# Complements of hypersurfaces in projective spaces

Jérémy Blanc  Jérémy Blanc, Universität Basel, Departement Mathematik und
Informatik, Spiegelgasse $1$, CH-$4051$ Basel, Switzerland.
<EMAIL_ADDRESS>http://algebra.dmi.unibas.ch/blanc ,  Pierre-Marie
Poloni  Pierre-Marie Poloni, Universität Basel, Departement Mathematik und
Informatik, Spiegelgasse $1$, CH-$4051$ Basel, Switzerland.
<EMAIL_ADDRESS>http://algebra.dmi.unibas.ch/poloni and  Immanuel
Van Santen  Immanuel Van Santen, Universität Basel, Departement Mathematik und
Informatik, Spiegelgasse $1$, CH-$4051$ Basel, Switzerland.
<EMAIL_ADDRESS>http://algebra.dmi.unibas.ch/vansanten

###### Abstract.

We study the complement problem in projective spaces ${\mathbb{P}}^{n}$ over
any algebraically closed field: If $H,H^{\prime}\subseteq{\mathbb{P}}^{n}$ are
irreducible hypersurfaces of degree $d$ such that the complements
${\mathbb{P}}^{n}\setminus H$, ${\mathbb{P}}^{n}\setminus H^{\prime}$ are
isomorphic, are the hypersurfaces $H$, $H^{\prime}$ isomorphic?

For $n=2$, the answer is positive if $d\leq 7$ and there are counterexamples
when $d=8$. In contrast we provide counterexamples for all $n,d\geq 3$ with
$(n,d)\neq(3,3)$. Moreover, we show that the complement problem has an
affirmative answer for $d=2$ and give partial results in case $(n,d)=(3,3)$.
In the course of the exposition, we prove that rational normal projective
surfaces admitting a desingularisation by trees of smooth rational curves are
piecewise isomorphic if and only if they coincide in the Grothendieck ring,
answering affirmatively a question posed by Larsen and Lunts for such
surfaces.

###### Key words and phrases:

Complement problem, cylinder over Danielewski surfaces, piecewise isomorphisms

###### 2010 Mathematics Subject Classification:

14J70, 14R05 (primary), 14E07, 14F43, (secondary)

The first author acknowledges support by the Swiss National Science Foundation
Grant “Geometrically ruled surfaces”200020–192217.

###### Contents

  1. 1 Introduction
  2. 2 Lift to affine hypersurfaces
  3. 3 Non-isomorphic hypersurfaces having isomorphic complements
    1. 3.1 Explicit isomorphisms between complements of projective cones
    2. 3.2 Non-isomorphic hypersurfaces
  4. 4 Topological Euler characteristic and piecewise isomorphisms
  5. 5 Partial answers in low degree
    1. 5.1 Degree two
    2. 5.2 Degree three and beyond in dimension three
    3. 5.3 Automorphisms of complements of singular cubic surfaces

##  1\. Introduction

Let ${\mathbf{k}}$ be a field, let $n\geq 2$ be an integer and let
$f\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ be an irreducible homogeneous
polynomial of degree $d\geq 1$. The variety

${\mathbb{P}}^{n}_{f}=\\{\,[x_{0}:\cdots:x_{n}]\in{\mathbb{P}}^{n}\ |\
f(x_{0},\ldots,x_{n})\neq 0\,\\}$

is the complement in ${\mathbb{P}}^{n}$ of the hypersurface given by $f=0$. It
is an affine open subset of ${\mathbb{P}}^{n}$ with Picard group
${\mathbb{Z}}/d{\mathbb{Z}}$. Moreover, any open subset of ${\mathbb{P}}^{n}$
isomorphic to it is of the form ${\mathbb{P}}^{n}_{g}$ for some irreducible
homogeneous polynomial $g\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ of degree $d$
(see Lemma 2.6). One may then ask the following natural questions, for all
irreducible homogeneous polynomials $f,g\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$
of the same degree $d\geq 1$.

  1. (1)

Does any isomorphism
${\mathbb{P}}^{n}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{n}_{g}$
extend to an element of $\mathrm{Aut}({\mathbb{P}}^{n})$?

  2. (2)

If ${\mathbb{P}}^{n}_{f}$ and ${\mathbb{P}}^{n}_{g}$ are isomorphic, is there
an element of $\mathrm{Aut}({\mathbb{P}}^{n})$ that sends
${\mathbb{P}}^{n}_{f}$ onto ${\mathbb{P}}^{n}_{g}$?

  3. (3)

If ${\mathbb{P}}^{n}_{f}$ and ${\mathbb{P}}^{n}_{g}$ are isomorphic, are the
zero loci $V_{{\mathbb{P}}^{n}}(f)$ and $V_{{\mathbb{P}}^{n}}(g)$ of $f$ and
$g$ isomorphic?

Of course, any positive answer to one of the questions also gives a positive
answer to the next one(s).

If $d=1$, the first question has a negative answer, as there are many
automorphisms of ${\mathbb{P}}^{n}_{f}\simeq{\mathbb{A}}^{n}$ that do not
extend to automorphisms of ${\mathbb{P}}^{n}$. However, the two other
questions have a positive answer.

For $d=2$, we can obtain similar results, in any dimension, when
${\mathbf{k}}$ is algebraically closed. Indeed, we will answer the first
question in the negative for any irreducible quadric, see Example 5.6, and
provide positive answers to the two other questions, see Theorem B below.

The situation for $d=3$ is already more complicated, at least in dimension
$n=3$. The first question has a negative answer for each singular cubic with
Du Val singularities over an algebraically closed field of characteristic zero
[CDP18, Theorem C and Theorem 4.3], and more generally for each singular cubic
over any algebraically closed field (Proposition 5.22), but is wide open for
any smooth cubic surface [CDP18, Conjecture 1.1]. The two other questions are
open as well.

In dimension $n=2$, it was conjectured in [Yos84] that the answer to the
second question should be always positive for algebraically closed fields of
characteristic zero. This conjecture was proved for all algebraically closed
fields if the curve $C\subseteq{\mathbb{P}}^{2}$ given by $f=0$ is such that
$C\setminus L\simeq{\mathbb{A}}^{1}$ for some line (see [Yos84, Theorem] for a
proof in characteristic zero and [Hem19, Theorem 1] for a proof in any
characteristic). The conjecture was however proven to be false in general,
with a first example of degree $39$ given in [Bla09]. A second family of
counterexamples of degree $9+4m$ for each $m\geq 0$ was given later in
[Cos12]. Finally, an example of degree $8$ and a proof that the conjecture is
true for curves of degree $\leq 7$ appeared in [Hem19, Corollary 1.2, Theorem
3] for algebraically closed fields of any characteristic. Moreover, the
examples given in [Hem19] and [Bla09] also give a negative answer to the third
question.

In dimension $n\geq 3$, there was no negative answer to Questions (2) and (3)
until now. In this article, we show by constructing explicit examples, that
the answer of the third question, and thus of the two others, is negative for
large degree and large dimension. More precisely, in any dimension $n\geq 3$,
the smallest degree for counterexamples is $3$, except maybe for $n=3$, where
the smallest degree might be $4$. This contrasts the case of dimension $n=2$
where counterexamples for algebraically closed fields always have degree $\geq
8$.

######  Theorem A.

Let $\mathbf{k}$ be a field and let $d,n\geq 3$ be integers such that $d$ is
not a multiple of $\mathrm{char}(\mathbf{k})$ and such that $(d,n)\neq(3,3)$.

There are two non-isomorphic irreducible hypersurfaces
$H,H^{\prime}\subseteq{\mathbb{P}}^{n}_{\mathbf{k}}$ of degree $d$, that have
isomorphic complements.

We will prove Theorem A at the end of §3.2. If $d=2$, then Question (2) has an
affirmative answer, at least when ${\mathbf{k}}$ is algebraically closed.

######  Theorem B.

Let $n\geq 2$ and let $\mathbf{k}$ be an algebraically closed field. If the
complements of two irreducible quadric hypersurfaces $H,H^{\prime}$ in
${\mathbb{P}}^{n}_{\mathbf{k}}$ are isomorphic, then there exists an
automorphism of ${\mathbb{P}}^{n}_{\mathbf{k}}$ that sends $H$ onto
$H^{\prime}$.

We will prove Thereom B at the end of §5.1. In the case where $n=3$, we have
the following partial positive result concerning Question (2):

######  Theorem C.

Let $\mathbf{k}$ be an algebraically closed field and let
$H,H^{\prime}\subseteq{\mathbb{P}}^{3}_{\mathbf{k}}$ be irreducible
hypersurfaces such that their complements in ${\mathbb{P}}^{3}_{\mathbf{k}}$
are isomorphic. Then we have:

  1. (1)

If both $H$ and $H^{\prime}$ are normal, rational and each admits a
desingularisation by trees of smooth rational curves, then $H,H^{\prime}$ are
piecewise isomorphic, see Def. 4.6;

  2. (2)

If $H$ or $H^{\prime}$ is a cubic, then $H,H^{\prime}$ are piecewise
isomorphic. Moreover, $H$ is smooth if and only if $H^{\prime}$ is smooth;

  3. (3)

If $H$ or $H^{\prime}$ is a non-rational cubic, then there exists
$\varphi\in\mathrm{Aut}({\mathbb{P}}^{3}_{\mathbf{k}})$ with
$\varphi(H)=H^{\prime}$.

We will prove Theorem C at the end of §5.2. For $\mathbf{k}={\mathbb{C}}$, it
is proven in [LS12] that if ${\mathbb{P}}^{3}_{f}\simeq{\mathbb{P}}^{3}_{g}$
for irreducible polynomials $f,g$, then $V_{{\mathbb{P}}^{2}}(f)$ and
$V_{{\mathbb{P}}^{2}}(g)$ are piecewise isomorphic, which gives essentially
Theorem C over the field of complex numbers.

To prove Theorem C, we study piecewise isomorphisms between surfaces. We prove
in particular that every normal rational projective surface admitting a
desingularisation by trees of smooth rational curves (e.g. Du Val
singularities) is piecewise isomorphic to the disjoint union of
${\mathbb{A}}^{2}$, one point and $n$ copies of ${\mathbb{A}}^{1}$ for some
$n\geq 1$ (Proposition 4.19). A similar result does not hold for all rational
normal projective surfaces (Example 4.20). Using the Grothendieck ring
$K_{0}(\operatorname{Var}_{\mathbf{k}})$ and the topological Euler
characteristic $($see §4 for the definition$)$, one sees that the integer $n$
determines the piecewise isomorphism class. As a consequence of this, we give
an affirmative answer to a question posed by Larsen and Lunts (see [LL03,
Question 1.2]) for rational projective normal surfaces admitting a
desingularisation by trees of smooth rational curves (see Proposition 5.14).
This also partially generalises [LS10, Theorem 4].

