $g.$ Assume that $l_{1}$ and $l_{2}$ intersect. Then by Lemma 3.5
there is a unique line $l_{3}$ such that $l_{3}$ intersects both $l_{1}$ and
$l_{2}.$ However, since $l_{1}$ and $l_{2}$ are fixed by $g,$ then $l_{3}$ is
also fixed by $g.$ This contradicts Lemma 3.4(i).

∎

######  Lemma 3.7.

Let $\mathbf{F}$ be a separably closed field of characteristic different from
$2$ and $3.$ Let $S$ be a smooth cubic surface over $\mathbf{F}.$ Let $g$ be
an automorphism of $S$ of order $3.$ Denote by $\omega\in\mathbf{F}$ a non-
trivial cube root of unity. Then for the automorphism $g$ one of the
followings holds.

  1. (i)

The automorphism $g$ lies in the conjugacy class $A_{2}$ and up to conjugation
the action on $\mathbb{P}^{3}$ is given by

$\begin{pmatrix}\omega^{2}&0&0&0\\\ 0&\omega^{2}&0&0\\\ 0&0&\omega&0\\\
0&0&0&\omega\end{pmatrix}.$

  2. (ii)

The automorphism $g$ lies in the conjugacy class $A_{2}^{2}$ and up to
conjugation the action on $\mathbb{P}^{3}$ is given by

$\begin{pmatrix}\omega^{2}&0&0&0\\\ 0&\omega&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{pmatrix}.$

  3. (iii)

The automorphism $g$ lies in the conjugacy class $A_{2}^{3}$ and up to
conjugation the action on $\mathbb{P}^{3}$ is given by

$\begin{pmatrix}\omega&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{pmatrix}.$

###### Proof.

For an algebraically closed field this lemma follows from [7, Theorem 10.4].
Applying Lemma 2.5 we get the statement for a separably closed field.

∎

######  Corollary 3.8.

If $g$ is an automorphism of order $3$ of a smooth cubic surface over a field
$\mathbf{F}$ of characteristic different from $2$ and $3,$ such that
$\mathbf{F}$ does not contain non-trivial cube root of unity, then $g$ does
not lie in the conjugacy class $A_{2}^{3}.$

###### Proof.

Assume that $g$ lies in the conjugacy class $A_{2}^{3}.$ By Lemma 3.7(iii) we
get that over $\mathbf{F}^{sep}$ up to conjugation the element $g$ is given by
the matrix

$A=\begin{pmatrix}\omega&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{pmatrix}.$

Let us consider the characteristic polynomial of $aA,$ where
$a\in\overline{\mathbf{F}}^{*}.$ This polynomial is of the form

$F(\lambda)=(\lambda-a\omega)(\lambda-a)^{3}=\lambda^{4}-a\lambda^{3}(3+\omega)+a^{2}\lambda^{2}(3+\omega)-a^{3}\lambda(3+\omega)+a^{4}\omega.$

Since $g$ is an automorphism of a smooth cubic surface over the field
$\mathbf{F},$ we get that there is $a\in\overline{\mathbf{F}}^{*}$ such that
all coefficient of $F(\lambda)$ lie in $\mathbf{F}.$ In particular, we get
that quotient of the coefficient of $\lambda^{2}$ and the square of the
coefficient of $\lambda^{3}$

$\frac{a^{2}(3+\omega)}{a^{2}(3+\omega)^{2}}=\frac{1}{3+\omega}\in\mathbf{F}.$

But by the assumption this is not true. This contradiction concludes the
proof.

∎

The contrapositive statement of Corollary 3.8 is the following.

######  Corollary 3.9.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different form $2$ and $3.$ Assume that there is an element
$g\in\mathrm{Aut}(S)$ which lies in the conjugacy class $A^{3}_{2}.$ Then
$\mathbf{F}$ contains a non-trivial cube root of unity
$\omega\in\mathbf{F}^{sep}.$

Now we remind the structure of the automorphism group of the Fermat cubic
surface.

######  Example 3.10.

Let us consider the Fermat cubic surface $S$ over a field $\mathbf{F}$ of
characteristic different from $2$ and $3.$ Assume that a non-trivial cube root
of unity $\omega\in\mathbf{F}^{sep}$ lies in $\mathbf{F}.$ It is not hard to
see that

$\mathrm{Aut}(S)\supseteq\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4},$

where $\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}$ acts as a multiplication by
cube roots of unity on the homogeneous coordinates $x,y,z,t$ on
$\mathbb{P}^{3}$ and $\mathfrak{S}_{4}$ acts as a permutation of homogeneous
coordinates $x,y,z,t.$ By Theorem 1.1 (iii)′ and (iv)′ and Remark 2.2 we get
that

$\mathrm{Aut}(S_{\mathbf{F}^{sep}})\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}.$

Therefore, we obtain
$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}.$

######  Lemma 3.11.

There are elements of the conjugacy class $A_{2}^{3}$ in the group

$\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}\subset\mathrm{PSU}_{4}(\mathbb{F}_{2})\subset
W(\mathrm{E}_{6}).$

Moreover, if

$\mathrm{pr}\colon\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}\to\mathfrak{S}_{4}$

is the projection homomorphism to $\mathfrak{S}_{4}$, then all elements from
the conjugacy class $A_{2}^{3}$ lie in $\mathrm{Ker}(\mathrm{pr}).$

###### Proof.

By Lemma 2.16 the subgroup
$\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$ in the group
$\mathrm{PSU}_{4}(\mathbb{F}_{2})$ is the automorphism group of the Fermat
cubic surface over a separably closed field of characteristic different from
$2$ and $3.$ So by Lemma 3.7 and Example 3.10 we get that the elements of the
group $\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}\rtimes\mathfrak{S}_{4}$ from
the conjugacy class $A_{2}^{3}$ lie in $\mathrm{Ker}(\mathrm{pr}).$

∎

######  Lemma 3.12.

Let $G=(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ be a subgroup in
$\mathrm{PSU}_{4}(\mathbb{F}_{2}).$ Let $g$ be a non-trivial element of the
normal subgroup $(\mathbb{Z}/3\mathbb{Z})^{3}$ in $G.$ Then the centralizer of
$g$ in the group $G$ is isomorphic to $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes D,$
where $D$ is isomorphic either to $\mathbb{Z}/2\mathbb{Z},$ or to
$(\mathbb{Z}/2\mathbb{Z})^{2},$ or to $\mathfrak{S}_{3}.$

###### Proof.

By Lemma 2.16 the subgroup $G$ in the group $\mathrm{PSU}_{4}(\mathbb{F}_{2})$
is the automorphism group of the Fermat cubic surface over a separably closed
field of characteristic different from $2$ and $3.$ By Example 3.10 this means
that the normal subgroup $(\mathbb{Z}/3\mathbb{Z})^{3}$ in $G$ consists of the
elements

(3.3) $\begin{pmatrix}\omega^{a}&0&0&0\\\ 0&\omega^{b}&0&0\\\
0&0&\omega^{c}&0\\\
0&0&0&\omega^{d}\end{pmatrix}\in\mathrm{PGL}_{4}(\mathbf{F})$

where $\omega\in\mathbf{F}$ is a non-trivial cube root of unity in a field
$\mathbf{F}$ of characteristic different from $2$ and $3$ and
$a,b,c,d\in\\{0,1,2\\}.$ The subgroup $\mathfrak{S}_{4}$ acts on
$(\mathbb{Z}/3\mathbb{Z})^{3}$ as a permutation of the diagonal elements in
the representation of the elements in $(\mathbb{Z}/3\mathbb{Z})^{3}$ as
elements (3.3) in $\mathrm{PGL}_{4}(\mathbf{F}).$

Now let us consider all possible centralizers of a non-trivial element
$g\in(\mathbb{Z}/3\mathbb{Z})^{3}$ of the group $G.$ Up to conjugation we have
the following cases.

  * •

Let $g$ be an element such that $a=b,$ $c\neq d$ and $c,d\neq a.$ Then the
centralizer of $g$ of the group $G$ is
$\mathrm{C}_{G}(g)\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathbb{Z}/2\mathbb{Z}.$

  * •

Let $g$ be an element such that $a=b,$ $c=d$ and $c\neq a.$ Then the
centralizer of $g$ of the group $G$ is
$\mathrm{C}_{G}(g)\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes(\mathbb{Z}/2\mathbb{Z})^{2}.$

  * •

Let $g$ be an element such that $a=b=c$ and $d\neq a.$ Then the centralizer of
$g$ of the group $G$ is
$\mathrm{C}_{G}(g)\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{3}.$

∎

##  4\. The group $\mathfrak{S}_{5}$ and the Clebsch cubic surface

In this section we consider the group $\mathfrak{S}_{5}$ as the automorphism
group of a smooth cubic surface over an arbitrary field $\mathbf{F}$ of
characteristic different from $5.$ First of all, let us recall the structure
of the automorphism group of the Clebsch cubic surface.

######  Example 4.1.

It is obvious that the automorphism group of the Clebsch cubic surface over an
arbitrary field $\mathbf{F}$ of characteristic different from $5$ contains the
group $\mathfrak{S}_{5}$ which acts on $\mathbb{P}^{4}$ as a permutation of
its homogeneous coordinates $x$, $y$, $z$, $t$, $w.$ Note that by Lemma 2.12
the image of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6})$ in this
case lies in the group isomorphic to $\mathbb{Z}/2\mathbb{Z}.$

######  Definition 4.2.

We say that the points $P_{1},\ldots,P_{i}\in\mathbb{P}^{2}$ for $i\leqslant
6$ are in general position if no $3$ points of this set lie on a line and no
$6$ points of this set lie on a conic. Note that this means that any of the
points $P_{1},\ldots,P_{i}\in\mathbb{P}^{2}$ for $i\leqslant 6$ is defined
over the considered field.

######  Remark 4.3.

Let $P_{1}$, $P_{2}$, $P_{3}$, $P_{4}$ be $4$ points in general position on
$\mathbb{P}^{2}.$ Then this quadruple of points is unique up to a
collineation.

######  Lemma 4.4.

Let $F$ be a smooth del Pezzo surface of degree $5$ over a field $\mathbf{F}.$
Assume that all lines on $F_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F}.$
Then such $F$ is unique up to isomorphism.

###### Proof.

As all lines on $F_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F},$ we can
blow down $4$ skew lines and get $\mathbb{P}^{2}.$ The surface $F$ is a blowup
of $4$ points in general position on $\mathbb{P}^{2}.$ By Remark 4.3 four
points in general position on $\mathbb{P}^{2}$ are uniquely defined up to an
automorphism of $\mathbb{P}^{2}.$ Therefore, a del Pezzo surface of degree $5$
such that all lines on it over $\mathbf{F}^{sep}$ are defined over
$\mathbf{F}$ is unique up to isomorphism.