######  Proposition D.

Let $\mathbf{k}$ be algebraically closed and let $X,Y$ be rational normal
projective surfaces admitting a desingularisation by trees of smooth rational
curves. If the classes $[X]$ and $[Y]$ coincide in the Grothendieck ring
$K_{0}(\operatorname{Var}_{\mathbf{k}})$, then $X$ and $Y$ are piecewise
isomorphic.

##  2\. Lift to affine hypersurfaces

Let $n\geq 1$. For each homogeneous polynomial
$f\in\mathbf{k}[x_{0},\ldots,x_{n}]$ of degree $d\geq 1$ and each
$\mu\in{\mathbf{k}}^{*}$, we denote by $X_{f,\mu}\subseteq{\mathbb{A}}^{n+1}$
the hypersurface given by

$X_{f,\mu}=\mathrm{Spec}({\mathbf{k}}[x_{0},\ldots,x_{n}]/(f-\mu))\subseteq{\mathbb{A}}^{n+1}\,,$

and obtain a canonical finite morphism

$\begin{array}[]{cccc}\pi_{f,\mu}\colon&X_{f,\mu}&\to&{\mathbb{P}}^{n}_{f}\\\
&(x_{0},\ldots,x_{n})&\mapsto&[x_{0}:\cdots:x_{n}]\end{array}$

of degree $d$, which is an étale covering if $\operatorname{char}(\mathbf{k})$
does not divide $d$.

If $\mathbf{k}$ is algebraically closed, all the varieties $X_{f,\mu}$ are
isomorphic to $X_{f,1}$ by homotheties.

If ${\mathbf{k}}$ is equal to the field of complex numbers ${\mathbb{C}}$,
then $\pi_{f,\mu}$ is the universal abelian covering, as the abelianisation of
the fundamental group of ${\mathbb{P}}^{n}_{f}$ with respect to the Euclidean
topology is equal to ${\mathbb{Z}}/d{\mathbb{Z}}$ [Lib05, Proposition 2.3].

We now prove that for each field $\mathbf{k}$, the isomorphisms between
complements of hypersurfaces lift to these coverings. This will be useful in
the sequel, in order to study isomorphisms between varieties
${\mathbb{P}}^{n}_{f}$ and ${\mathbb{P}}^{n}_{g}$.

######  Proposition 2.1.

Let $\mathbf{k}$ be a field, let $n\geq 1$ be an integer and let
$f,g\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ be irreducible homogeneous
polynomials of degree $d\geq 1$. Let
$\Phi_{0},\ldots,\Phi_{n}\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ be homogeneous
polynomials of degree $\ell\geq 1$. Then, the following statements are
equivalent:

  1. (1)

The rational map

$\begin{array}[]{rcl}\Phi\colon{\mathbb{P}}^{n}_{f}&\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}&{\mathbb{P}}^{n}_{g}\,,\\\
{[x_{0}:\cdots:x_{n}]}&\mapsto&[\Phi_{0}(x_{0},\ldots,x_{n}):\cdots:\Phi_{n}(x_{0},\ldots,x_{n})]\,,\end{array}$

is an isomorphism and the gcd of $\Phi_{0},\ldots,\Phi_{n}$ is a power of $f$.

  2. (2)

There exists $\mu\in{\mathbf{k}}^{*}$ such that the following map is an
isomorphism

$\begin{array}[]{rcl}\varphi\colon
X_{f,1}&\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}&X_{g,\mu}\,,\\\
(x_{0},\ldots,x_{n})&\mapsto&(\Phi_{0}(x_{0},\ldots,x_{n}),\ldots,\Phi_{n}(x_{0},\ldots,x_{n}))\,.\end{array}$

Moreover, if these statements hold, then $\ell$ is invertible modulo $d$.

###### Proof.

“(1) $\Rightarrow$ (2)”: By assumption there exists $r\geq 0$ and
$\Phi_{0}^{\prime},\ldots,\Phi_{n}^{\prime}$ in
$\mathbf{k}[x_{0},\ldots,x_{n}]$ without a common factor such that
$\Phi_{i}=f^{r}\Phi_{i}^{\prime}$ for all $i$. The domain $\textrm{dom}(\Phi)$
of the rational self-map
$\Phi\colon{\mathbb{P}}^{n}\dashrightarrow{\mathbb{P}}^{n}$ is thus equal to
${\mathbb{P}}^{n}\setminus
V_{{\mathbb{P}}^{n}}(\Phi^{\prime}_{0},\ldots,\Phi^{\prime}_{n})$ and contains
${\mathbb{P}}^{n}_{f}$. As $g$ is nowhere zero on ${\mathbb{P}}^{n}_{g}$, the
polynomial
$g(\Phi_{0},\ldots,\Phi_{n})=f^{dr}g(\Phi_{0}^{\prime},\ldots,\Phi_{n}^{\prime})$
is nowhere zero on ${\mathbb{P}}^{n}_{f}$ and is thus equal to $\mu f^{t}$ for
some $\mu\in{\mathbf{k}}^{*}$ and some integer $t\geq 1$. We then obtain a
morphism

$\varphi\colon X_{f,1}\to
X_{g,\mu}\,,\quad(x_{0},\ldots,x_{n})\mapsto(\Phi_{0}(x_{0},\ldots,x_{n}),\ldots,\Phi_{n}(x_{0},\ldots,x_{n}))\,.$

It remains to see that $\varphi$ is an isomorphism.

The inverse of $\Phi$ is given by

$\Psi\colon{\mathbb{P}}^{n}_{g}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{n}_{f}\,,\quad[x_{0}:\cdots:x_{n}]\mapsto[\Psi_{0}(x_{0},\ldots,x_{n}):\cdots:\Psi_{n}(x_{0},\ldots,x_{n})]\,,$

where $\Psi_{0},\ldots,\Psi_{n}\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ are
homogeneous polynomials of the same degree $\ell^{\prime}\geq 1$, without
common factor.

As $\Psi\circ\Phi=\mathrm{id}_{{\mathbb{P}}^{n}_{f}}$ and
$\Phi\circ\Psi=\mathrm{id}_{{\mathbb{P}}^{n}_{g}}$, there exist homogeneous
polynomials $A,B\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ of degree
$\ell\ell^{\prime}-1$ such that

$\Psi_{i}(\Phi_{0},\ldots,\Phi_{n})=Ax_{i}\text{ and
}\Phi_{i}(\Psi_{0},\ldots,\Psi_{n})=Bx_{i}\text{ for all
}i\in\\{0,\ldots,n\\}.$

Moreover, as the zero locus of $A$ is sent by $\Phi$ into the base-locus of
$\Psi$ and the zero locus of $B$ is sent by $\Psi$ into the base-locus of
$\Phi$, we find that $A$ is nowhere vanishing on ${\mathbb{P}}^{n}_{f}$ and
$B$ is nowhere vanishing on ${\mathbb{P}}^{n}_{g}$. We get $A=\lambda f^{s}$
and $B=\lambda^{\prime}g^{s}$ where
$\lambda,\lambda^{\prime}\in{\mathbf{k}}^{*}$ and where $s\geq 1$ is such that
$ds=\ell\ell^{\prime}-1$. In particular, $\ell$ is invertible modulo $d$. We
replace $\Psi_{i}$ with $\Psi_{i}/\lambda$ for each $i\in\\{0,\ldots,n\\}$ and
reduce to the case where $\lambda=1$.

Consider now the morphism

$\psi\colon
X_{g,\mu}\to{\mathbb{A}}^{n+1}\,,\quad(x_{0},\ldots,x_{n})\mapsto(\Psi_{0}(x_{0},\ldots,x_{n}),\ldots,\Psi_{n}(x_{0},\ldots,x_{n}))\,.$

Using that $\Psi_{i}(\Phi_{0},\ldots,\Phi_{n})=f^{s}x_{i}$ for each
$i\in\\{0,\ldots,n\\}$, we obtain that $\psi\circ\varphi\colon
X_{f,1}\to{\mathbb{A}}^{n+1}$ is the natural closed embedding. Hence,
$(\psi\circ\varphi)^{\ast}=\varphi^{\ast}\circ\psi^{\ast}\colon\mathbf{k}[{\mathbb{A}}^{n+1}]\to\mathbf{k}[X_{f,1}]$
is surjective and $\varphi$ is dominant, i.e.
$\varphi^{\ast}\colon\mathbf{k}[X_{g,\mu}]\to\mathbf{k}[X_{f,1}]$ is
injective. This implies that $\varphi^{\ast}$ and thus $\varphi$ is an
isomorphism.

“(2) $\Rightarrow$ (1)”: As $\Phi_{0},\ldots,\Phi_{n}$ are homogeneous of
degree $\ell$, we may define a rational self-map
$\Phi\colon[x_{0}:\cdots:x_{n}]\mapsto[\Phi_{0}(x_{0},\ldots,x_{n}):\cdots:\Phi_{n}(x_{0},\ldots,x_{n})]$
of ${\mathbb{P}}^{n}$. We now prove that $\Phi$ restricts to an isomorphism
${\mathbb{P}}^{n}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{n}_{g}$
and that the gcd of $\Phi_{0},\ldots,\Phi_{n}$ is a power of $f$.

By assumption $g(\Phi_{0},\ldots,\Phi_{n})=(f-1)p+\mu$ for some polynomial
$p\in\mathbf{k}[x_{0},\ldots,x_{n}]$ of degree $r=d(\ell-1)\geq 0$. Write
$p=\sum_{i=0}^{r}p_{i}$ where $p_{i}$ is homogeneous of degree $i$. Since
$g(\Phi_{0},\ldots,\Phi_{n})$ is homogeneous, it follows that $p_{0}=\mu$ and
$p_{di}=fp_{d(i-1)}$ for all $1\leq i\leq\ell-1$. This implies that

$g(\Phi_{0},\ldots,\Phi_{n})=\mu f^{\ell}\,.$

Since $g$ is homogeneous and $f$ is irreducible the gcd of
$\Phi_{0},\ldots,\Phi_{n}$ is a power of $f$. Moreover, $\Phi$ restricts to a
morphism $\Phi\colon{\mathbb{P}}^{n}_{f}\to{\mathbb{P}}^{n}_{g}$.

Using that $\pi_{g,\mu}\circ\varphi=\Phi\circ\pi_{f,1}$, we get the following
commutative diagram

$\textstyle{\mathbf{k}(X_{g,\mu})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi^{\ast}}$$\textstyle{\mathbf{k}(X_{f,1})}$$\textstyle{\mathbf{k}({\mathbb{P}}^{n}_{g})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{g,\mu}^{\ast}}$$\scriptstyle{\Phi^{\ast}}$$\textstyle{\mathbf{k}({\mathbb{P}}^{n}_{f})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{f,1}^{\ast}}$

and that $\Phi\colon{\mathbb{P}}^{n}_{f}\to{\mathbb{P}}^{n}_{g}$ is a quasi-
finite surjection. Note that $\pi_{f,1}^{\ast}$ and $\pi_{g,\mu}^{\ast}$ are
field extensions of degree $d$. Since $\varphi^{\ast}$ is an isomorphism,
$\Phi^{\ast}$ is an isomorphism as well, i.e.
$\Phi\colon{\mathbb{P}}^{n}_{f}\to{\mathbb{P}}^{n}_{g}$ is birational.
Zariski’s Main Theorem [Gro61, Corollaire 4.4.9] gives that
$\Phi\colon{\mathbb{P}}^{n}_{f}\to{\mathbb{P}}^{n}_{g}$ is an isomorphism. ∎

######  Remark 2.2.

If there exists an isomorphism
$\Phi\colon{\mathbb{P}}^{n}_{f}\to{\mathbb{P}}^{n}_{g}$, there are homogeneous
polynomials $\Phi_{0},\ldots,\Phi_{n}\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ of
degree $\ell\geq 1$ without common factor such that
$\Phi([x_{0}:\cdots:x_{n}])=[\Phi_{0}(x_{0},\ldots,x_{n}):\cdots:\Phi_{n}(x_{0},\ldots,x_{n})]$.
Proposition 2.1 then gives the existence of $\mu\in{\mathbf{k}}^{*}$ such that

$\begin{array}[]{rcl}\varphi\colon
X_{f,1}&\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}&X_{g,\mu}\,,\\\
(x_{0},\ldots,x_{n})&\mapsto&(\Phi_{0}(x_{0},\ldots,x_{n}),\ldots,\Phi_{n}(x_{0},\ldots,x_{n}))\end{array}$

is an isomorphism. If $\mathbf{k}$ is moreover algebraically closed, then we
may multiply each $\Phi_{i}$ with a root of $\mu^{-1}$ and assume that
$\mu=1$. Hence, $X_{f,1}$ and $X_{g,1}$ are isomorphic.

######  Remark 2.3.

Let $\mathbf{k}$ be a field, $n\geq 1$ be an integer and let
$f,g\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ be irreducible homogeneous
polynomials of degree $d\geq 1$. Let $e\geq 1$ be an integer and let
$\Phi_{0},\ldots,\Phi_{n}\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ be polynomials
such that the degree of each of their monomials is congruent to $e$ modulo
$d$. If

$X_{f,1}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}X_{g,1}\,,\quad(x_{0},\ldots,x_{n})\mapsto(\Phi_{0}(x_{0},\ldots,x_{n}),\ldots,\Phi_{n}(x_{0},\ldots,x_{n}))$

is an isomorphism, then we may multiply the monomials of $\Phi_{i}$ with some
powers of $f$ and assume that $\Phi_{0},\ldots,\Phi_{n}$ are homogeneous of
the same degree $\ell\geq 1$, and then apply Proposition 2.1 to obtain an
isomorphism
${\mathbb{P}}^{n}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{n}_{g}$.

######  Lemma 2.4.

Let $\mathbf{k}$ be a field, $n\geq 1$ be an integer,
$f,g\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ be irreducible homogeneous
polynomials of degree $d\geq 1$, and let
$\Phi\colon{\mathbb{P}}^{n}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{n}_{g}$
be an isomorphism given by homogeneous polynomials of degree $\ell$ such that
their gcd is a power of $f$. If $\ell$ is congruent to $1$ modulo $d$, then
${\mathbb{P}}^{m}_{f}$ and ${\mathbb{P}}^{m}_{g}$ are isomorphic for each
$m\geq n$.

###### Proof.

We write $\Phi$ as
$[x_{0}:\cdots:x_{n}]\mapsto[\Phi_{0}(x_{0},\ldots,x_{n}):\cdots:\Phi_{n}(x_{0},\ldots,x_{n})]$,
where $\Phi_{0},\ldots,\Phi_{n}\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ are
homogeneous of degree $\ell$ and the $\mathrm{gcd}$ of
$\Phi_{0},\ldots,\Phi_{n}$ is a power of $f$. By Proposition 2.1, the morphism

$\varphi\colon
X_{f,1}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}X_{g,\mu}\,,\quad(x_{0},\ldots,x_{n})\mapsto(\Phi_{0}(x_{0},\ldots,x_{n}),\ldots,\Phi_{n}(x_{0},\ldots,x_{n}))$

is an isomorphism, where $X_{f,1},X_{g,\mu}\subseteq{\mathbb{A}}^{n+1}$ are
defined as before. If $\ell$ is congruent to $1$ modulo $d$, we can write
$\ell=1+dr$ for some integer $r\geq 0$. For each $m\geq n$, we thus get an
isomorphism
$X_{f,1}\times{\mathbb{A}}^{m-n}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}X_{g,\mu}\times{\mathbb{A}}^{m-n}$
given by

$(x_{0},\ldots,x_{m})\mapsto(\varphi(x_{0},\ldots,x_{m}),x_{n+1}f(x_{0},\ldots,x_{n})^{r},\ldots,x_{m}f(x_{0},\ldots,x_{n})^{r}))\,.$

Applying again Proposition 2.1, now in the converse direction, we get that the
homogeneous polynomials
$\Phi_{0},\ldots,\Phi_{n},x_{n+1}f^{r},\ldots,x_{m}f^{r}$ of degree $\ell$
also define an isomorphism
${\mathbb{P}}^{m}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{m}_{g}$.
∎

The next examples shows that $\ell\neq 1$ in Proposition 2.1 is possible.

######  Example 2.5.

Let ${\mathbf{k}}$ be a field and let
$f=g=xyz+x^{3}+y^{3}\in{\mathbf{k}}[x,y,z]$, which is the equation of a nodal
cubic curve $\Gamma\subset{\mathbb{P}}^{2}$. By blowing-up the singular point,
then blowing-up sucessively points on the exceptional divisor created lying on
the curve, we obtain a birational morphism $X\to{\mathbb{P}}^{2}$ with a
strict transform $\tilde{\Gamma}\subseteq X$ of $\Gamma$ being a $(-1)$-curve.
We then contract this curve and contract the exceptional divisors created,
except the last one, to get an automorphism of
${\mathbb{P}}^{2}\setminus\Gamma$ that does not extend to ${\mathbb{P}}^{2}$
and has degree $8$. Explicitely, we define
$\Phi_{0},\Phi_{1},\Phi_{2}\in{\mathbf{k}}[x,y,z]$, homogeneous of degree $8$,
by

$\displaystyle\Phi_{0}$ $\displaystyle=$
$\displaystyle(-x^{4}z+2x^{3}y^{2}-2x^{2}yz^{2}+2xy^{3}z+y^{5}-y^{2}z^{3})\cdot
f$ $\displaystyle\Phi_{1}$ $\displaystyle=$ $\displaystyle(x^{2}+yz)\cdot
f^{2}$ $\displaystyle\Phi_{2}$ $\displaystyle=$ $\displaystyle
x^{7}y-x^{6}z^{2}+6x^{5}y^{2}z-x^{4}y^{4}-3x^{4}yz^{3}+9x^{3}y^{3}z^{2}$
$\displaystyle+x^{2}y^{5}z-3x^{2}y^{2}z^{4}-xy^{7}+4xy^{4}z^{3}+2y^{6}z^{2}-y^{3}z^{5}$

and calculate that these define an involution

${\mathbb{P}}^{2}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{2}_{f}\,,\quad[x:y:z]\mapsto[\Phi_{0}(x,y,z):\Phi_{1}(x,y,z):\Phi_{2}(x,y,z)]\,.$

Here, the common degree is $8$, that is not congruent to $1$ modulo $3$, but
invertible in ${\mathbb{Z}}/3{\mathbb{Z}}$, as Proposition 2.1 says.

######  Lemma 2.6.

Let $\mathbf{k}$ be a field, $n\geq 1$, $f\in\mathbf{k}[x_{0},\ldots,x_{n}]$
an irreducible homogeneous polynomial of degree $d\geq 1$ and let
$U\subseteq{\mathbb{P}}^{n}$ be an open subvariety. If there exists an
isomorphism
${\mathbb{P}}^{n}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}U$,
then there exists an irreducible homogeneous polynomial
$g\in\mathbf{k}[x_{0},\ldots,x_{n}]$ of degree $d$ with
$U={\mathbb{P}}^{n}_{g}$.

Before we start the proof we recall the following fact: If $X$ is an affine
irreducible normal variety over a field $\mathbf{k}$ and if $U\subseteq X$ is
a big open subset, i.e. the codimension of $X\setminus U$ in $X$ is at least
$2$, then $\mathbf{k}[U]=\mathbf{k}[X]$ (see [Mat86, Theorem 11.5]). In
particular, if $U$ is affine as well, then $U=X$.

###### Proof of Lemma 2.6.

If $U$ is big open in ${\mathbb{P}}^{n}$, then for any open affine subvariety
$V\subseteq{\mathbb{P}}^{n}$, the affine subvariety $U\cap V$ is big open in
$V$ and hence $U\cap V=V$. This would imply that $U={\mathbb{P}}^{n}$ is
affine, a contradiction since $n\geq 1$. Hence the union $Z$ of the
$(n-1)$-dimensional irreducible components of ${\mathbb{P}}^{n}\setminus U$ is
non-empty. As ${\mathbb{P}}^{n}\setminus Z$ is an affine big open subset of
$U$ we get ${\mathbb{P}}^{n}\setminus Z=U$. Moreover,
$\mathbf{k}[{\mathbb{P}}^{n}_{f}]^{\ast}={\mathbf{k}}^{\ast}$ yields
$\mathbf{k}[U]^{\ast}={\mathbf{k}}^{\ast}$, which implies that $Z$ is
irreducible. Hence, there exists an irreducible homogeneous polynomial
$g\in\mathbf{k}[x_{0},\ldots,x_{n}]$ such that $Z$ is the zero locus of $g$,
i.e. $U={\mathbb{P}}^{n}_{g}$. Finally,
${\mathbb{Z}}/\deg(g){\mathbb{Z}}\simeq\operatorname{Pic}({\mathbb{P}}^{n}_{g})\simeq\operatorname{Pic}({\mathbb{P}}^{n}_{f})\simeq{\mathbb{Z}}/d{\mathbb{Z}}$
implies that $\deg(g)=d$. ∎

##  3\. Non-isomorphic hypersurfaces having isomorphic complements

###  3.1. Explicit isomorphisms between complements of projective cones

The following formulas are inspired by isomorphisms between cylinders over
Danielewski surfaces given in [MJP21].