∎

######  Example 4.5.

Let $\mathbf{F}$ be a separably closed field of characteristic different from
$5.$ Let us construct a smooth del Pezzo surface of degree $5$ with a non-
trivial regular action of $\mathbb{Z}/5\mathbb{Z}$ over the field
$\mathbf{F}.$ Let $C\subset\mathbb{P}^{2}$ be a smooth conic with a non-
trivial action of $\mathbb{Z}/5\mathbb{Z}.$ The action of
$\mathbb{Z}/5\mathbb{Z}$ on $C$ extends to the action of $\mathbb{P}^{2}.$ The
group $\mathbb{Z}/5\mathbb{Z}$ fixes $2$ points $P_{1}$ and $P_{2}$ on $C.$
Therefore, it fixes the line $\mathcal{L}\subset\mathbb{P}^{2}$ which passes
through $P_{1}$ and $P_{2}.$ Note that $\mathbb{Z}/5\mathbb{Z}$ acts on
$\mathcal{L}$ non-trivially, because otherwise $\mathbb{Z}/5\mathbb{Z}$ is
generated up to conjugation by the element

$\begin{pmatrix}\xi&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}\in\mathrm{PGL}_{3}(\mathbf{F}),$

where $\xi\in\mathbf{F}$ is a non-trivial fifth root of unity. In this case
there are no smooth conics on $\mathbb{P}^{2}$ fixed by
$\mathbb{Z}/5\mathbb{Z}.$ Thus, there is a point $P_{3}$ out of $C$ which is
fixed by $\mathbb{Z}/5\mathbb{Z}.$

Let us blow up the orbit of $5$ points on $C$ under the action of
$\mathbb{Z}/5\mathbb{Z}.$ We get $\mathbb{Z}/5\mathbb{Z}$-equivariant morphism

$p\colon X\to\mathbb{P}^{2}$

such that $p^{-1}C$ is a $(-1)$-curve. So we can
$\mathbb{Z}/5\mathbb{Z}$-equivariantly blow down $p^{-1}C$ and get the smooth
del Pezzo surface of degree $5$ with $2$ fixed points.

Now we formulate the direct consequence of [16, Theorem 1.5].

######  Lemma 4.6.

Let $F$ be a smooth del Pezzo surface of degree $5$ over a field $\mathbf{F}$
such that all lines on $F_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F}.$
Then

$\mathrm{Aut}(F)\simeq\mathfrak{S}_{5}.$

######  Corollary 4.7.

Let $F$ be a smooth del Pezzo surface of degree $5$ with a non-trivial action
of $\mathbb{Z}/5\mathbb{Z}$ over an arbitrary field $\mathbf{F}$ of
characteristic different from $5.$ Then this action fixes exactly $2$ points
on $F_{\mathbf{F}^{sep}}.$

###### Proof.

By Lemma 4.4 over $\mathbf{F}^{sep}$ a smooth del Pezzo surface of degree $5$
is unique up to isomorphism. By Lemma 4.6 we have

$\mathrm{Aut}(F_{\mathbf{F}^{sep}})\simeq\mathfrak{S}_{5}.$

Therefore, as up to conjugation there is a unique element of order $5$ in
$\mathfrak{S}_{5},$ by Example 4.5 we get that $\mathbb{Z}/5\mathbb{Z}$ fixes
exactly $2$ points on $F_{\mathbf{F}^{sep}}.$

∎

######  Lemma 4.8.

Let $F$ be a smooth del Pezzo surface of degree $5$ over a field $\mathbf{F}$
of characteristic different from $5.$ Assume that all lines on
$F_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F}.$ Then such del Pezzo
surface is unique up to isomorphism. Moreover, $F$ possesses the non-trivial
action of $\mathbb{Z}/5\mathbb{Z},$ and such an action is unique up to
conjugation.

###### Proof.

By Lemma 4.4 such $F$ is unique up to isomorphism. By Lemma 4.6 the
automorphism group of $F$ is isomorphic to $\mathfrak{S}_{5}.$ So up to
conjugation there is a unique action of $\mathbb{Z}/5\mathbb{Z}$ on $F.$

∎

######  Lemma 4.9.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ with the regular
action of $G\simeq\mathbb{Z}/5\mathbb{Z}$ such that the image $\Gamma$ of the
absolute Galois group $\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in the group
$W(\mathrm{E}_{6})$ lies in the group isomorphic to $\mathbb{Z}/2\mathbb{Z}.$
Then $G$-equivariant blowdown of $2$ skew lines

$\pi\colon S\to F$

gives a del Pezzo surface $F$ of degree $5$ such that any line on
$F_{\mathbf{F}^{sep}}$ is defined over $\mathbf{F}.$

###### Proof.

By Lemmas 3.4(i) and 3.6 there are two skew lines $l_{1}$ and $l_{2}$ on $S$
which are fixed by $G.$ Suppose that $\Gamma$ is a trivial group. The blowdown
of them gives a $G$-equivariant morphism from $S$ to a del Pezzo surface $F$
of degree $5$ such that any line on $F_{\mathbf{F}^{sep}}$ is defined over
$\mathbf{F}.$

Now assume that $\Gamma\simeq\mathbb{Z}/2\mathbb{Z}.$ Let $g$ be a generator
of $G$ and $\tau$ be a generator of $\Gamma.$ Then $g$ and $\tau$ commute. So
the element $\sigma=g\tau$ is of order $10.$ By Lemma 2.13 the element $\tau$
lies in the conjugacy class $A_{1}.$ By Lemma 3.4 the lines $l_{1}$ and
$l_{2}$ are interchanged by the element $\sigma.$ This means that they form
$\Gamma$-orbit. This means that we can blowdown this $\Gamma$-orbit and get a
del Pezzo surface of degree $5$

$\pi\colon S\to F.$

Since $l_{1}$ and $l_{2}$ are fixed by $G,$ we get that $\pi$ is
$G$-equivariant blowdown.

By the property of blowup we have

$\mathrm{Pic}(S_{\mathbf{F}^{sep}})=\mathrm{Pic}(F_{\mathbf{F}^{sep}})\oplus\mathbb{Z}\langle
l_{1},l_{2}\rangle.$

By Lemma 2.7 the eigenvalues of the action of $\tau$ on
$\mathrm{Pic}(S_{\mathbf{F}^{sep}})$ are

$1,1,1,1,1,1,-1.$

Moreover, the element $\tau$ acts non-trivial on the sublattice
$\mathbb{Z}\langle l_{1},l_{2}\rangle,$ because $\\{l_{1},l_{2}\\}$ forms
$\Gamma$-orbit of length $2.$ This means that $\tau$ acts trivially on
$\mathrm{Pic}(F_{\mathbf{F}^{sep}}).$ Therefore, all lines on
$F_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F}.$

∎

######  Lemma 4.10.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different from $5$ such that $\mathrm{Aut}(S)\supset\mathbb{Z}/5\mathbb{Z}$
and the image $\Gamma$ of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6})$ lies in the
group isomorphic to $\mathbb{Z}/2\mathbb{Z}.$ Then $S$ is isomorphic to the
Clebsch cubic surface.

###### Proof.

Let us consider the element $g\in\mathrm{Aut}(S)$ of order $5.$ In the Weyl
group $W(\mathrm{E}_{6})$ this element is unique up to conjugation by Lemma
2.7. By Lemmas 3.4(i) and 3.6 there are $2$ skew lines $l_{1}$ and $l_{2}$
which are invariant under the action of $g.$ After blowing them down we get a
del Pezzo surface $F$ of degree $5$ with the regular action of
$\mathbb{Z}/5\mathbb{Z}.$

By Lemma 4.9 there is $\mathbb{Z}/5\mathbb{Z}$-equivariant blowdown of $S$ to
a del Pezzo surface $F$ of degree $5$

$\pi\colon S\to F$

such that all lines on $F_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F}.$
By Lemma 4.8 such $F$ is unique up to isomorphism and on $F$ there is a non-
trivial regular action of $\mathbb{Z}/5\mathbb{Z}$ which is unique up to
conjugation. By Corollary 4.7 the number of fixed points of the action of
$\mathbb{Z}/5\mathbb{Z}$ on $F$ is equal to $2.$ So the set of two points on
$F$ which are fixed by a regular action of $\mathbb{Z}/5\mathbb{Z}$ is unique
up to automorphism, since the regular action of $\mathbb{Z}/5\mathbb{Z}$ is
unique up to conjugation. Therefore, $\mathbb{Z}/5\mathbb{Z}$-equivariant
blowup of two points is unique up to isomorphism. So we get the uniqueness of
a smooth cubic surface with the required properties. By Example 4.1 the group
$\mathbb{Z}/5\mathbb{Z}$ acts regularly on the Clebsch cubic surface. This
means that $S$ is isomorphic to the Clebsch cubic surface.

∎

Now we give the direct corollary of this lemma which we need in the future for
proving the uniqueness of smooth cubic surface with automorphism group
isomorphic to $\mathfrak{S}_{5}$ over a field of characteristic different from
$2$ and $5.$

######  Corollary 4.11.

Let $\mathbf{F}$ be a field of characteristic different from $2$ and $5.$ Let
$S$ be a smooth cubic surface over $\mathbf{F}$ such that
$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}.$ Then $S$ is isomorphic to the Clebsch
cubic surface.

###### Proof.

Since $\mathrm{Aut}(S)\simeq\mathfrak{S}_{5},$ by Lemma 2.12 the image
$\Gamma$ of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6})$ lies in
$\mathbb{Z}/2\mathbb{Z}.$ So we can apply Lemma 4.10 and get that the smooth
cubic surface $S$ such that $\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$ is
isomorphic to the Clebsch cubic surface.

∎

Before we give the consequences of Lemma 4.10 for characteristic $2,$ let us
first formulate the following two lemmas.

######  Lemma 4.12.

Let $\mathbf{F}$ be a field of characteristic $2.$ Then if there is a non-
trivial cube root of unity in $\mathbf{F},$ then
$\mathbb{F}_{4}\subset\mathbf{F}.$

###### Proof.

Since the characteristic of $\mathbf{F}$ is equal to $2$ we get that
$\mathbb{F}_{2}\subset\mathbf{F}.$ As the non-trivial cube root of unity
$\omega\in\mathbf{F}^{sep}$ lies in $\mathbf{F},$ we get

$\mathbf{F}\supset\mathbb{F}_{2}(\omega)=\mathbb{F}_{2}[t]/(t^{2}+t+1)=\mathbb{F}_{4}.$

∎

######  Lemma 4.13 ([15, Lemmas 5.13 and 5.15]).

Let $\mathbf{F}$ be a finite field of characteristic $2.$ Let $S$ be a smooth
cubic surface over $\mathbf{F}.$

  1. (i)

If $\mathbf{F}=\mathbb{F}_{4^{k}}$ for some $k\in\mathbb{N}$ and
$\mathrm{Aut}(S)\simeq\mathrm{PSU}_{4}(\mathbb{F}_{2}),$ then the image of the
absolute Galois group $\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in
$W(\mathrm{E}_{6})$ is trivial.

  2. (ii)

If $\mathbf{F}=\mathbb{F}_{2^{2k+1}}$ for some $k\in\mathbb{Z}_{\geqslant 0}$
and $\mathrm{Aut}(S)\simeq\mathfrak{S}_{6},$ then the image of the absolute
Galois group $\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in
$W(\mathrm{E}_{6})$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

######  Corollary 4.14.

Let $\mathbf{F}$ be a field of characteristic $2.$ Then the following holds.

  1. (i)

If $\mathbf{F}$ contains a non-trivial cube root of unity, the Fermat cubic
surface is isomorphic to the Clebsch cubic surface.