######  Lemma 3.1.

Let $\mathbf{k}$ be a field, let $s,m,n\geq 0$ be integers and let
$S={\mathbf{k}}[x_{0},x_{1},\ldots,x_{s}]$ be a polynomial ring in $s+1$
variables over $\mathbf{k}$. Let $P,Q\in S[z]$ and denote by $H_{P}$ and
$H_{Q}$ the hypersurfaces of
${\mathbb{A}}^{s+3}_{{\mathbf{k}}}=\mathrm{Spec}(S[y,z])$ defined by the
equations

$x_{0}^{n}y=P(x_{0},\ldots,x_{r},z)\quad\text{and}\quad
x_{0}^{n}y=Q(x_{0},\ldots,x_{r},z)\,,$

respectively. Suppose that there exist $A,B\in S[z]$ such that

  1. (1)

$z-A(B(z))$ and $z-B(A(z))$ both belong to the ideal $x_{0}^{m}S[z]$,

  2. (2)

$Q\left(A(z)+x_{0}^{m}w\right)\in x_{0}^{n}S[z,w]+P(z)S[z,w]$ and
$P\left(B(z)+x_{0}^{m}w\right)\in x_{0}^{n}S[z,w]+Q(z)S[z,w]$, where $w$ is
transcendental over $S[y,z]$.

Then, the following are inverse isomorphisms.

$\begin{array}[]{rcl}\Phi\colon
H_{P}\times{\mathbb{A}}^{1}_{{\mathbf{k}}}&\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}&H_{Q}\times{\mathbb{A}}^{1}_{{\mathbf{k}}}\\\
(x_{0},\ldots,x_{s},y,z,w)&\mapsto&(x_{0},\ldots,x_{s},\frac{Q\left(A(z)+x_{0}^{m}w\right)}{x_{0}^{n}},\\\
&&A(z)+x_{0}^{m}w,\frac{z-B\left(A(z)+x_{0}^{m}w\right)}{x_{0}^{m}})\,,\\\
\Psi\colon
H_{Q}\times{\mathbb{A}}^{1}_{{\mathbf{k}}}&\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}&H_{P}\times{\mathbb{A}}^{1}_{{\mathbf{k}}}\\\
(x_{0},\ldots,x_{s},y,z,w)&\mapsto&(x_{0},\ldots,x_{s},\frac{P\left(B(z)+x_{0}^{m}w\right)}{x_{0}^{n}},\\\
&&B(z)+x_{0}^{m}w,\frac{z-A\left(B(z)+x_{0}^{m}w\right)}{x_{0}^{m}}).\end{array}$

###### Proof.

The hypotheses on $A$ and $B$ imply that $\Phi$ and $\Psi$ are morphisms.
Moreover, the compositions are given by

$(\Psi\circ\Phi)(x_{0},\ldots,x_{s},y,z,w)=(x_{0},\ldots,x_{s},P(z)/x_{0}^{n},z,w),$
$(\Phi\circ\Psi)(x_{0},\ldots,x_{s},y,z,w)=(x_{0},\ldots,x_{s},Q(z)/x_{0}^{n},z,w).$

∎

######  Proposition 3.2.

Let $\mathbf{k}$ be a field, $d\geq 4$ be integers and let
$f,g\in{\mathbf{k}}[x,y,z]$ be the homogeneous polynomials of degree $d$
defined by

$f=x^{d-1}y+z^{d}\text{ and }g=x^{d-1}y+z^{d}+dx^{d-2}z^{2}.$

Then, for each $r\geq 1$, the open subvarieties
${\mathbb{P}}^{r+2}_{f},{\mathbb{P}}^{r+2}_{g}\subseteq{\mathbb{P}}^{r+2}$ are
isomorphic.

######  Remark 3.3.

The open subvarieties
${\mathbb{P}}^{2}_{f},{\mathbb{P}}^{2}_{g}\subseteq{\mathbb{P}}^{2}$ are not
isomorphic when $\mathbf{k}=\mathbb{C}$ is the field of complex numbers. This
follows from Proposition 2.1 and from the fact that, for every
$\mu\in{\mathbb{C}}^{*}$, the (Danielewski) hypersurfaces
$X_{f,1},X_{g,\mu}\subseteq{\mathbb{A}}^{3}_{\mathbb{C}}$ defined by the
equations $f=1$ and $g=\mu$, respectively, are not isomorphic (see [Pol11,
Theorem 9]).

###### Proof.

We define $s=0$, $x_{0}=x$, $S=\mathbf{k}[x]$, $m=d$, $n=d-1$,

$\begin{array}[]{rclrcl}P(z)&=&1-z^{d},&Q(z)&=&1-z^{d}-dx^{d-2}z^{2}\,,\\\
A(z)&=&z-x^{d-2}z^{3},&B(z)&=&z+x^{d-2}z^{3}\,,\end{array}$

and prove that the hypotheses of Lemma 3.1 are satisfied for the above
polynomials in $S[z]$.

Firstly, we have $A(z)\equiv z\equiv B(z)\pmod{x^{d-2}}$. Thus, since $d\leq
2(d-2)$, as $d\geq 4$, we get $x^{d-2}A(z)^{3}\equiv x^{d-2}z^{3}\equiv
x^{d-2}B(z)^{3}\pmod{x^{d}}$ and we obtain

$\displaystyle z-A(B(z))$ $\displaystyle=$ $\displaystyle
z-B(z)+x^{d-2}B(z)^{3}\equiv 0\pmod{x^{d}}\,,$ $\displaystyle z-B(A(z))$
$\displaystyle=$ $\displaystyle z-A(z)-x^{d-2}A(z)^{3}\equiv 0\pmod{x^{d}}\,.$

We then check that $P\left(B(z)+x^{d}w\right)\in x^{d-1}S[z,w]+Q(z)S[z,w]$:

$\displaystyle P(B(z)+x^{d}w)$ $\displaystyle\equiv P(B(z))$
$\displaystyle\equiv 1-(z+x^{d-2}z^{3})^{d}$ $\displaystyle\equiv
1-z^{d}-dx^{d-2}z^{d+2}$
$\displaystyle\equiv(1-z^{d}-dx^{d-2}z^{2})\cdot(1+dx^{d-2}z^{2})$
$\displaystyle\equiv Q\cdot(1+dx^{d-2}z^{2})\,\pmod{x^{d-1}}.$

Similarly, we have $Q(A(z)+x^{d}w)\in x^{d-1}S[z,w]+P(z)S[z,w]$:

$\displaystyle Q(A(z)+x^{d}w)$ $\displaystyle\equiv Q\left(A(z)\right)$
$\displaystyle\equiv 1-(z-x^{d-2}z^{3})^{d}-dx^{d-2}(z-x^{d-2}z^{3})^{2}$
$\displaystyle\equiv 1-z^{d}+dx^{d-2}z^{d+2}-dx^{d-2}z^{2}$
$\displaystyle\equiv(1-z^{d})\cdot(1-dx^{d-2}z^{2})$ $\displaystyle\equiv
P\cdot(1-dx^{d-2}z^{2})\,\pmod{x^{d-1}}.$

Noting that $f-1=x^{d-1}y-P(z)$ and $g-1=x^{d-1}y-Q(z)$, we may now apply
Lemma 3.1 to obtain an isomorphism

$\begin{array}[]{rcl}\varphi\colon\mathrm{Spec}({\mathbf{k}}[x,y,z,w]/(f-1))&\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}&\mathrm{Spec}({\mathbf{k}}[x,y,z,w]/(g-1))\\\
(x,y,z,w)&\mapsto&(x,\frac{Q(A(z)+x^{d}w)}{x^{d-1}},A(z)+x^{d}w,\\\
&&\frac{z-B(A(z)+x^{d}w)}{x^{d}}).\end{array}$

Arguing as in Remark 2.3, we may then use this isomorphism to construct an
isomorphism
${\mathbb{P}}^{3}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{3}_{g}$.
Indeed, the second component of $\varphi$ can be replaced by a polynomial,
since

$Q\left(A(z)+x^{d}w\right)\equiv P\cdot(1-dx^{d-2}z^{2})\pmod{x^{d-1}}$

and $P\equiv x^{d-1}y\pmod{f-1}$. Moreover, the fourth component of $\varphi$
is a polynomial, since $z-A(B(z))\equiv 0\pmod{x^{d}}$. Doing this, we obtain
an expression for $\varphi$ given by four polynomials. It remains to check
that each monomial appearing in that polynomial expression of $\varphi$ has
degree congruent to $1$ modulo $d$. Working modulo $d$ with the degrees, we
observe that $A,B,P,Q$ are homogeneous of degree $1,1,0,0$, respectively,
hence that the polynomials $z-B(A(z)+x^{d}w)$, $Q(A(z)+x^{d}w)$ and
$P-x^{d-1}y$ are homogeneous of degree $1$, $0$ and $0$, respectively.
Therefore, we can conclude that
${\mathbb{P}}^{3}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{3}_{g}$
as in Remark 2.3 and furthermore that
${\mathbb{P}}^{r+2}_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{r+2}_{g}$
for all $r\geq 1$ by Lemma 2.4. ∎

######  Proposition 3.4.

Let $\mathbf{k}$ be a field, $d\geq 3$ be an integer and let
$f,g\in{\mathbf{k}}[x_{0},x_{1},y,z]$ be the homogeneous polynomials of degree
$d$ defined by

$f=x_{0}^{d-1}y+z^{d}\quad\textrm{and}\quad
g=x_{0}^{d-1}y+z^{d}+dx_{0}^{d-2}x_{1}^{2}\,.$

Then, for each $r\geq 1$, the open subvarieties
${\mathbb{P}}^{r+3}_{f},{\mathbb{P}}^{r+3}_{g}\subseteq{\mathbb{P}}^{r+3}$ are
isomorphic.