  2. (ii)

Cubic surface (1.2) is isomorphic to the Clebsch cubic surface.

###### Proof.

Let us prove assertion (i). Let $S$ be the Fermat cubic surface over
$\mathbb{F}_{4}.$ By Theorem 1.3(i) we have
$\mathrm{Aut}(S)\simeq\mathrm{PSU}_{4}(\mathbb{F}_{2}),$ so
$\mathrm{Aut}(S)\supset\mathbb{Z}/5\mathbb{Z}.$ By Lemma 4.13(i) the image
$\Gamma$ of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6})$ is trivial.
So by Lemma 4.10 the cubic surface $S$ is isomorphic to the Clebsch cubic
surface over $\mathbb{F}_{4}.$ By Lemma 4.12 the field $\mathbf{F}$ contains
$\mathbb{F}_{4}.$ Therefore, the Fermat cubic surface is isomorphic to the
Clebsch cubic surface over any extension $\mathbf{F}$ of $\mathbb{F}_{4}.$

Let us prove assertion (ii). Let $S$ be cubic surface (1.2) over
$\mathbb{F}_{2}.$ By Theorem 1.2(ii) we have
$\mathrm{Aut}(S)\simeq\mathfrak{S}_{6},$ so
$\mathrm{Aut}(S)\supset\mathbb{Z}/5\mathbb{Z}.$ By Lemma 4.13(ii) the image
$\Gamma$ of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6})$ is
isomorphic to $\mathbb{Z}/2\mathbb{Z}$. So by Lemma 4.10 the cubic surface $S$
is isomorphic to the Clebsch cubic surface over $\mathbb{F}_{2}.$ Therefore,
cubic surface (1.2) is isomorphic to the Clebsch cubic surface over any
extension $\mathbf{F}$ of $\mathbb{F}_{2}.$

∎

##  5\. Characteristic $3$

In this section we study smooth cubic surfaces with the largest automorphism
group over fields of characteristic $3.$ First of all, we formulate three
obvious lemmas about fields of characteristic $3.$

######  Lemma 5.1.

Let $\mathbf{F}$ be a field of characteristic $3.$ Then $\mathbf{F}$ contains
a primitive fourth root of unity if and only if $\mathbf{F}$ contains the
subfield $\mathbb{F}_{9}.$

###### Proof.

Since the characteristic of $\mathbf{F}$ is equal to $3$ we get that
$\mathbb{F}_{3}\subset\mathbf{F}.$ Assume that a primitive fourth root of
unity lies in $\mathbf{F}.$ Denote it by $\mathbf{i}.$ Then we have

$\mathbf{F}\supset\mathbb{F}_{3}(\mathbf{i})=\mathbb{F}_{3}[t]/(t^{2}+1)=\mathbb{F}_{9}.$

Assume that $\mathbf{F}\supset\mathbb{F}_{9}.$ Then there are elements in
$\mathbf{F}$ satisfying the equation $t^{8}=1.$ In particular, there are
elements in $\mathbf{F}$ satisfying the equation $t^{4}=1.$ This means that
there is a primitive fourth root of unity in $\mathbf{F}.$

∎

######  Lemma 5.2.

Let $\mathbf{F}$ be a field of characteristic $3.$ Then $\mathbf{F}$ contains
a primitive fourth root of unity if and only if $\mathbf{F}$ contains a
primitive eighth root of unity.

###### Proof.

If $\mathbf{F}$ contains a primitive fourth root of unity, then by Lemma 5.1
we have $\mathbf{F}\supset\mathbb{F}_{9}.$ Therefore, there are elements in
$\mathbf{F}$ which satisfy

$t^{8}=1\;\text{and}\;t^{i}\neq 1\;\text{for any integer}\;1\leqslant
i\leqslant 7.$

This means that there is a primitive eighth root of unity in $\mathbf{F}.$ If
$\mathbf{F}$ does not contain a primitive fourth root of unity, then it is
obvious that it does not contain a primitive eighth root of unity.

∎

######  Lemma 5.3.

Let $\mathbf{F}$ be a field of characteristic $3$ which does not contain a
primitive eighth root of unity $\varepsilon\in\mathbf{F}^{sep}.$ Then
$\varepsilon\pm\varepsilon^{-1}\notin\mathbf{F}.$

###### Proof.

We have

$\left(\varepsilon\pm\varepsilon^{-1}\right)^{9}=\varepsilon\pm\varepsilon^{-1},$

so $\varepsilon\pm\varepsilon^{-1}\in\mathbb{F}_{9}\subset\mathbf{F}^{sep}.$
However,

$\left(\varepsilon\pm\varepsilon^{-1}\right)^{3}=\varepsilon^{3}\pm\varepsilon^{-3}\neq\varepsilon\pm\varepsilon^{-1}.$

This means that
$\mathbb{F}_{3}\left(\varepsilon\pm\varepsilon^{-1}\right)=\mathbb{F}_{9}\subset\mathbf{F}^{sep}.$
By Lemmas 5.1 and 5.2 we get that $\mathbb{F}_{9}\not\subset\mathbf{F}.$
Therefore, $\varepsilon\pm\varepsilon^{-1}\notin\mathbf{F}.$

∎

Let $\mathbf{F}$ be a field of characteristic $3.$ We divide this section into
two parts: in the first part we consider a field $\mathbf{F}$ which contains a
primitive fourth root of unity and in the second part we consider a field
$\mathbf{F}$ which does not contain a primitive fourth root of unity.

We start from the case when $\mathbf{F}$ contains a primitive fourth root of
unity. By Theorem 1.1(ii)′ the automorphism group of a smooth cubic surface of
maximal order over an algebraically closed field of characteristic $3$ is
isomorphic to $\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$
Let us prove that over $\mathbf{F}$ this group is realized.

######  Lemma 5.4.

Let $\mathbf{F}$ be a field of characteristic $3$ such that it contains a
primitive fourth root of unity. The automorphism group of smooth cubic surface
(1.1) over $\mathbf{F}$ is isomorphic to
$\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$

###### Proof.

By Lemma 5.2 a primitive eighth root of unity $\varepsilon\in\mathbf{F}^{sep}$
lies in $\mathbf{F}.$ Let us consider the element of order $8$

$h=\begin{pmatrix}\varepsilon^{6}&0&0&0\\\ 0&\varepsilon&0&0\\\
0&0&\varepsilon^{4}&0\\\ 0&0&0&1\end{pmatrix}\in\mathrm{PGL}_{4}(\mathbf{F}).$

Consider the group $G$ which consists of the elements

$\begin{pmatrix}1&\alpha^{3}&-\alpha^{6}&0\\\ 0&1&\alpha^{3}&0\\\ 0&0&1&0\\\
0&\alpha&c&1\end{pmatrix}\in\mathrm{PGL}_{4}(\overline{\mathbf{F}}),$

where $\alpha$ and $c$ lie in $\overline{\mathbf{F}}$ and satisfy the
equations

(5.1) $\displaystyle\alpha^{9}-\alpha=0$ (5.2) $\displaystyle
c^{3}+c-\alpha^{4}=0.$

One can see that $\alpha,c\in\mathbf{F}.$ Indeed, by Lemma 5.1 we have
$\alpha\in\mathbf{F}.$ Moreover, by the same lemma the primitive fourth root
of unity $\mathbf{i}\in\mathbf{F}^{sep}$ lies in $\mathbf{F}.$ If $\alpha=0,$
then the equation (5.2) has three solutions: $0$ and $\pm\mathbf{i}.$ If
$\alpha\neq 0,$ then from (5.1) we get that $\alpha^{4}=\pm 1.$ Therefore,
equation (5.2) can be written as $c^{3}+c\pm 1=0.$ The roots of these
equations are $\pm 1$ and $\pm 1\pm\mathbf{i}$ which lie in $\mathbf{F}.$

By straightforward computations one can check that $G$ is of order $27,$ any
non-trivial element in $G$ is of order $3$ and $G$ is non-abelian. So we get
$G\simeq\mathcal{H}_{3}(\mathbb{F}_{3}).$ So the automorphism group of smooth
cubic surface (1.1) contains the subgroup of order $216,$ which is generated
by $G$ and $h.$ By Theorem 1.1(ii)′ and Remark 2.2 over the field
$\mathbf{F}^{sep}$ the automorphism group of smooth cubic surface (1.1) is
isomorphic to $\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$
which is of order $216.$ So this means that the automorphism group of smooth
cubic surface (1.1) over $\mathbf{F}$ is isomorphic to
$\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$

∎

######  Lemma 5.5.

Let $\mathbf{F}$ be a field of characteristic different from $2.$ Assume that
it contains a primitive eighth root of unity $\varepsilon\in\mathbf{F}^{sep}.$
Let $S$ be a smooth cubic surface over a field $\mathbf{F}.$ Let $H$ be a
cyclic subgroup of order $8$ in the automorphism group of $S.$ Then up to
conjugation $H$ is generated by the element

(5.3) $h=\begin{pmatrix}\varepsilon^{6}&0&0&0\\\ 0&\varepsilon&0&0\\\
0&0&\varepsilon^{4}&0\\\ 0&0&0&1\end{pmatrix}\in\mathrm{PGL}_{4}(\mathbf{F}).$

###### Proof.

For an algebraically closed field of characteristic different from $2$ this
directly follows from [6, §9.5.1]. By Lemma 2.6 the lemma holds for a field
$\mathbf{F}$ of characteristic different from $2$ which contains a primitive
eighth root of unity $\varepsilon\in\mathbf{F}^{sep}.$

∎

######  Lemma 5.6.

Let $\mathbf{F}$ be a field of characteristic $3$ such that it contains a
primitive eighth root of unity $\varepsilon\in\mathbf{F}^{sep}.$ Let $S$ be a
smooth cubic surface over $\mathbf{F}$ such that

$\mathrm{Aut}(S)\supset\mathbb{Z}/8\mathbb{Z}.$

Then $S$ is isomorphic to a smooth cubic surface defined by the equation

(5.4) $\alpha t^{3}+tz^{2}-xy^{2}+x^{2}z=0$

for some $\alpha\in\mathbf{F}^{*}.$

###### Proof.

Let $h\in\mathrm{Aut}(S)$ be an automorphism of order $8.$ Then by Lemma 5.5
up to conjugation the element $h$ is of the form (5.3). So all smooth cubic
surfaces which are fixed by $h$ are isomorphic to some cubic surface defined
by the equation

(5.5) $at^{3}+btz^{2}+cxy^{2}+dx^{2}z=0$

for $a,b,c,d\in\mathbf{F}^{*}.$ Note that if at least one of the coefficient
$a,b,c$ or $d$ is zero, then (5.5) is singular. By the change of coordinates

$x\mapsto x,\;\;y\mapsto y,\;\;-\frac{d}{c}z\mapsto
z,\;\;-\frac{bc}{d^{2}}t\mapsto t$

smooth cubic surface (5.5) transforms to smooth cubic surface (5.4) with
$\alpha=\frac{ad^{6}}{b^{3}c^{4}}.$

∎

######  Lemma 5.7.

Let $\mathbf{F}$ be a field of characteristic $3$ such that it contains a
primitive eighth root of unity $\varepsilon\in\mathbf{F}^{sep}.$ Let $S$ be a
smooth cubic surface over $\mathbf{F}$ isomorphic to smooth cubic surface
(5.4) for some $\alpha\in\mathbf{F}^{*}.$ Assume that all lines on
$S_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F}.$ Then
$\alpha\in\mathbf{F}^{8}.$

###### Proof.

Let us fix $\alpha\in\mathbf{F}^{*}$ such that $S$ is isomorphic to
$S^{\prime}$ defined by the equation

$\alpha t^{3}+tz^{2}-xy^{2}+x^{2}z=0.$

The line $l$ which is defined by the system of equations in $\mathbb{P}^{3}$

$\begin{cases}x=0\\\ t=0\end{cases}$

lies on $S^{\prime}.$ Consider the pencil of hyperplanes in
$\mathbb{P}^{3}_{\mathbf{F}^{sep}}$ passing through $l.$ Any hyperplane in
this pencil is defined by the equation

(5.6) $x=\lambda
t\;\;\text{for}\;\;\lambda\in\mathbf{F}^{sep}\;\;\text{or}\;\;t=0.$

Let us find all such hyperplanes $T$ among (5.6) so that the intersection of
$T$ and $S^{\prime}$ consists of $3$ lines. If $T$ is defined by the equation
$t=0,$ then $T\cap S^{\prime}$ is a line and a smooth conic. Assume that $T$
is defined by the equation $x=\lambda t$ for some
$\lambda\in\mathbf{F}^{sep}.$ Then $T\cap S^{\prime}$ is defined by the
equation

$t(\alpha t^{2}+\lambda^{2}tz+z^{2}-\lambda y^{2})=0.$

Let

$f=\alpha t^{2}+\lambda^{2}tz+z^{2}-\lambda y^{2}.$

It is enough to find all $\lambda$ such that $f$ is a product of two different
linear functions. By direct computations one can find that it happens if and
only if either $\lambda=0,$ or

(5.7) $\alpha=\lambda^{4}.$

Therefore, by (5.7) we get

$f=(\lambda^{2}t-z)^{2}-\lambda y^{2}=(\lambda^{2}t-z-\mu
y)(\lambda^{2}t-z+\mu y),$

where $\mu\in\mathbf{F}^{sep}$ and $\mu^{2}=\lambda.$ Since any line on
$S^{\prime}_{\mathbf{F}^{sep}}$ is defined over $\mathbf{F},$ we get, in
particular, that the lines

$\begin{cases}x=\lambda t\\\ \lambda^{2}t-z\pm\mu y=0\end{cases}$

is defined over $\mathbf{F}.$ This means that $\mu\in\mathbf{F}$ or,
equivalently, $\lambda\in\mathbf{F}^{2}.$ Therefore, by (5.7) we have that
$\alpha=\gamma^{8}$ for some $\gamma\in\mathbf{F}^{*}.$

∎

######  Lemma 5.8.

Let $\mathbf{F}$ be a field of characteristic $3$ such that it contains a
primitive eighth root of unity $\varepsilon.$ Let $S$ be a smooth cubic
surface over $\mathbf{F}$ isomorphic to smooth cubic surface (5.4) for some
$\alpha\in\mathbf{F}^{*}.$ Assume that all lines on $S_{\mathbf{F}^{sep}}$ are
defined over $\mathbf{F}.$ Then $S$ is isomorphic to smooth cubic surface
(1.1).

###### Proof.

By Lemma 5.7 the smooth cubic surface $S$ is isomorphic to smooth cubic
surface (5.4) for some non-zero $\alpha\in\mathbf{F}^{8}.$ So we have
$\alpha=\gamma^{8}$ for some $\gamma\in\mathbf{F}.$ By the change of
coordinates

$x\mapsto\gamma^{2}x,\;\;y\mapsto\gamma^{3}y,\;\;z\mapsto\gamma^{4}z,\;\;t\mapsto
t$

equation (5.4) transforms to (1.1) and we get the desired.

∎

######  Lemma 5.9 (see [7, Table 7]).

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
$3.$ Assume that
$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$
Then if $g\in\mathrm{Aut}(S)$ is an element of order $6,$ then it lies in the
conjugacy class $E_{6}(a_{2}).$ If $g\in\mathrm{Aut}(S)$ is an element of
order $3,$ then it lies either in the conjugacy class $A_{2}^{2},$ or in the
conjugacy class $A_{2}^{3}.$

######  Lemma 5.10.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
$3.$ Assume that
$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$
Then the centre of $\mathrm{Aut}(S)$ is trivial.

###### Proof.

Assume the converse. Let
$(a,b)\in\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$ be an
element of the centre of the group
$\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$ Then for any

$(c,d)\in\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$

we have

(5.8)
$(a,b)\cdot(c,d)=(a\leftindex^{b}{c},bd)=(c\leftindex^{d}{a},db)=(c,d)\cdot(a,b),$

where $\leftindex^{b}{c}$ and $\leftindex^{d}{a}$ are the results of the
action of $\mathbb{Z}/8\mathbb{Z}$ on $\mathcal{H}_{3}(\mathbb{F}_{3}).$ In
particular, we have $a\leftindex^{b}{c}=c\leftindex^{d}{a}$ for any
$c\in\mathcal{H}_{3}(\mathbb{F}_{3})$ and $d\in\mathbb{Z}/8\mathbb{Z}.$ So
taking $c=e$ the identity element, we obtain $a=\leftindex^{d}{a}$ for any
$d\in\mathbb{Z}/8\mathbb{Z},$ which means that
$a\in\mathcal{H}_{3}(\mathbb{F}_{3})$ commutes with any
$d\in\mathbb{Z}/8\mathbb{Z}.$ Assume that $a$ is not trivial. Then taking as
$d$ a generator of $\mathbb{Z}/8\mathbb{Z},$ we get that $ad$ is an element of
order $24$ which does not exist in $W(\mathrm{E}_{6})$ by Lemma 2.14.

So we get that $a$ is the identity element in
$\mathcal{H}_{3}(\mathbb{F}_{3}).$ Assume that $b\in\mathbb{Z}/8\mathbb{Z}$ is
not the identity element. Then by (5.8) be obtain the equality
$\leftindex^{b}{c}=c$ for any $c\in\mathcal{H}_{3}(\mathbb{F}_{3}).$ This
means that $b\in\mathbb{Z}/8\mathbb{Z}$ commutes with any
$c\in\mathcal{H}_{3}(\mathbb{F}_{3}).$ Since the order of $b$ is even, take
the element $g\in\mathbb{Z}/8\mathbb{Z}$ of order $2$ such that $g=b^{i}$ for
some integer number $i.$ By Lemma 2.9 the element $g$ lies in the conjugacy
class $A^{4}_{1}.$

The element $h=gc$ is of order $6$ for any non-trivial
$c\in\mathcal{H}_{3}(\mathbb{F}_{3}).$ By Lemma 5.9 the element $h$ lies in
the conjugacy class $E_{6}(a_{2})$ and $c$ lies in the conjugacy class
$A_{2}^{2}$ or $A_{2}^{3}.$ By Lemma 2.10 the element $c$ has to lie only in
the conjugacy class $A_{2}^{3},$ which contradicts the assumption that $g$ and
$c$ commute for any $c\in\mathcal{H}_{3}(\mathbb{F}_{3}).$ This contradiction
concludes the triviality of the centre of $\mathrm{Aut}(S).$

∎

######  Lemma 5.11.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
$3.$ Assume that
$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$
Then the image $\Gamma$ of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6})$ is trivial.

###### Proof.

By Theorem 1.1(ii)′ and Theorem 2.1(i) we get that

$\mathrm{Aut}(S_{\mathbf{F}^{sep}})\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$

and $S_{\mathbf{F}^{sep}}$ is isomorphic to smooth cubic surface (1.1). This
means that $\mathrm{Aut}(S)$ is isomorphic to the automorphism group of smooth
cubic surface (1.1). Consider the element $g\in\mathrm{Aut}(S)$ of order $8.$
Since $\Gamma$ commutes with any element of $\mathrm{Aut}(S),$ we get that
$\Gamma$ lies in the centralizer $\mathrm{C}_{W(\mathrm{E}_{6})}(g)$ of $g$ in
$W(\mathrm{E}_{6}).$ By Lemma 2.7 there is a unique conjugacy class of
elements of order $8$ in $W(\mathrm{E}_{6})$ and its centralizer
$\mathrm{C}_{W(\mathrm{E}_{6})}(g)$ is of order $8.$ So
$\mathrm{C}_{W(\mathrm{E}_{6})}(g)$ is a cyclic group of order $8$ generated
by $g.$ So we have

$\Gamma\subset\mathrm{C}_{W(\mathrm{E}_{6})}(g)\subset\mathrm{Aut}(S).$

However, by Lemma 5.10 the centre of $\mathrm{Aut}(S)$ is trivial. This means
that $\Gamma$ is trivial.

∎

######  Lemma 5.12.

Let $\mathbf{F}$ be a field of characteristic $3$ such that it contains a
primitive eighth root of unity $\varepsilon\in\mathbf{F}^{sep}.$ Let $S$ be a
smooth cubic surface over $\mathbf{F}$ such that

$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$

Then $S$ is isomorphic to smooth cubic surface (1.1).

###### Proof.

By Lemma 5.6 the smooth cubic surface $S$ is isomorphic to the smooth cubic
surface $S^{\prime}$ defined by equation (5.4) for some
$\alpha\in\mathbf{F}^{*}.$ By Lemma 5.11 the image $\Gamma$ of the Galois
group $\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in the Weyl group
$W(\mathrm{E}_{6})$ is trivial. This means that all lines on
$S^{\prime}_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F}.$ So by Lemma 5.8
the smooth cubic surface $S^{\prime}$ is isomorphic to smooth cubic surface
(1.1).

∎

Now we consider smooth cubic surfaces over a field $\mathbf{F}$ of
characteristic $3$ which does not contain a primitive fourth root of unity.
Let us formulate some auxiliary lemmas.

######  Lemma 5.13.

Let $\mathbf{F}$ be a separably closed field of characteristic different from
$2$ and $3.$ Then the automorphism group of cubic surface (1.1) over
$\mathbf{F}$ is isomorphic to $\mathbb{Z}/8\mathbb{Z}.$

###### Proof.

For characteristic zero the result follows from [6, Theorem 9.5.8]. For
positive characteristic the result follows from [7, Lemma 12.12] applying
Theorem 2.1(i).