######  Remark 3.5.

We don’t know, whether the open subvarieties
${\mathbb{P}}^{3}_{f},{\mathbb{P}}^{3}_{g}\subseteq{\mathbb{P}}^{3}$ are
isomorphic, or even if the hypersurfaces
$X_{f,1},X_{g,1}\subseteq{\mathbb{A}}^{3}$ given by $f=1$ and $g=1$,
respectively, are isomorphic or not.

###### Proof of Proposition 3.4.

We define $s=1$, $S=\mathbf{k}[x_{0},x_{1}]$, $m=d$, $n=d-1$,
$\Delta=x_{0}^{d-2}x_{1}^{2}$,

$\begin{array}[]{rclrcl}P(z)&=&1-z^{d},&Q(z)&=&1-z^{d}-d\Delta,\\\
A(z)&=&z(1-\Delta+\Delta^{2}),&B(z)&=&z(1+\Delta)\,,\end{array}$

and prove that the hypotheses of Lemma 3.1 are satisfied for the above
polynomials in $S[z]$.

Firstly, we calculate
$A(B(z))=B(A(z))=z(1-\Delta+\Delta^{2})(1+\Delta)=z(1+\Delta^{3})$. As $d\geq
3$, $\Delta^{3}$ is divisible by $x_{0}^{3d-6}$. Hence, we find that
$A(B(z))=B(A(z))$ is congruent to $z$ modulo $x_{0}^{d}$.

We then check that $P(B(z)+x_{0}^{d}w)\in x_{0}^{d-1}S[z,w]+Q(z)S[z,w]$. As
$d\geq 3$, $\Delta^{2}$ is divisible by $x_{0}^{d-1}$. Hence:

$\displaystyle P(B(z)+x_{0}^{d}w)$ $\displaystyle\equiv P(B(z))$
$\displaystyle\equiv 1-z^{d}(1+\Delta)^{d}$ $\displaystyle\equiv
1-z^{d}(1+d\Delta)$ $\displaystyle\equiv(1-z^{d}-d\Delta)\cdot(1+d\Delta)$
$\displaystyle\equiv Q\cdot(1+d\Delta)\,\pmod{x_{0}^{d-1}}.$

Similarly, we have $Q(A(z)+x_{0}^{d}w)\in x_{0}^{d-1}S[z,w]+P(z)S[z,w]$:

$\displaystyle Q(A(z)+x_{0}^{d}w)$ $\displaystyle\equiv Q(A(z))$
$\displaystyle\equiv 1-z^{d}(1-\Delta)^{d}-d\Delta$ $\displaystyle\equiv
1-z^{d}(1-d\Delta)-d\Delta$ $\displaystyle\equiv(1-z^{d})\cdot(1-d\Delta)$
$\displaystyle\equiv P\cdot(1-d\Delta)\,\pmod{x_{0}^{d-1}}.$

Since $f-1=x_{0}^{d-1}y-P(z)$ and $g-1=x_{0}^{d-1}y-Q(z)$, we may now apply
Lemma 3.1 to obtain an isomorphism

$\begin{array}[]{rcl}\varphi\colon\mathrm{Spec}({\mathbf{k}}[x_{0},x_{1},y,z,w]/(f-1))&\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}&\mathrm{Spec}({\mathbf{k}}[x_{0},x_{1},y,z,w]/(g-1))\\\
(x_{0},x_{1},y,z,w)&\mapsto&(x_{0},x_{1},\frac{Q(A(z)+x_{0}^{d}w)}{x_{0}^{d-1}},A(z)+x_{0}^{d}w,\\\
&&\frac{z-B(A(z)+x_{0}^{d}w)}{x_{0}^{d}})\,.\end{array}$

We again proceed as in Remark 2.3. The third component of $\varphi$ can be
expressed by a polynomial, since

$Q(A(z)+x_{0}^{d}w)\equiv P\cdot(1-dx_{0}^{d-2}x_{1}^{2})\pmod{x_{0}^{d-1}}$

and $P\equiv x_{0}^{d-1}y\pmod{f-1}$. The last component of $\varphi$ is
already a polynomial, as its numerator is a multiple of its denominator. It
remains to check that each monomial appearing in the expression of $\varphi$
has degree congruent to $1$ modulo $d$. For this, we simply work with the
degrees modulo $d$ and observe that $\Delta$ is homogeneous of degree $0$, so
that $A,B,P,Q$ are homogeneous of degree $1,1,0,0$, respectively. Hence,
$z-B(A(z)+x_{0}^{d}w)$, $Q(A(z)+x_{0}^{d}w)$ and $P-x_{0}^{d-1}y$ are
homogeneous of degree $1$, $0$ and $0$, respectively. Therefore, we can
conclude by Remark 2.3 and Lemma 2.4. ∎

###  3.2. Non-isomorphic hypersurfaces

We now want to prove that the hypersurfaces of Propositions 3.2 and 3.4 are
not isomorphic (except when the characteristic divides $d$, in which case both
are in fact equal).

Recall that the multiplicity $\mathrm{mult}_{p}(X)$ of a variety $X$ at a
point $p$ is the multiplicity of the maximal ideal $\mathfrak{m}_{X,p}$ in the
local ring $\mathcal{O}_{X,p}$.

######  Lemma 3.6.

Let $n\geq 1$ be an integer, let $H\subseteq{\mathbb{P}}^{n}$ be a
hypersurface of degree $d\geq 1$. The set $M=\\{p\in
H\mid\mathrm{mult}_{p}(H)=d\\}$ is either empty or a linear subspace of
${\mathbb{P}}^{n}$.

###### Proof.

It suffices to take two different points $p,q\in M$ and to prove that the line
containing $p$ and $q$ is contained in $M$. Changing coordinates, we may
assume that $p=[0:\cdots:0:1]$ and $q=[0:\cdots:0:1:0]$. The equation of $H$
is then given by a non-zero homogeneous polynomial
$P\in{\mathbf{k}}[x_{0},\ldots,x_{n-2}]$ of degree $d\geq 1$ (and thus we have
$n\geq 2$). Hence the line $V_{{\mathbb{P}}^{n}}(x_{0},\ldots,x_{n-2})$ (which
passes through $p$ and $q$) is contained in $M$. ∎

######  Lemma 3.7.

Let $\mathbf{k}$ be a field, let $n\geq 1$ and let
$f,g\in\mathbf{k}[x_{0},\ldots,x_{n}]$ be homogeneous polynomials of the same
degree $d\geq 1$. If $V_{{\mathbb{P}}^{n+1}}(f)$ and
$V_{{\mathbb{P}}^{n+1}}(g)$ are isomorphic, then $V_{{\mathbb{P}}^{n}}(f)$ and
$V_{{\mathbb{P}}^{n}}(g)$ are isomorphic.

###### Proof.

We write $H=V_{{\mathbb{P}}^{n+1}}(f)$,
$H^{\prime}=V_{{\mathbb{P}}^{n+1}}(g)$. Let $\varphi\colon
H\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}H^{\prime}$ be an
isomorphism. It then sends $M$ onto $M^{\prime}$, where $M=\\{\,x\in H\ |\
\mathrm{mult}_{p}(H)=d\,\\}$ and $M^{\prime}=\\{\,x\in H^{\prime}\ |\
\mathrm{mult}_{p}(H^{\prime})=d\,\\}$. As
$f,g\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$, the point
$q=[0:\cdots:0:1]\in{\mathbb{P}}^{n+1}$ belongs to both $M$ and $M^{\prime}$.
By Lemma 3.6, $M$ and $M^{\prime}$ are linear subspaces of
${\mathbb{P}}^{n+1}$ given by $r\geq 1$ linearly independent linear
polynomials in the variables $x_{0},\ldots,x_{n}$ (and $r\leq n+1$). Applying
a change of coordinates on $x_{0},\ldots,x_{n}$, we may assume that
$M=M^{\prime}=V_{{\mathbb{P}}^{n+1}}(x_{0},\ldots,x_{r-1})$. The restriction
of $\varphi$ then becomes a linear automorphism of $M=M^{\prime}$, and
$f,g\in{\mathbf{k}}[x_{0},\ldots,x_{r-1}]$. Replacing $\varphi$ with
$\varphi\circ\alpha$, where $\alpha\in\mathrm{Aut}({\mathbb{P}}^{n+1})$ does
nothing on $x_{0},\ldots,x_{r-1}$ and is a linear automorphism on
$x_{r},\ldots,x_{n+1}$, we may assume that $\varphi$ fixes the point
$q=[0:0:\cdots:0:1]$. It then induces an isomorphism
$\hat{H}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}\hat{H}^{\prime}$,
where $\hat{H}$, $\hat{H}^{\prime}$ are the blow-ups of $H$ and $H^{\prime}$
at the point $q$. The exceptional divisors are isomorphic to
$V_{{\mathbb{P}}^{n}}(f)$ and $V_{{\mathbb{P}}^{n}}(g)$, respectively, so we
find that these hypersurfaces are isomorphic. ∎

######  Lemma 3.8.

Let $\mathbf{k}$ be a field, let $d\geq 4$ be an integer that is not a
multiple of $\mathrm{char}(\mathbf{k})$ and let

$f=x^{d-1}y+z^{d}\text{ and }g=x^{d-1}y+z^{d}+\mu x^{d-2}z^{2},$

where $\mu\in{\mathbf{k}}$. Suppose that $\alpha\colon
V_{{\mathbb{P}}^{2}}(f)\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}V_{{\mathbb{P}}^{2}}(g)$
is an isomorphism. Then, $\mu=0$ and $\alpha$ is equal to
$[x:y:z]\mapsto[a^{d}x:y:a^{d-1}z]$ for some $a\in{\mathbf{k}}^{*}$.

###### Proof.

Expressing $y$ in terms of $x,z$, we obtain two birational morphisms

$\begin{array}[]{rrcl}\tau\colon&{\mathbb{P}}^{1}&\to&V_{{\mathbb{P}}^{2}}(f)\\\
&{[u:v]}&\mapsto&[u^{d}:-v^{d}:u^{d-1}v]\end{array}$

and

$\begin{array}[]{rrcl}\tau^{\prime}\colon&{\mathbb{P}}^{1}&\to&V_{{\mathbb{P}}^{2}}(g)\\\
&{[u:v]}&\mapsto&[u^{d}:-v^{d}-\mu u^{d-2}v^{2}:u^{d-1}v]\end{array}$

whose inverses are given by $[x:y:z]\mapsto[x:z]$. They induce isomorphisms
${\mathbb{P}}^{1}\setminus\\{[0:1]\\}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}V_{{\mathbb{P}}^{2}}(f)\setminus\\{[0:1:0]\\}$
and
${\mathbb{P}}^{1}\setminus\\{[0:1]\\}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}V_{{\mathbb{P}}^{2}}(g)\setminus\\{[0:1:0]\\}$.