∎

######  Lemma 5.14.

Let $S$ be a smooth cubic surface over separabely closed field $\mathbf{F}$ of
characteristic different from $2.$ Let $h\in\mathrm{Aut}(S)$ be an
automorphism of order $8.$ Then $h$ has $3$ fixed points on $S.$

###### Proof.

By Lemma 5.5 up to conjugation $h$ acts on $\mathbb{P}^{3}$ as

$\begin{pmatrix}\varepsilon^{6}&0&0&0\\\ 0&\varepsilon&0&0\\\
0&0&\varepsilon^{4}&0\\\ 0&0&0&1\end{pmatrix}\in\mathrm{PGL}_{4}(\mathbf{F}).$

The element $h$ has exactly $4$ fixed points on $\mathbb{P}^{3},$ namely,

(5.9) $[1:0:0:0],\;[0:1:0:0],\;[0:0:1:0],\;[0:0:0:1].$

Any invariant under the action of $h$ smooth cubic surface is of the form

(5.10) $axy^{2}+bx^{2}z+cz^{2}t+dt^{3}=0,$

where $a,b,c,d\in\mathbf{F}^{*}.$ The points among (5.9) which lies on a
smooth cubic surface (5.10) are

$[1:0:0:0],\;[0:1:0:0],\;[0:0:1:0].$

∎

######  Lemma 5.15.

Let $\mathbf{F}$ be a field of characteristic $3$ which does not contain a
primitive eighth root of unity. Then $\mathrm{GL}_{2}(\mathbf{F})$ does not
contain a subgroup isomorphic to $\mathbb{Z}/8\mathbb{Z}.$

###### Proof.

Assume the converse. Let $h\in\mathrm{GL}_{2}(\mathbf{F})$ be an element of
order $8.$ Then its characteristic polynomial is

$F(\lambda)=\lambda^{2}-(\varepsilon^{i}+\varepsilon^{j})\lambda+\varepsilon^{i+j},$

where $\varepsilon\in\mathbf{F}^{sep}$ is a primitive eighth root of unity and
$i,j\in\mathbb{Z}.$ So both $\varepsilon^{i}+\varepsilon^{j}$ and
$\varepsilon^{i+j}$ belong to $\mathbf{F}.$ By Lemma 5.2 we get that

(5.11) $i+j=4k\;\;\text{for}\;\;k\in\mathbb{Z}.$

Note that at least one of the integer numbers $i$ and $j$ is odd, since $h$ is
of order $8.$ By (5.11) we get that both the integer numbers $i$ and $j$ are
odd. Therefore, $\varepsilon^{i}$ is a primitive eighth root of unity. By
(5.11) we get

$\varepsilon^{i}+\varepsilon^{j}=\varepsilon^{i}+(-1)^{k}\varepsilon^{-i}$

which is not contained in $\mathbf{F}$ by Lemma 5.3. This contradiction
concludes the proof.

∎

######  Lemma 5.16.

Let $\mathbf{F}$ be a field of characteristic different from $2$ which does
not contain a non-trivial eighth root of unity. Let $S$ be a smooth cubic
surface over $\mathbf{F}.$ Then $\mathrm{Aut}(S)$ does not contain elements of
order $8.$

###### Proof.

Assume the converse. Let $h\in\mathrm{Aut}(S)$ be an element of order $8.$
Then by Lemma 5.14 the element $h,$ considered over $\mathbf{F}^{sep},$ has
$3$ fixed points on $S_{\mathbf{F}^{sep}}.$ By Lemma 2.7 the centralizer
$\mathrm{C}_{W(\mathrm{E}_{6})}(h)$ of the element $h,$ considered as an
element in $W(\mathrm{E}_{6}),$ is of order $8.$ Since
$h\in\mathrm{C}_{W(\mathrm{E}_{6})}(h)$ we get that
$\mathrm{C}_{W(\mathrm{E}_{6})}(h)$ is a cyclic group generated by $h.$
Therefore, the image $\Gamma$ of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ is a subgroup of the cyclic group
of order $8.$ Therefore, at least one of the fixed points of $h$ on
$S_{\mathbf{F}^{sep}}$ is defined over $\mathbf{F}.$

Let $P\in S$ be an $h$-fixed point. Let us consider the action of $h$ on the
tangent space $T_{P}S$ of $S$ in the point $P.$ By [4, Theorem 3.7] we get
that

$\mathbb{Z}/8\mathbb{Z}\subset\mathrm{GL}\left(T_{P}(S)\right)=\mathrm{GL}_{2}(\mathbf{F}).$

By Lemma 5.15 this is impossible. This contradiction concludes the proof.

∎

######  Lemma 5.17.

Let $\mathbf{F}$ be a field of characteristic $3$ which does not contain a
non-trivial eighth root of unity. Let $S$ be a smooth cubic surface over
$\mathbf{F}.$ Then we have the inequality $|\mathrm{Aut}(S)|\leqslant 120.$
Moreover, the equality holds if and only if
$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$ and $S$ is isomorphic to the Clebsch
cubic surface.

###### Proof.

By Theorem 1.1(ii)′ and Remark 2.2 the largest automorphism group of a smooth
cubic surface over $\mathbf{F}^{sep}$ is isomorphic to
$\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$ and has order
$216.$ By Lemma 5.16 an element of order $8$ does not lie in
$\mathrm{Aut}(S).$ This means that if

$\mathrm{Aut}(S_{\mathbf{F}^{sep}})\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z},$

then $|\mathrm{Aut}(S)|\leqslant 108.$

So let us consider a smooth cubic surface $S$ over $\mathbf{F}$ such that
$\mathrm{Aut}(S_{\mathbf{F}^{sep}})$ is the next largest automorphism group of
a smooth cubic surface over $\mathbf{F}^{sep}.$ By Theorem (ii)′′ this means
that $\mathrm{Aut}(S_{\mathbf{F}^{sep}})\simeq\mathfrak{S}_{5}.$ So we have
that the order of $\mathrm{Aut}(S)$ is at most $120.$ By Example 4.1 the group
$\mathfrak{S}_{5}$ is implemented on the Clebsch cubic surface over
$\mathbf{F}.$ By Corollary 4.11 the smooth cubic surface $S$ is unique up to
isomorphism, and hence isomorphic to the Clebsch cubic surface.

∎

##  6\. The Fermat cubic surface

In this section we study the Fermat cubic surface and the group
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ as the automorphism
group of a smooth cubic surface. First of all, note that in the sequel we
consider the group $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ as a
group which is isomorphic to the automorphism group of the Fermat cubic
surface over a separably closed field of characteristic different from $2$ and
$3.$ This means that up to conjugation $(\mathbb{Z}/3\mathbb{Z})^{3}$ acts as
a multiplication by cube roots of unity on the homogeneous coordinates
$x,y,z,t$ on $\mathbb{P}^{3}$ and $\mathfrak{S}_{4}$ acts as a permutation of
the homogeneous coordinates $x,y,z,t.$ Moreover, we have the following result.

######  Lemma 6.1.

Let $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ be a subgroup in
$\mathrm{PSU}_{4}(\mathbb{F}_{2}).$ Then this subgroup is isomorphic to the
automorphism group of the Fermat cubic surface over a separably closed field
of characteristic different from $2$ and $3$.

###### Proof.

By Lemma 2.15 there is only one subgroup isomorphic to
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ in the group
$\mathrm{PSU}_{4}(\mathbb{F}_{2})$ up to conjugation. By Example 3.10 up to
conjugation the subgroup $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$
can be realized as an automorphism group of the Fermat cubic surface over a
separably closed field of characteristic different from $2$ and $3$. By
Theorem 2.1(i) the same holds for separably closed field.

∎

######  Lemma 6.2.

Let $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ be a subgroup in
$\mathrm{PSU}_{4}(\mathbb{F}_{2}).$ Then its centre is trivial.

###### Proof.

Assume the converse. Let
$(a,b)\in(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ be an element of
the centre of the group $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$
Then for any

$(c,d)\in(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$

we have

(6.1)
$(a,b)\cdot(c,d)=(a\leftindex^{b}{c},bd)=(c\leftindex^{d}{a},db)=(c,d)\cdot(a,b),$

where $\leftindex^{b}{c}$ and $\leftindex^{d}{a}$ are the results of the
action of $\mathfrak{S}_{4}$ on $(\mathbb{Z}/3\mathbb{Z})^{3}.$ As the centre
of $\mathfrak{S}_{4}$ is trivial, by (6.1) we get that $b$ is the identity
element. This means that for any $d\in\mathfrak{S}_{4}$ we have
$a=\leftindex^{d}{a}.$ This also means by Example 3.10 and Lemma 6.1 that the
element $a$ is the identity element.

∎

Now we are ready to study smooth cubic surfaces over a field $\mathbf{F}$ of
characteristic different from $2$ and $3$ with automorphism group
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$

######  Lemma 6.3.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different from $2$ and $3.$ Assume that
$\mathrm{Aut}(S)\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$
Then there is an element of order $9$ in $\mathrm{Aut}(S).$

###### Proof.

By Theorem 1.1(iv)′ and (iii)′ and Theorem 2.1(i) we get that

$\mathrm{Aut}(S_{\mathbf{F}^{sep}})\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$

and $S_{\mathbf{F}^{sep}}$ is isomorphic to the Fermat cubic surface. This
means that $\mathrm{Aut}(S)$ is isomorphic to the automorphism group of the
Fermat cubic surface. By the structure of the automorphism group of the Fermat
cubic surface over a separabely closed field of characteristic different form
$2$ and $3$ there is an element $g$ of order $9$ in $\mathrm{Aut}(S)$ (see
Example 3.10).

∎

######  Lemma 6.4.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different from $2$ and $3.$ Assume that
$\mathrm{Aut}(S)\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$
Then the image $\Gamma$ of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6})$ is trivial.

###### Proof.

By Lemma lemma:autorder9 there is an element $g$ of order $9$ in
$\mathrm{Aut}(S).$ Since $\Gamma$ commutes with any element of
$\mathrm{Aut}(S),$ we get that $\Gamma$ lies in the centralizer
$\mathrm{C}_{W(\mathrm{E}_{6})}(g)$ of $g$ in $W(\mathrm{E}_{6}).$ By Lemma
2.7 there is a unique conjugacy class of elements of order $9$ in
$W(\mathrm{E}_{6})$ and its centralizer $\mathrm{C}_{W(\mathrm{E}_{6})}(g)$ is
of order $9.$ So $\mathrm{C}_{W(\mathrm{E}_{6})}(g)$ is a cyclic group of
order $9$ generated by $g.$ So we have

$\Gamma\subset\mathrm{C}_{W(\mathrm{E}_{6})}(g)\subset\mathrm{Aut}(S).$

However, by Lemma 6.2 the centre of $\mathrm{Aut}(S)$ is trivial. This means
that $\Gamma$ is trivial.

∎

######  Lemma 6.5.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different from $2$ and $3.$ Assume that its automorphism group is isomorphic
to

$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$

Then $\mathbf{F}$ contains a non-trivial cube root of unity
$\omega\in\mathbf{F}^{sep}.$

###### Proof.

By Lemma lemma:autorder9 there is an element $g$ of order $9$ in
$\mathrm{Aut}(S).$ By Lemma 2.7 the collection of eigenvalues of the
representation $g$ on the root lattice $\mathrm{E}_{6}$ is

$\zeta,\zeta^{2},\zeta^{4},\zeta^{5},\zeta^{7},\zeta^{8},$

where $\zeta\in\mathbb{C}$ is a primitive ninth root of unity. By Lemma 2.7
this means that the cube $g^{3}$ lies in the conjugacy class $A_{2}^{3}.$
Therefore, by Corollary 3.9 there is a non-trivial cube root of unity in
$\mathbf{F}.$

∎

######  Lemma 6.6.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different from $2$ and $3.$ Assume that its automorphism group is isomorphic
to

$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$

Then $S$ is isomorphic to the Fermat cubic surface.

###### Proof.

Since
$\mathrm{Aut}(S)\simeq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4},$ by
Theorem 1.1(iii)′ and (iv)′ we have

$\mathrm{Aut}(S_{\mathbf{F}^{sep}})\simeq\mathrm{Aut}(S)$

and $S_{\mathbf{F}^{sep}}$ isomorphic to the Fermat cubic surface. By Lemma
6.5 there is a non-trivial cube root of unity $\omega\in\mathbf{F}.$ Moreover,
by Lemma 6.4 the image $\Gamma$ of the absolute Galois group
$\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in $W(\mathrm{E}_{6})$ is trivial.
Therefore, all lines on $S_{\mathbf{F}^{sep}}$ are defined over $\mathbf{F}.$

Let us consider the normal subgroup $G$ in $\mathrm{Aut}(S)$ isomorphic to
$\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}.$ By Example 3.10 up to conjugation
over $\mathbf{F}^{sep}$ this subgroup acts as a multiplication by cube roots
of unity on the homogeneous coordinates $x,y,z,t$ on $\mathbb{P}^{3}$ By Lemma
2.6 this realization holds up to conjugation over $\mathbf{F}.$ Smooth cubic
surfaces which are fixed by $G$ are defined by the equations

(6.2) $x^{3}+ay^{3}+bz^{3}+ct^{3}=0,$

where $a,b,c\in\mathbf{F}^{*}.$

Consider the hyperplane $T$ in $\mathbb{P}^{3}_{\mathbf{F}^{sep}}$ defined by
the equation

$x+\alpha y=0,$

where $\alpha\in\mathbf{F}^{sep}$ satisfies the equation $\alpha^{3}=a.$ There
are $3$ lines in the intersection of the hyperplane $T$ and the smooth cubic
surface $S_{\mathbf{F}^{sep}}.$

Assume that $\alpha\notin\mathbf{F}.$ Then there is an element
$\tau\in\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ which satisfies

(6.3) $\tau(\alpha)\neq\alpha.$

The intersection of $\tau(T)$ and $S_{\mathbf{F}^{sep}}$ again consists of $3$
lines. However, the intersection of $T$ and $\tau(T)$ is a line by (6.3). This
means that not all lines on the intersection $T$ and $S_{\mathbf{F}^{sep}}$
are defined over $\mathbf{F}.$ However, by Lemma 6.4 the image $\Gamma$ of the
absolute Galois group $\mathrm{Gal}(\mathbf{F}^{sep}/\mathbf{F})$ in
$W(\mathrm{E}_{6})$ is trivial. Therefore, all lines on $S_{\mathbf{F}^{sep}}$
are defined over $\mathbf{F}.$ Thus, $\alpha\in\mathbf{F}.$

The same works for the hyperplanes $x+\beta z=0$ and $x+\gamma t=0,$ where
$\beta^{3}=b$ and $\gamma^{3}=c,$ respectively. So we obtain
$a,b,c\in\mathbf{F}^{3}.$ Therefore, by change of coordinates

$x\mapsto x,\;\;\sqrt[3]{a}y\to y,\;\;\sqrt[3]{b}z\to z,\;\;\sqrt[3]{c}t\to t$

smooth cubic surface (6.2) transforms to

$x^{3}+y^{3}+z^{3}+t^{3}=0$

which is the Fermat cubic surface.

∎

##  7\. Fields without cube root of unity

In this section we study smooth cubic surfaces over fields of characteristic
different from $3$ which do not contain a non-trivial cube root of unity.
First of all, we need the following auxiliary lemma.

######  Lemma 7.1.

Let $G$ be a subgroup in
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\subset\mathrm{PSU}_{4}(\mathbb{F}_{2}).$
Let us consider the homomorphism

$\mathrm{pr}\colon(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\to\mathfrak{S}_{4}$

and denote by $\mathrm{pr}\mid_{G}$ the restriction of $\mathrm{pr}$ on the
subgroup $G.$ Assume that $\mathrm{Ker}(\mathrm{pr}\mid_{G})$ is neither
trivial, nor isomorphic to $\left(\mathbb{Z}/3\mathbb{Z}\right)^{3}.$ Then
$\mathrm{Coker}(\mathrm{pr}\mid_{G})$ is isomorphic neither to
$\mathfrak{S}_{4},$ nor to $\mathfrak{A}_{4}.$

###### Proof.

Assume the converse. Then we have the following exact sequence

$1\to(\mathbb{Z}/3\mathbb{Z})^{i}\to
G\xrightarrow{\mathrm{pr}\mid_{G}}\mathrm{Coker}(\mathrm{pr}\mid_{G})\to 1,$

where $i=1$ or $2$ and $\mathrm{Coker}(\mathrm{pr}\mid_{G})$ isomorphic either
to $\mathfrak{S}_{4},$ or to $\mathfrak{A}_{4}.$ Then as
$\mathrm{Ker}(\mathrm{pr}\mid_{G})$ is an abelian group, we have the following
homomorphism

$f:\mathrm{Coker}(\mathrm{pr}\mid_{G})\to\mathrm{Aut}\left((\mathbb{Z}/3\mathbb{Z})^{i}\right).$

Note that we have the following isomorphisms

$\mathrm{Aut}\left(\mathbb{Z}/3\mathbb{Z}\right)\simeq\mathbb{Z}/2\mathbb{Z}\quad\text{and}\quad\mathrm{Aut}\left((\mathbb{Z}/3\mathbb{Z})^{2}\right)\simeq\mathfrak{D}_{4}.$

This means that $\mathrm{Ker}(f)$ is isomorphic either to $\mathfrak{S}_{4},$
or to $\mathfrak{A}_{4}.$ Therefore, either $\mathfrak{S}_{4},$ or
$\mathfrak{A}_{4}$ act trivially by conjugations on
$(\mathbb{Z}/3\mathbb{Z})^{i}.$ Recall that by Lemma 2.15 we get that up to
conjugation the group $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ is
a unique subgroup in $\mathrm{PSU}_{4}(\mathbb{F}_{2}).$ Thus, by Lemma 3.12
the centralizer of any non-trivial element in the normal subgroup
$(\mathbb{Z}/3\mathbb{Z})^{3}$ of
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ is isomorphic to
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes D,$ where $D$ is a subgroup of
$\mathfrak{S}_{4}$ and $D$ is isomorphic either to $\mathfrak{S}_{3},$ or to
$(\mathbb{Z}/2\mathbb{Z})^{2},$ or to $\mathbb{Z}/2\mathbb{Z}.$ It is obvious,
that neither $\mathfrak{S}_{4},$ nor $\mathfrak{A}_{4}$ lie in these groups.
This contradiction concludes the proof.

∎

######  Lemma 7.2.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different from $2$ and $3.$ Assume that $\mathbf{F}$ does not contain a non-
trivial cube root of unity. Then $|\mathrm{Aut}(S)|\leqslant 120.$

###### Proof.

Assume that $|\mathrm{Aut}(S)|>120.$ Then by Theorem 1.1 (iii)′, (iii)′′,
(iv)′ and (iv)′′ we have
$\mathrm{Aut}(S)\subseteq(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$
Let us consider the homomorphism

$\mathrm{pr}\colon(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\to\mathfrak{S}_{4}$

which is a natural projection on $\mathfrak{S}_{4}.$ Let us consider the
restriction $\mathrm{pr}\mid_{\mathrm{Aut}(S)}$ of the homomorphism
$\mathrm{pr}$ on the subgroup $\mathrm{Aut}(S).$ The kernel of
$\mathrm{pr}\mid_{\mathrm{Aut}(S)}$ is a subgroup in the normal subgroup
$(\mathbb{Z}/3\mathbb{Z})^{3}$ of the group
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$ Since $\mathbf{F}$ does
not contain a non-trivial cube root of unity, by Corollary 3.9 and Lemma 3.11
we get that

(7.1)
$\mathrm{Ker}\left(\mathrm{pr}\mid_{\mathrm{Aut}(S)}\right)\subsetneq(\mathbb{Z}/3\mathbb{Z})^{3}.$

Since $648\geqslant|\mathrm{Aut}(S)|>120,$ we get that the order of the group
$\mathrm{Aut}(S)$ is either $162,$ or $216,$ or $324,$ or $648.$ So by (7.1)
the only possibility for the order of $\mathrm{Aut}(S)$ is $216,$ i.e. for
$\mathrm{Aut}(S)$ the following exact sequence

$1\to(\mathbb{Z}/3\mathbb{Z})^{2}\to\mathrm{Aut}(S)\xrightarrow{\mathrm{pr}\mid_{\mathrm{Aut}(S)}}\mathfrak{S}_{4}\to
1$

holds. By Lemma 7.1 this possibility is in fact the impossibility. This
contradiction concludes the proof that

$|\mathrm{Aut}(S)|\leqslant 120.$

∎

######  Lemma 7.3.

Let $S$ be a smooth cubic surface over a field $\mathbf{F}$ of characteristic
different from $2,$ $3$ and $5,$ such that $\mathbf{F}$ does not contain a
non-trivial cube root of unity. Then

$|\mathrm{Aut}(S)|\leqslant 120.$

Moreover, the equality holds if and only if
$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}$ and $S$ is isomorphic to the Clebsch
cubic surface.

###### Proof.

By Lemma 7.2 we have the inequality $|\mathrm{Aut}(S)|\leqslant 120.$ By
Theorem 1.1(iv)′ and (iv)′′ the equality holds if and only if
$\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}.$ By Example 4.1 the group
$\mathfrak{S}_{5}$ is realized over $\mathbf{F}$ as the automorphism group of
the Clebsch cubic surface.

Let $\mathrm{Aut}(S)\simeq\mathfrak{S}_{5}.$ Then by Corollary 4.11 the smooth
cubic surface $S$ is isomorphic to the Clebsch cubic surface.