Both $\tau$ and $\tau^{\prime}$ are bijective and they send $[0:1]$ onto
$[0:1:0]$, which is the unique singular point of $V_{{\mathbb{P}}^{2}}(f)$ and
$V_{{\mathbb{P}}^{2}}(g)$, respectively. Since the isomorphism $\alpha\colon
V_{{\mathbb{P}}^{2}}(f)\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}V_{{\mathbb{P}}^{2}}(g)$
must fix the point $[0:1:0]$, the birational map
$\hat{\alpha}=(\tau^{\prime})^{-1}\alpha\tau$ is an automorphism of
${\mathbb{P}}^{1}$ that fixes the point $[0:1]$. Thus, $\hat{\alpha}$ is of
the form $[u:v]\mapsto[u:av+bu]$ for some $a\in{\mathbf{k}}^{*}$,
$b\in{\mathbf{k}}$.

Note that $s_{a}\colon[x:y:z]\mapsto[a^{d}x:y:a^{d-1}z]$ is an automorphism of
$V_{{\mathbb{P}}^{2}}(f)$, that lifts to
$\hat{s}_{a}=(\tau)^{-1}s_{a}\tau=[u:v]\mapsto[au:v]$. Replacing $\alpha$ with
$\alpha s_{a}$, we replace $\hat{\alpha}$ with $\hat{\alpha}\hat{s}_{a}$, and
we may assume that $a=1$. It remains to see that $b=\mu=0$. We calculate

$\displaystyle\alpha([x:y:z])$ $\displaystyle=$
$\displaystyle\tau^{\prime}\hat{\alpha}\tau^{-1}([x:y:z])$ $\displaystyle=$
$\displaystyle\tau^{\prime}\hat{\alpha}([x:z])$ $\displaystyle=$
$\displaystyle\tau^{\prime}([x:z+bx])$ $\displaystyle=$
$\displaystyle[x^{d}:-(z+bx)^{d}-\mu x^{d-2}(z+bx)^{2}:x^{d-1}(z+bx)]\,.$

As $\alpha$ fixes $[0:1:0]$, the pull-back by $\alpha$ of $x/y$ and $z/y$ are
rational functions

$\frac{x^{d}}{-(z+bx)^{d}-\mu
x^{d-2}(z+bx)^{2}}\quad\text{and}\quad\frac{x^{d-1}}{-(z+bx)^{d-1}-\mu
x^{d-2}(z+bx)}\,,$

whose restrictions to $V_{{\mathbb{P}}^{2}}(f)$ are regular at $[0:1:0]$. For
each $j\in\\{d-1,d\\}$, there are thus two homogeneous polynomials
$P_{j},Q_{j}\in{\mathbf{k}}[x,y,z]$, of the same degree, such that
$Q_{j}(0,1,0)\neq 0$ and such that

$\frac{P_{j}(x,y,z)}{Q_{j}(x,y,z)}=\frac{x^{j}}{-(z+bx)^{j}-\mu
x^{d-2}(z+bx)^{j+2-d}}$

on the curve $V_{{\mathbb{P}}^{2}}(f)$. Using the morphism

${\mathbb{A}}^{1}\to V_{{\mathbb{P}}^{2}}(f)\,,\quad
t\mapsto\tau([t:1])=[t^{d}:-1:t^{d-1}]=[t:-t^{1-d}:1]\,,$

we find

$\frac{P_{j}(t^{d},-1,t^{d-1})}{Q_{j}(t^{d},-1,t^{d-1})}=\frac{t^{j}}{B_{j}}\,,$

where $B_{j}=-(1+bt)^{j}-\mu t^{d-2}(1+bt)^{j+2-d}\in\mathbf{k}[t]$. As
$Q_{j}(0,1,0)\neq 0$, the polynomial
$Q_{j}(t^{d},-1,t^{d-1})\in{\mathbf{k}}[t]$ is not divisible by $t$. As
$B_{j}(0)\neq 0$ (we use here $d\geq 4$), $t^{j}$ divides
$P_{j}(t^{d},-1,t^{d-1})$. There is thus $A_{j}\in{\mathbf{k}}[t]$ such that

$P_{j}(t^{d},-1,t^{d-1})=A_{j}\cdot t^{j}\quad\text{and}\quad
Q_{j}(t^{d},-1,t^{d-1})=A_{j}\cdot B_{j}\,.$

We consider now the case $j=d$ and prove that $b=0$. We shall afterwards prove
$\mu=0$ by considering the case $j=d-1$.

As $d\geq 4$, the equality $P_{d}(t^{d},-1,t^{d-1})=A_{d}\cdot t^{d}$ gives
the existence of $\epsilon\in{\mathbf{k}}$ such that
$A_{d}\equiv\epsilon\pmod{t^{2}}$. We moreover have
$B_{d}\equiv-1-bdt\pmod{t^{2}}$. This gives

$Q_{d}(t^{d},-1,t^{d-1})=A_{d}\cdot B_{d}\equiv-\epsilon-\epsilon
bdt\pmod{t^{2}}.$

The polynomial $Q(t^{d},-1,t^{d-1})$ is not divisible by $t$, so $\epsilon\neq
0$, and its coefficient of $t$ is zero, as $d\geq 4$. Since
$\mathrm{char}({\mathbf{k}})$ does not divide $d$, this gives $b=0$.

We now use $j=d-1$. As $d\geq 4$, the equality
$P_{d-1}(t^{d},-1,t^{d-1})=A_{d-1}\cdot t^{d-1}$ gives the existence of
$\xi,\xi^{\prime}\in{\mathbf{k}}$ such that
$A_{d-1}\equiv\xi+t\xi^{\prime}\pmod{t^{d-1}}$. As $b=0$, we find
$B_{d-1}=B_{d}=-1-\mu t^{d-2}$. This gives

$Q_{d-1}(t^{d},-1,t^{d-1})=A_{d-1}\cdot B_{d-1}\equiv-\xi-t\xi^{\prime}-\mu\xi
t^{d-2}\pmod{t^{d-1}}.$

As $t$ does not divide this polynomial, $\xi\neq 0$. Moreover, we obtain
$\xi^{\prime}=0$ and $\mu=0$, as $d\geq 4$. ∎

######  Proposition 3.9.

Let $\mathbf{k}$ be a field, let $d\geq 4$ be an integer that is not a
multiple of $\mathrm{char}(\mathbf{k})$ and let $f,g\in\mathbf{k}[x,y,z]$ be
defined by

$f=x^{d-1}y+z^{d}\text{ and }g=x^{d-1}y+z^{d}+\mu x^{d-2}z^{2},$

where $\mu\in{\mathbf{k}}^{*}$. Then, for each $r\geq 0$, the two
hypersurfaces $V_{{\mathbb{P}}^{r+2}}(f)$, $V_{{\mathbb{P}}^{r+2}}(g)$ in
${\mathbb{P}}^{r+2}$ are not isomorphic.

###### Proof.

By Lemma 3.8, the result is true when $r=0$. Using Lemma 3.7, we can then
argue by induction and obtain the result for every integer $r$. ∎

######  Proposition 3.10.

Let $\mathbf{k}$ be a field, $d\geq 3$ be an integer and let
$f,g\in\mathbf{k}[x_{0},x_{1},y,z]$ be defined by

$f=x_{0}^{d-1}y+z^{d}\,,\quad g=x_{0}^{d-1}y+z^{d}+\mu
x_{0}^{d-2}x_{1}^{2}\,,$

where $\mu\in{\mathbf{k}}^{*}$. Then, for each $r\geq 0$, the two
hypersurfaces $V_{{\mathbb{P}}^{r+3}}(f)$, $V_{{\mathbb{P}}^{r+3}}(g)$ in
${\mathbb{P}}^{r+3}$ are not isomorphic.

###### Proof.

By Lemma 3.7, it suffices to consider the case $r=0$ and to prove that
$V_{{\mathbb{P}}^{3}}(f)$ and $V_{{\mathbb{P}}^{3}}(g)$ are not isomorphic.
Looking at the derivative with respect to $y$, the singular locus of both
hypersurfaces is contained in the line $\ell\subseteq{\mathbb{P}}^{3}$ given
by $x_{0}=z=0$.

The surface $V_{{\mathbb{P}}^{3}}(f)$ has multiplicity $d$ at the point where
$x_{0}=y=z=0$.

It remains to see that $V_{{\mathbb{P}}^{3}}(g)$ has multiplicity $<d$ at
every point. For this, write $g=z^{d}+x_{0}^{d-2}(x_{0}y+\mu x_{1}^{2})$ and
observe that $V_{{\mathbb{P}}^{3}}(x_{0}y+\mu x_{1}^{2})$ is smooth outside
$x_{0}=x_{1}=y=0$ and thus on $\ell$. ∎

We may now prove Theorem A, which is a direct consequence of Propositions 3.2,
3.4, 3.9 and 3.10.

###### Proof of Theorem A.

As in the statement, we take a field $\mathbf{k}$, and integers $d,n\geq 3$
such that $d$ is not a multiple of $\mathrm{char}(\mathbf{k})$ and such that
$(d,n)\neq(3,3)$. We are looking for hypersurfaces
$H=V_{{\mathbb{P}}^{n}}(f),H^{\prime}=V_{{\mathbb{P}}^{n}}(g)\subseteq{\mathbb{P}}^{n}_{\mathbf{k}}$
that are not isomorphic but have isomorphic complements.

If $d\geq 4$, we may choose $f=x^{d-1}y+z^{d}$ and
$g=x^{d-1}y+z^{d}+dx^{d-2}z^{2}$, that we see in
${\mathbf{k}}[x,y,z,w_{1},\ldots,w_{r}]$ with $r=n-2\geq 1$. By Propositions
3.2, the complements of $H$ and $H^{\prime}$ are isomorphic, and by
Proposition 3.9, the hypersurfaces $H$ and $H^{\prime}$ are not isomorphic.