∎

######  Example 7.4.

Let $\mathbf{F}$ be a field such that it does not contain a non-trivial cube
root of unity. Let us consider the group
$(\mathbb{Z}/3\mathbb{Z})^{2}\rtimes\mathfrak{D}_{4}$ as a subgroup in
$\mathrm{PGL}_{4}(\mathbf{F}),$ such that $(\mathbb{Z}/3\mathbb{Z})^{2}$ is
generated by the elements

(7.2) $\begin{pmatrix}-1&1&0&0\\\ -1&0&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\
0&0&-1&1\\\ 0&0&-1&0\end{pmatrix}$

and $\mathfrak{D}_{4}$ is a normalizer of the group
$(\mathbb{Z}/3\mathbb{Z})^{2}$ in the group of permutations of the
coordinates. In other words, if $x,y,z,t$ are homogeneous coordinates on
$\mathbb{P}^{3},$ then $\mathfrak{D}_{4}$ is generated by the permutations
$(12)$ and $(1324).$

######  Lemma 7.5.

Consider the subgroup $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ in
the group $\mathrm{PSU}_{4}(\mathbb{F}_{2}).$ Let us consider the homomorphism

$\mathrm{pr}\colon(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\to\mathfrak{S}_{4}$

which is a natural projection on $\mathfrak{S}_{4}.$ Let us fix the subgroup
$\mathfrak{D}_{4}\subset\mathrm{pr}\left((\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\right).$
Then $\mathfrak{D}_{4}$ normalizes the unique subgroup $K$ in
$\mathrm{Ker}\left(\mathrm{pr}\right)$ and
$K\simeq(\mathbb{Z}/3\mathbb{Z})^{2}.$

###### Proof.

By Lemma 6.1 the subgroup
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ in the group
$\mathrm{PSU}_{4}(\mathbb{F}_{2})$ is isomorphic to the automorphism group of
the Fermat cubic surface over a separably closed field of characteristic
different from $2$ and $3$. By Example 3.10 and since
$\mathfrak{D}_{4}\subset\mathfrak{S}_{4}$ is unique up to conjugation, without
loss of generality we can assume that $\mathfrak{D}_{4}$ acts as permutation
of the diagonal elements in the automorphisms of the Fermat cubic surface of
the form

(7.3) $\begin{pmatrix}\omega^{a}&0&0&0\\\ 0&\omega^{b}&0&0\\\
0&0&\omega^{c}&0\\\
0&0&0&\omega^{d}\end{pmatrix}\in\mathrm{PGL}_{4}(\mathbf{F})$

for separably closed field $\mathbf{F}$ of characteristic different from $2$
and $3$ and generated by the permutations $(12)$ and $(1324).$ The subgroup in
the group, generated by the elements (7.3), which is invariant under the
action of the permutation $(12)$ generated by the elements satisfy the
equation $a=b.$ The subgroup in the group, generated by the elements (7.3)
which satisfy the equation $a=b,$ which is invariant under the action of the
permutation $(1324),$ generated by the elements, satisfy the equation, also
satisfies the equation $c=d.$ This group is isomorphic to
$(\mathbb{Z}/3\mathbb{Z})^{2}.$

∎

######  Lemma 7.6.

Let $G$ be a subgroup of order $72$ in
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\subset\mathrm{PSU}_{4}(\mathbb{F}_{2}).$
Let us consider the projection homomorphism

$\mathrm{pr}\colon(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\to\mathfrak{S}_{4}$

Denote by $\mathrm{pr}\mid_{G}$ the restriction of $\mathrm{pr}$ on the
subgroup $G.$ Assume that for $G$ the following exact sequence holds

(7.4) $1\to(\mathbb{Z}/3\mathbb{Z})^{2}\to
G\xrightarrow{\mathrm{pr}\mid_{G}}\mathfrak{D}_{4}\to 1.$

Then $G\simeq(\mathbb{Z}/3\mathbb{Z})^{2}\rtimes\mathfrak{D}_{4}$ and such
subgroup is unique up to conjugation.

###### Proof.

Since the orders of the kernel and the cokernel in the exact sequence (7.4)
are coprime, by the Schur–Zassenhaus theorem (see [8, Section 17.4, Theorem
39]) we get

(7.5) $G\simeq(\mathbb{Z}/3\mathbb{Z})^{2}\rtimes_{\phi}\mathfrak{D}_{4}$

for some homomorphism

(7.6)
$\phi:\mathfrak{D}_{4}\to\mathrm{Aut}\left((\mathbb{Z}/3\mathbb{Z})^{2}\right)\simeq\mathfrak{D}_{4}.$

By Lemma 7.5 the fixed subgroup $\mathfrak{D}_{4}$ in
$\mathrm{pr}\left((\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\right)$
normalizes the unique subgroup in $\mathrm{Ker}\left(\mathrm{pr}\right)$
isomorphic to $(\mathbb{Z}/3\mathbb{Z})^{2}.$ By Example 7.4 there is a
subgroup

$(\mathbb{Z}/3\mathbb{Z})^{2}\rtimes_{\phi}\mathfrak{D}_{4}$

in $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}$ such that $\phi$ is
an automorphism. Therefore, since $\mathfrak{D}_{4}$ is unique subgroup up to
conjugation in $\mathfrak{S}_{4},$ we get that all $\phi$ in (7.6) are
automorphisms. So $G$ is unique up to conjugation.

∎

######  Lemma 7.7.

Let $G$ be a subgroup of the group
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\subset
W(\mathrm{E}_{6}).$ Assume that $G$ does not contain elements which lie in the
conjugacy class $A^{3}_{2}.$ If $|G|\leqslant 120,$ then $|G|\leqslant 72$ and
the equality holds if and only if
$G\simeq(\mathbb{Z}/3\mathbb{Z})^{2}\rtimes\mathfrak{D}_{4}.$ Moreover, in the
case of equality the group $G$ with such properties is unique up to
conjugation.

###### Proof.

Let us consider the projection homomorphism

$\mathrm{pr}\colon(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}\to\mathfrak{S}_{4}.$

Denote by $\mathrm{pr}\mid_{G}$ the restriction of the homomorphism
$\mathrm{pr}$ on the group $G.$ By Lemma 3.11 the kernel
$\mathrm{Ker}(\mathrm{pr}\mid_{G})$ lies in $(\mathbb{Z}/3\mathbb{Z})^{2}.$

If $|G|>72,$ then $|G|$ is either $108,$ or $81.$ Now we prove that these two
values are impossible. Assume that the order of $G$ is equal to $108.$ Then
since the order of $\mathrm{Ker}(\mathrm{pr}\mid_{G})$ is at most $9$ and the
order of $\mathrm{Im}(\mathrm{pr}\mid_{G})$ is at most $24,$ we get that in
the case $|G|=108$ the only possibility is

$1\to(\mathbb{Z}/3\mathbb{Z})^{2}\to
G\xrightarrow{\mathrm{pr}\mid_{G}}\mathfrak{A}_{4}\to 1.$

By Lemma 7.1 this possibility is in fact the impossibility.

The case $|G|=81$ is also impossible, because the kernel of
$\mathrm{pr}\mid_{G}$ has to be isomorphic to $(\mathbb{Z}/3\mathbb{Z})^{3},$
but by Lemma 3.11 there are elements from the conjugacy class $A_{2}^{3}$ in
the normal subgroup $(\mathbb{Z}/3\mathbb{Z})^{3}$ of
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$

Let us consider the case $|G|=72.$ We have two possibilities for $G$

(7.7) $\displaystyle 1\to\mathbb{Z}/3\mathbb{Z}\to
G\xrightarrow{\mathrm{pr}\mid_{G}}\mathfrak{S}_{4}\to 1,$ (7.8) $\displaystyle
1\to(\mathbb{Z}/3\mathbb{Z})^{2}\to
G\xrightarrow{\mathrm{pr}\mid_{G}}\mathfrak{D}_{4}\to 1.$

By Lemma 7.1 the case (7.7) is not realized. Let us consider the case (7.8).
By Example 7.4 this case is realized. By Lemma 7.6 the group $G$ is isomorphic
to $(\mathbb{Z}/3\mathbb{Z})^{2}\rtimes\mathfrak{D}_{4}$ and is unique up to
conjugation.

∎

######  Lemma 7.8.

Let $\mathbf{F}$ be a field of characteristic $5$ such that it does not
contain a non-trivial cube root of unity. Let $S$ be a smooth cubic surface
over $\mathbf{F}$ such that

$\mathrm{Aut}(S_{\mathbf{F}^{sep}})\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/4\mathbb{Z}.$

Then
$\mathrm{Aut}(S)\subsetneq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/4\mathbb{Z}.$

###### Proof.

By Corollary 3.8 in $\mathrm{Aut}(S)$ there are no elements lying in the
conjugacy class $A^{3}_{2}.$ But by [7, Table 8] and Remark 2.2 the group
$\mathrm{Aut}(S_{\mathbf{F}^{sep}})$ contains elements lying in the conjugacy
class $A^{3}_{2}.$ Therefore, we have strict inclusion

$\mathrm{Aut}(S)\subsetneq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/4\mathbb{Z}.$

∎

######  Corollary 7.9.

Let $\mathbf{F}$ be a field of characteristic $5$ which does not contain a
non-trivial cube root of unity. Let $S$ be a smooth cubic surface over
$\mathbf{F}$. Then we have the inequality $|\mathrm{Aut}(S)|\leqslant 72.$
Moreover, the equality holds if and only if

$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{2}\rtimes\mathfrak{D}_{4}.$

###### Proof.

By Lemma 7.2 we have the inequality $|\mathrm{Aut}(S)|\leqslant 120.$ By
Theorem 1.1(iii)′′ this implies the inequality $|\mathrm{Aut}(S)|\leqslant
108.$ Assume that
$\mathrm{Aut}(S)\subset(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$
Then by Lemma 7.7 we obtain $|\mathrm{Aut}(S)|\leqslant 72,$ and the equality
holds if and only if

$\mathrm{Aut}(S)\simeq\left(\mathbb{Z}/3\mathbb{Z}\right)^{2}\rtimes\mathfrak{D}_{4}.$

Otherwise, assume that $\mathrm{Aut}(S)$ lies in the second largest
automorphism group of a smooth cubic surface over an algebraically closed
field of characteristic $5.$ By Theorem 1.1(iii)′′ this means that
$\mathrm{Aut}(S)\subset\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/4\mathbb{Z}.$
However, by Lemma 7.8 this means that the order of $\mathrm{Aut}(S)$ is at
most $54$ which is less than $72.$ By Theorem 1.1(iii)′′ the third largest
automorphism group of a smooth cubic surface over an algebraically closed
field of characteristic $5$ is the group
$\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/2\mathbb{Z}.$ Its order is
equal to $54$ and, again, it is less than $72.$

∎

######  Lemma 7.10.

Let $\mathbf{F}$ be a field of characteristic different from $2$ and $3$ such
that it does not contain non-trivial cube root of unity. Let $S$ be a smooth
cubic surface over $\mathbf{F}$ such that its automorphism group is isomorphic
to $(\mathbb{Z}/3\mathbb{Z})^{2}\rtimes\mathfrak{D}_{4}.$ Then $S$ is
isomorphic to the cubic surface

$2(x^{3}+y^{3}+z^{3}+t^{3})-3(x^{2}y+xy^{2}+z^{2}t+zt^{2})=0.$

###### Proof.

Let us consider all invariant cubic surfaces under the action of
$\mathfrak{D}_{4},$ where $\mathfrak{D}_{4}$ acts on the coordinates
$[x:y:z:t]\in\mathbb{P}^{3}_{\mathbf{F}}$ as permutations. Assume that the
group $\mathfrak{D}_{4}$ is generated by $(1324)$ and $(12).$ Then cubic
surfaces, which are invariant under the action of the element $(12),$ are of
the form

(7.9)
$\begin{gathered}a(x^{3}+y^{3})+b(x^{2}y+xy^{2})+l_{1}(z,t)(x^{2}+y^{2})+l_{2}(z,t)xy+\\\
+q(z,t)(x+y)+c(z,t)=0,\end{gathered}$

where $a,b\in\mathbf{F},$ $l_{1}(z,t)$ and $l_{2}(z,t)$ are linear homogeneous
polynomials, $q(z,t)$ is a quadratic homogeneous polynomial and $c(z,t)$ is a
cubic homogeneous polynomial. Those cubic surfaces among (7.9), which are
invariant under the action of the element $(1324),$ are of the form

$\displaystyle
a(x^{3}+y^{3})+b(x^{2}y+xy^{2})+(z+t)\left(u(x^{2}+y^{2})+vxy\right)\pm$
$\displaystyle\pm(x+y)\left(u(z^{2}+t^{2})+vzt\right)\pm a(z^{3}+t^{3})\pm
b(z^{2}t+zt^{2})=0,$

where $a,b,u,v\in\mathbf{F}.$ Now we act on these cubic surfaces by the group
$(\mathbb{Z}/3\mathbb{Z})^{2}$ which is generated by the elements (7.2) and
get two smooth cubic surfaces

$2(x^{3}+y^{3})-3(x^{2}y+xy^{2})\pm(2(z^{3}+t^{3})-3(z^{2}t+zt^{2}))=0.$

However, by change of coordinates

$x\mapsto x,\quad y\mapsto y,\quad z\mapsto-z,\quad t\mapsto-t$

we get that these two surfaces are isomorphic.