If $n\geq 4$, we may choose $f=x_{0}^{d-1}y+z^{d}$ and
$g=x_{0}^{d-1}y+z^{d}+dx_{0}^{d-2}x_{1}^{2}$, that we see in
${\mathbf{k}}[x_{0},x_{1},y,z,w_{1},\ldots,w_{r}]$ with $r=n-3\geq 1$. By
Propositions 3.4, the complements of $H$ and $H^{\prime}$ are isomorphic, and
by Proposition 3.10, the hypersurfaces $H$ and $H^{\prime}$ are not
isomorphic. ∎

##  4\. Topological Euler characteristic and piecewise isomorphisms

Throughout this section we assume that $\mathbf{k}$ is algebraically closed.

In the sequel we recall the definition and some basic facts of the topological
Euler characteristic. In order to do this, we also recall some facts from
étale $\ell$-adic cohomology with compact support. As a reference we take
[Mil13] and [Mil80].

Let $\ell$ be a prime number that is different from the characteristic of the
ground field $\mathbf{k}$. For a variety $X$, the group
$H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})$ denotes the $i$-th étale
$\ell$-adic cohomology with compact support, i.e.

$H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})\coloneqq\left(\varprojlim
H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Z}}/\ell^{n}{\mathbb{Z}})\right)\otimes_{{\mathbb{Z}}_{\ell}}{\mathbb{Q}}_{\ell}\,,$

where
${\mathbb{Z}}_{\ell}\coloneqq\varprojlim{\mathbb{Z}}/\ell^{n}{\mathbb{Z}}$ and
${\mathbb{Q}}_{\ell}$ is the quotient field of ${\mathbb{Z}}_{\ell}$, see e.g.
[Mil13, §18, §19]. We have:

######  Lemma 4.1.

Let $X$ be a variety. Then the ${\mathbb{Q}}_{\ell}$-vector space
$H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})$ has finite dimension and
vanishes for $i>2\dim X$.

###### Proof.

Let $k\colon X\to\bar{X}$ be a completion. By definition (see e.g. [Mil13,
Definition 18.1]) we have

$H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Z}}/\ell^{n}{\mathbb{Z}})=H^{i}(\bar{X}_{\textrm{\'{e}t}},k_{!}({\mathbb{Z}}/\ell^{n}{\mathbb{Z}}))\,.$

By [Mil13, Theorem 19.2] the limit $\varprojlim
H^{i}(\bar{X}_{\textrm{\'{e}t}},k_{!}({\mathbb{Z}}/\ell^{n}{\mathbb{Z}}))$ is
a finitely generated ${\mathbb{Z}}_{\ell}$-module; this implies the first
statement. By [Mil13, Theorem 15.1], we have that
$H^{i}(\bar{X}_{\textrm{\'{e}t}},k_{!}({\mathbb{Z}}/\ell^{n}{\mathbb{Z}}))$
vanishes for all $i>2\dim X$ and hence the second statement follows. ∎

The topological Euler-characteristic of a variety $X$ is defined by

$\chi(X)\coloneqq\sum_{i=0}^{2\dim
X}(-1)^{i}\dim_{{\mathbb{Q}}_{\ell}}H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})\,.$

The following properties are very useful in order to compute the topological
Euler characteristic. For lack of a reference with proof, we give an argument
here:

######  Lemma 4.2.

We have:

  1. (1)

Let $X$ be a variety and let $Z\subseteq X$ be a closed subvariety. Then
$\chi(X)=\chi(X\setminus Z)+\chi(Z)$.

  2. (2)

Let $X,Y$ be varieties. Then $\chi(X\times Y)=\chi(X)\cdot\chi(Y)$.

###### Proof.

(1): Let $U\coloneqq X\setminus Z$. By [AGV73, Ch.XVII, §5.1.16] we have a
long exact sequence

$\displaystyle\ldots\to
H^{i}_{c}(U_{\textrm{\'{e}t}},{\mathbb{Z}}/\ell^{n}{\mathbb{Z}})\to
H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Z}}/\ell^{n}{\mathbb{Z}})\to
H^{i}_{c}(Z_{\textrm{\'{e}t}},{\mathbb{Z}}/\ell^{n}{\mathbb{Z}})$
$\displaystyle\to
H^{i+1}_{c}(U_{\textrm{\'{e}t}},{\mathbb{Z}}/\ell^{n}{\mathbb{Z}})\to\ldots\,.$

Taking the limit over $n$ and tensoring with ${\mathbb{Q}}_{\ell}$ over
${\mathbb{Z}}_{\ell}$ gives a long exact sequence

$\ldots\to H^{i}_{c}(U_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})\to
H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})\to
H^{i}_{c}(Z_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})\to
H^{i+1}_{c}(U_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})\to\ldots\,.$

The statement follows now by using Lemma 4.1.

(2): By the Künneth-formula (see [Mil80, Ch. VI, Corollary 8.23]) we get for
$0\leq k\leq\dim X+\dim Y$

$\dim_{{\mathbb{Q}}_{\ell}}H^{k}_{c}((X\times
Y)_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})=\sum_{i=0}^{k}\dim_{{\mathbb{Q}}_{\ell}}H^{i}_{c}(X_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})\otimes_{{\mathbb{Q}}_{\ell}}H^{k-i}_{c}(Y_{\textrm{\'{e}t}},{\mathbb{Q}}_{\ell})\,,$

which gives $\chi(X\times Y)=\chi(X)\cdot\chi(Y)$. ∎

######  Example 4.3.

Let $C$ be a smooth irreducible projective curve of genus $g$. Then

$H^{0}(C,{\mathbb{Q}}_{\ell})=H^{2}(C,{\mathbb{Q}}_{\ell})={\mathbb{Q}}_{\ell}\quad\textrm{and}\quad
H^{1}(C,{\mathbb{Q}}_{\ell})={\mathbb{Q}}_{\ell}^{2g}\,,$

see [Mil13, Proposition 14.2]. Using that $H^{i}(C,{\mathbb{Q}}_{\ell})=0$ for
all $i>2$ (see Lemma 4.1), it follows that $\chi(C)=2-2g$.

######  Example 4.4.

Let $m\geq 0$. Then

$H^{i}({\mathbb{P}}^{m},{\mathbb{Q}}_{\ell})=\left\\{\begin{array}[]{rl}{\mathbb{Q}}_{\ell}\,,&\textrm{$0\leq
i\leq 2m$ and $i$ is even}\\\ 0\,,&\textrm{otherwise}\end{array}\right.\,,$

see [Mil13, Example 16.3]. Hence, $\chi({\mathbb{P}}^{m})=m+1$. Using Lemma
4.2 and that
${\mathbb{A}}^{m}\simeq{\mathbb{P}}^{m}\setminus{\mathbb{P}}^{m-1}$ for all
$m\geq 0$ (where ${\mathbb{P}}^{m-1}$ is linearly embedded in
${\mathbb{P}}^{m}$ and ${\mathbb{P}}^{-1}=\varnothing$), it follows that
$\chi({\mathbb{A}}^{m})=1$ for all $m\geq 0$.

######  Example 4.5.

Let $n\geq 0$ and let $\pi\colon X\to B$ be a locally trivial
${\mathbb{A}}^{n}$-bundle with respect to the Zariski topology, where $B$ is
any variety. Then $\chi(X)=\chi(B)$. Indeed, we proceed by induction on $\dim
B$ and note that the case $\dim B=0$ is clear. By assumption there exists a
closed subvariety $B^{\prime}\subseteq B$ such that $\pi$ is trivial over
$B\setminus B^{\prime}$ and $\dim B^{\prime}<\dim B$. Hence,

$\chi(X)=\chi({\mathbb{A}}^{n}\times(B\setminus
B^{\prime}))+\chi(\pi^{-1}(B^{\prime}))=\chi({\mathbb{A}}^{n}\times
B)=\chi({\mathbb{A}}^{n})\chi(B)=\chi(B)\,,$

where for the second equality we used the induction hypothesis
$\chi(\pi^{-1}(B^{\prime}))=\chi(B^{\prime})=\chi({\mathbb{A}}^{n}\times
B^{\prime})$.

Let us denote by $\operatorname{Var}_{\mathbf{k}}$ the set of isomorphism
classes of varieties (over $\mathbf{k}$) and denote by
${\mathbb{Z}}\operatorname{Var}_{\mathbf{k}}$ the free abelian group over
$\operatorname{Var}_{\mathbf{k}}$. Moreover, let
$I\subseteq{\mathbb{Z}}\operatorname{Var}_{\mathbf{k}}$ be the subgroup that
is generated by

$[X]-[Z]-[X\setminus Z]$

for all varieties $X$ and closed subvarieties $Z$ (here $[W]$ denotes the
isomorphism class of a variety $W$). By bilinear extension of the operation
$[X]\cdot[Y]\coloneqq[X\times Y]$ we get a ring structure on
${\mathbb{Z}}\operatorname{Var}_{\mathbf{k}}$ and $I$ is an ideal of it. The
_Grothendieck ring_ is then the quotient of
${\mathbb{Z}}\operatorname{Var}_{\mathbf{k}}$ by $I$:

$K_{0}(\operatorname{Var}_{\mathbf{k}})\coloneqq{\mathbb{Z}}\operatorname{Var}_{\mathbf{k}}/I\,,$

see e.g. [LS10, §2.2.1]. By abuse of notation we denote by $[X]$ the class of
a variety $X$ in $K_{0}(\operatorname{Var}_{\mathbf{k}})$. Using Lemma 4.2,
the topological Euler characteristic gives a ring homomorphism

$\chi\colon
K_{0}(\operatorname{Var}_{\mathbf{k}})\to{\mathbb{Z}}\,,\quad\sum_{i=0}^{n}a_{i}[X_{i}]\mapsto\sum_{i=0}^{n}a_{i}\chi(X_{i}).$

######  Definition 4.6.

Let $X,Y$ be varieties. Then $X,Y$ are called _piecewise isomorphic_ if $X$
and $Y$ can be decomposed as disjoint unions of $n\geq 1$ locally closed
subsets $X_{1},\ldots,X_{n}$ and $Y_{1},\ldots,Y_{n}$, respectively, such that
$X_{i}\simeq Y_{i}$ for all $1\leq i\leq n$. Note that we take on $X_{i}$ the
reduced scheme structure.

Note that “piecewise isomorphic” defines an equivalence relation among the
varieties.