∎

######  Lemma 7.11.

Let $\mathbf{F}$ be a field of characteristic $5$ such that it does not
contain non-trivial cube root of unity. Let $S$ be a smooth cubic surface over
$\mathbf{F}$ such that its automorphism group is isomorphic to
$(\mathbb{Z}/3\mathbb{Z})^{2}\rtimes\mathfrak{D}_{4}.$ Then $S$ is isomorphic
to cubic surface (1.2).

###### Proof.

By Lemma 7.10 we get that $S$ is isomorphic to

(7.10) $2(x^{3}+y^{3}+z^{3}+t^{3})-3(x^{2}y+xy^{2}+z^{2}t+zt^{2})=0.$

Since $2=-3$ in the field $\mathbf{F}$ we get that cubic surface (7.10) is
defined by the equation

(7.11) $x^{3}+y^{3}+z^{3}+t^{3}+x^{2}y+xy^{2}+z^{2}t+zt^{2}=0.$

Let us consider the change of coordinates

$x\mapsto x+\mathbf{i}y,\quad y\mapsto\mathbf{i}x+y,\quad z\mapsto
z+\mathbf{i}t,\quad t\mapsto\mathbf{i}z+t,$

where $\mathbf{i}$ is a primitive fourth root of unity in $\mathbf{F}.$ After
this change of coordinates the equation (7.11) takes the form

(7.12) $x^{2}y+xy^{2}+z^{2}t+zt^{2}=0.$

By the change of coordinates $[x:y:z:t]\mapsto[x:t:z:y]$ the equation (7.12)
transforms to

$x^{2}t+y^{2}z+z^{2}y+t^{2}x=0.$

∎

Now we prove that the Fermat cubic surface is not isomorphic to cubic surface
(1.2) over fields without cube root of unity.

######  Lemma 7.12.

Let $\mathbf{F}$ be a field of characteristic different from $3$ such that it
does not contain a non-trivial cube root of unity. Then the Fermat cubic
surface is not isomorphic to cubic surface (1.2).

###### Proof.

Let us consider these cubic surfaces over $\mathbf{F}^{sep}.$ The lines of the
Fermat cubic surface over $\mathbf{F}^{sep}$ are defined by the systems of the
equations in $\mathbb{P}^{3}_{\mathbf{F}^{sep}}$

(7.13) $\begin{cases}x+\omega^{i}y=0\\\
z+\omega^{j}t=0\end{cases}\begin{cases}x+\omega^{i}z=0\\\
t+\omega^{j}y=0\end{cases}\begin{cases}x+\omega^{i}t=0\\\
z+\omega^{j}y=0,\end{cases}$

where $\omega\in\mathbf{F}^{sep}$ is a non-trivial cube root of unity and
$i,j\in\\{0,1,2\\}.$ Let $\mathbf{L}$ be the extension of degree $2$ of
$\mathbf{F}$ such that $\mathbf{L}=\mathbf{F}[t]/(t^{2}+t+1).$ Then from
(7.13) we immediately see that all lines on the Fermat cubic surface over
$\mathbf{F}^{sep}$ are defined over $\mathbf{L}.$ Since the non-trivial
element of the Galois group
$\mathrm{Gal}(\mathbf{L}/\mathbf{F})\simeq\mathbb{Z}/2\mathbb{Z}$ maps
$\omega$ to $\omega^{2}$ we get that there are only $3$ lines on the Fermat
cubic surface which are defined over $\mathbf{F},$ namely,

$\begin{cases}x+y=0\\\ z+t=0\end{cases}\begin{cases}x+z=0\\\
t+y=0\end{cases}\begin{cases}x+t=0\\\ z+y=0.\end{cases}$

However, on cubic surface (1.2) there are more than $3$ lines defined over
$\mathbf{F},$ for instance,

$\begin{cases}x=0\\\ z=0\end{cases}\begin{cases}y=0\\\
z=0\end{cases}\begin{cases}x=0\\\ t=0\end{cases}\begin{cases}x=0\\\
z+t=0\end{cases}\text{etc.}$

This means that these cubic surfaces are not isomorphic.

∎

After this lemma one wonders if the Fermat cubic surface is isomorphic to
cubic surface (1.2) over fields which contain non-trivial cube roots of unity?
The following lemma gives the positive answer on this question.

######  Lemma 7.13.

Let $\mathbf{F}$ be a field of characteristic different from $3$ such that it
contains a non-trivial cube root of unity. Then the Fermat cubic surface is
isomorphic to cubic surface (1.2).

###### Proof.

Let $\omega\in\mathbf{F}$ be a non-trivial cube root of unity. Then the
isomorphism between the Fermat cubic surface and cubic surface (1.2) is given
by the collineation

$[x:y:z:t]\mapsto[x+y:\omega x+\omega^{2}y:z+t:\omega z+\omega^{2}t]$

which maps cubic surface (1.2) to the Fermat cubic surface.

∎

##  8\. Proof of the main results

In this section we prove Theorem 1.3 and Proposition 1.5.

###### Proof of Theorem 1.3.

Let us prove (i)3. The inequality $|\mathrm{Aut}(S)|\leqslant 216$ follows
from Theorem 1.1(ii)′. The equality holds by Lemma 5.4. The group of order
$216$ which is an automorphism group of a smooth cubic surface is unique also
by Theorem 1.1(ii)′ and isomorphic to
$\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}.$ By Lemma 5.12
the smooth cubic surface $S$ such that
$\mathrm{Aut}(S)\simeq\mathcal{H}_{3}(\mathbb{F}_{3})\rtimes\mathbb{Z}/8\mathbb{Z}$
is unique up to isomorphism and is isomorphic to smooth cubic surface (1.1).

Assertion (ii)3 directly follows from Lemma 5.17.

Let us prove assertions (i)5 and (i)≠2,3,5. The inequality
$|\mathrm{Aut}(S)|\leqslant 648$ follows from Theorem 1.1(iii)′ and (iv)′. The
equality holds by Example 3.10. The group of order $648$ which is an
automorphism group of a smooth cubic surface is unique by Theorem 1.1(iii)′
and (iv)′ and isomorphic to
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$ So by Lemma 6.6 the
smooth cubic surface $S$ with such automorphism group is unique up to
isomorphism and isomorphic to the Fermat cubic surface.

Assertion (ii)5 follows from Lemma 7.3.

Let us prove assertion (ii)≠2,3,5. By Corollary 7.9 we get that
$|\mathrm{Aut}(S)|\leqslant 72$ and the equality holds if and only if
$\mathrm{Aut}(S)$ is isomorphic to
$\left(\mathbb{Z}/3\mathbb{Z}\right)^{2}\rtimes\mathfrak{D}_{4}.$ By Lemma
7.11 the smooth cubic surface $S$ is unique up to isomorphism and isomorphic
to cubic surface (1.2).

∎

Before we prove Proposition 1.5 let us give the following lemma.

######  Lemma 8.1.

Cubic surface (1.1) is isomorphic neither to the Fermat cubic surface, nor to
the Clebsch cubic surface, nor to cubic surface (1.2).

###### Proof.

Over a field of characteristic $2,$ cubic surface (1.1) has singularity at the
point $[0:0:1:1],$ while the other listed cubic surfaces are smooth.

Over a field of characteristic $3$ the Fermat cubic surface is non-reduced and
cubic surface (1.2) is singular, while both the Clebsch cubic surface and
cubic surface (1.1) are smooth. However, the last two cubic surface are not
isomorphic, since by Theorem 1.1(ii)′ and (ii)′′ they have different
automorphism groups over an algebraically closed field of characteristic $3.$

Let $\mathbf{F}$ be a field of characteristic different from $2$ and $3.$ By
Lemma 5.13 the automorphism group of cubic surface (1.1) over
$\mathbf{F}^{sep}$ is isomorphic to $\mathbb{Z}/8\mathbb{Z}.$ Meanwhile, by
Theorem 1.1(iii)′ and (iv)′ and Remark 2.2 we get that the automorphism group
of the Fermat cubic surface over $\mathbf{F}^{sep}$ is isomorphic to
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$ Since by Lemma 7.13 the
Fermat cubic surface is isomorphic to cubic surface (1.2) over
$\mathbf{F}^{sep}$, we get that the automorphism group of cubic surface (1.2)
is also isomorphic to $(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$
This means that cubic surface (1.1) is isomorphic neither to the Fermat cubic
surface, nor to cubic surface (1.2).

By Theorem 1.1(iv)′′ and Remark 2.2 the automorphism group of the Clebsch
cubic surface over a separably closed field of characteristic different from
$2,$ $3$ and $5$ is isomorphic to $\mathfrak{S}_{5},$ and by Lemma 5.13 the
automorphism group of cubic surface (1.1) over a separably closed field is
isomorphic to $\mathbb{Z}/8\mathbb{Z}.$ Over a field of characteristic $5$ the
Clebsch cubic surface is singular, while cubic surface (1.1) is smooth. This
means that cubic surface (1.1) is not isomorphic to the Clebsch cubic surface.

∎

###### Proof of Proposition 1.5.

One can easily check assertion (i). Assertion (ii) follows from Lemma 8.1.
Assertion (iii) directly follows from Corollary 4.14(ii).

Let us prove assertion (iv). If $\mathbf{F}$ contains a non-trivial cube root
of unity, then cubic surface (1.2) and the Fermat cubic surface are isomorphic
by Lemma 7.13. If $\mathbf{F}$ does not contain a non-trivial cube root of
unity, then cubic surface (1.2) and the Fermat cubic surface are not
isomorphic by Lemma 7.12.

Let us prove assertion (v). If $\mathbf{F}$ is of characteristic $3,$ then the
assertion follows from assertion (i). If $\mathbf{F}$ is of characteristic
$5,$ then the Clebsch cubic surface is isomorphic neither to cubic surfaces
(1.2), nor to the Fermat cubic surface, because in this case the Clebsch cubic
surface is singular, while the others are smooth. Assume that $\mathbf{F}$ is
of characteristic different from $5.$ Then by Theorem 1.1(iv)′′, Remark 2.2
and Example 4.1 the automorphism group of the Clebsch cubic surface over
$\mathbf{F}^{sep}$ is isomorphic to $\mathfrak{S}_{5}.$ By Theorem 1.1(iv)′,
Remark 2.2 and Lemma 7.13 we get that the automorphism groups of cubic surface
(1.2) and the Fermat cubic surface over $\mathbf{F}^{sep}$ are isomorphic to
$(\mathbb{Z}/3\mathbb{Z})^{3}\rtimes\mathfrak{S}_{4}.$ This means that over
$\mathbf{F}$ the Clebsch cubic surface is isomorphic neither to cubic surface
(1.2), nor to the Fermat cubic surface.

∎

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