######  Remark 4.7.

In the above definition, if some $X_{i}$ is not irreducible, we replace
$X_{i}$ in the decomposition by one of its irreducible components and its
complement, and do the same for their images in $Y_{i}$. After finitely many
such steps we may assume that all $X_{i}$ are irreducible, and thus all
$Y_{i}$ are irreducible.

######  Remark 4.8.

If $X,Y$ are piecewise isomorphic, then $\dim X=\dim Y$.

######  Example 4.9.

Let $\Gamma\subseteq{\mathbb{P}}^{2}$ be an irreducible cuspidal cubic curve.
Then $\Gamma$ is piecewise isomorphic to ${\mathbb{P}}^{1}$. Indeed, we choose
$X_{1},X_{2}\subseteq\Gamma$ as $X_{1}$ to be the singular locus (one point)
and $X_{2}=\Gamma\setminus X_{1}$. Then, $X_{1}$ is isomorphic to
$Y_{1}=[0:1]\in{\mathbb{P}}^{1}$ and $X_{2}$ is isomorphic to
$Y_{2}={\mathbb{P}}^{2}\setminus Y_{1}\simeq{\mathbb{A}}^{1}$.

######  Lemma 4.10.

Let $X,Y$ be varieties and consider the following statements:

  1. (1)

$X$ and $Y$ are piecewise isomorphic;

  2. (2)

There are open subsets $U\subseteq X$, $V\subseteq Y$ such that

$U\simeq V\,,\quad\dim X\setminus U=\dim Y\setminus V<\dim X=\dim Y$

and $X\setminus U$, $Y\setminus V$ are piecewise isomorphic. Moreover, if
$X^{\prime}\subseteq X$, $Y^{\prime}\subseteq Y$ are locally closed subsets
such that

$\dim X\setminus X^{\prime},\dim Y\setminus Y^{\prime}<\dim X=\dim Y$

we may assume that $U\subseteq X^{\prime}$ and $V\subseteq Y^{\prime}$.

  3. (3)

$[X]=[Y]$ inside $K_{0}(\operatorname{Var}_{\mathbf{k}})$;

  4. (4)

$\chi(X)=\chi(Y)$.

Then we have
$\ref{lemm.Grothendieckring1}\Leftrightarrow\ref{lemm.Grothendieckring1.5}\Rightarrow\ref{lemm.Grothendieckring3}\Rightarrow\ref{lemm.Grothendieckring4}$.

###### Proof.

“$\ref{lemm.Grothendieckring1}\Rightarrow\ref{lemm.Grothendieckring1.5}$”: Let
$X_{1},\ldots X_{n}$ and $Y_{1},\ldots Y_{n}$ be disjoint locally closed
subsets of $X$ and $Y$, respectively, such that their union is equal to $X$
and $Y$, respectively, and such that there are isomorphisms $\psi_{i}\colon
X_{i}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}Y_{i}$ for $1\leq i\leq
n$. By replacing $X_{i}$ with $X_{i}\cap X^{\prime}$ and $X_{i}\setminus
X^{\prime}$ and analogously for $Y_{i}$ we may assume that each $X_{i}$ is
either contained in $X^{\prime}$ or $X\setminus X^{\prime}$ and $Y_{i}$ is
either contained in $Y^{\prime}$ or $Y\setminus Y^{\prime}$. Using Remark 4.7,
we may moreover assume that all $X_{i}$ and $Y_{i}$ are irreducible.

Let $d\coloneqq\dim X=\dim Y$. After reordering, we may assume that the
closures $\overline{X_{1}},\ldots,\overline{X_{m}}$ in $X$ are the irreducible
components of $X$ of dimension $d$. Then
$\overline{Y_{1}},\ldots,\overline{Y_{m}}$ are irreducible closed subsets of
$Y$ of dimension $d$ and since $Y_{1},\ldots,Y_{m}$ are disjoint and locally
closed in $Y$, $\overline{Y_{1}},\ldots,\overline{Y_{m}}$ are mutually
different. This implies that $\overline{Y}_{1},\ldots,\overline{Y}_{m}$ are
the irreducible components of $Y$ of dimension $d$. For $1\leq i\leq m$, note
that $X_{i}$ and $Y_{i}$ is contained in $X^{\prime}$ and $Y^{\prime}$,
respectively, since $X\setminus X^{\prime}$ and $Y\setminus Y^{\prime}$ have
strictly lower dimension that $X$ and $Y$.

For each $i\in\\{1,\ldots,m\\}$, the locally closed subset $X_{i}\subseteq X$
is equal to $X_{i}=A\cap\overline{X_{i}}$, where $A\subseteq X_{i}$ is open.
Choosing $U_{i}^{\prime}\subseteq X$ to be the open subset $A\setminus I$,
where $I$ is the union of all irreducible components of $X$ different from
$\overline{X_{i}}$, we find a subset $U^{\prime}_{i}\subseteq X_{i}$ that is
open in $X$ with $\overline{U^{\prime}_{i}}=\overline{X}_{i}$. By
construction, $U_{1}^{\prime},\ldots,U^{\prime}_{m}$ are disjoint. Similarly,
there exist open disjoint subsets $V^{\prime}_{1},\ldots,V^{\prime}_{m}$ in
$Y$ such that for all $1\leq i\leq m$ we get $V^{\prime}_{i}\subseteq
Y^{\prime}_{i}$ and $\overline{V_{i}^{\prime}}=\overline{Y_{i}}$. For $1\leq
i\leq m$, let

$U_{i}\coloneqq U^{\prime}_{i}\cap\psi_{i}^{-1}(V^{\prime}_{i})\subseteq
X\quad\textrm{and}\quad V_{i}\coloneqq\psi_{i}(U^{\prime}_{i})\cap
V^{\prime}_{i}\subseteq Y\,.$

Then the restriction $\psi_{i}|_{U_{i}}\colon
U_{i}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}V_{i}$ is an
isomorphism. Let $U\coloneqq\bigcup_{i=1}^{m}U_{i}$ and
$V\coloneqq\bigcup_{i=1}^{m}V_{i}$. By construction, $U\subseteq X^{\prime}$
and $V\subseteq Y^{\prime}$. Since $U_{1},\ldots,U_{m}$ are disjoint open
subsets of $X$ and $V_{1},\ldots,V_{m}$ are disjoint open subsets of $Y$, we
have $U\simeq V$. Moreover, $X\setminus U$ and $Y\setminus V$ are piecewise
isomorphic (take the locally closed subsets $X_{1}\setminus
U_{1},\ldots,X_{m}\setminus U_{m},X_{m+1},\ldots,X_{n}$ and $Y_{1}\setminus
V_{1},\ldots,Y_{m}\setminus V_{m},Y_{m+1},\ldots,Y_{n}$ of $X$ and $Y$,
respectively). By construction, $\dim X\setminus U<\dim X$.

“$\ref{lemm.Grothendieckring1.5}\Rightarrow\ref{lemm.Grothendieckring1}$”: The
decomposition of $X\setminus U$, together with $U$, gives the decomposition of
$X$, and we do similarly for $Y$.

“$\ref{lemm.Grothendieckring1.5}\Rightarrow\ref{lemm.Grothendieckring3}$”: By
assumption we have $[U]=[V]$ inside $K_{0}(\operatorname{Var}_{\mathbf{k}})$.
As $X\setminus U$, $Y\setminus V$ are piecewise isomorphic and since $\dim
X\setminus U=\dim Y\setminus V<\dim X=\dim Y$ we may proceed by induction on
$d$ in order to get $[X\setminus U]=[Y\setminus V]$ inside
$K_{0}(\operatorname{Var}_{\mathbf{k}})$. This implies (3).

“$\ref{lemm.Grothendieckring3}\Rightarrow\ref{lemm.Grothendieckring4}$”:
Follows from Lemma 4.2. ∎

######  Corollary 4.11.

If $X$ and $Y$ are two varieties that are both equidimensional and piecewise
isomorphic, then $X$ and $Y$ are birational to each other.

###### Proof.

Follows from the implication (1) $\Rightarrow$ (2) in Lemma 4.10. ∎

######  Example 4.12.

Corollary 4.11 really needs both $X$ and $Y$ to be equidimensional. Indeed,
${\mathbb{A}}^{1}\amalg\\{pt\\}$ and ${\mathbb{P}}^{1}$ are piecewise
isomorphic, but not birational.

######  Lemma 4.13.

For $i\in\\{1,2\\}$, let $\Gamma_{i}$ be a variety of dimension $1$, and let
$\Gamma_{i}^{\prime}\subseteq\Gamma_{i}$ be the union of the irreducible
components of $\Gamma_{i}$ of dimension $1$ $($we remove isolated points$)$.
Then, the following are equivalent:

  1. (1)

$\Gamma_{1}$ and $\Gamma_{2}$ are piecewise isomorphic;

  2. (2)

$\Gamma_{1}^{\prime}$ and $\Gamma_{2}^{\prime}$ are birational, and
$\chi(\Gamma_{1})=\chi(\Gamma_{2})$.

###### Proof.

“$\ref{CurvepiecewiseBir1}\Rightarrow\ref{CurvepiecewiseBir2}$”: Suppose that
$\Gamma_{1}$ and $\Gamma_{2}$ are piecewise isomorphic. The implication (1)
$\Rightarrow$ (2) in Lemma 4.10 implies that $\Gamma_{1}^{\prime}$ and
$\Gamma_{2}^{\prime}$ are birational and the implication (1) $\Rightarrow$ (4)
yields $\chi(\Gamma_{1})=\chi(\Gamma_{2})$.

“$\ref{CurvepiecewiseBir2}\Rightarrow\ref{CurvepiecewiseBir1}$”: As
$\Gamma_{1}^{\prime}$ and $\Gamma_{2}^{\prime}$ are birational, there are open
dense subsets $U_{1}\subseteq\Gamma_{1}^{\prime}$ and
$U_{2}\subseteq\Gamma_{2}^{\prime}$ such that $U_{1}\simeq U_{2}$. For each
$i\in\\{1,2\\}$, the set $U_{i}$ is open in $\Gamma_{i}$, and
$F_{i}=\Gamma_{i}\setminus U_{i}$ is finite. As
$\chi(\Gamma_{1})=\chi(\Gamma_{2})$ and $\chi(U_{1})=\chi(U_{2})$, Lemma 4.2
implies that $\chi(F_{1})=\chi(F_{2})$. The two finite sets $F_{1}$ and
$F_{2}$ are thus isomorphic, which proves that $\Gamma_{1}$ and $\Gamma_{2}$
are piecewise isomorphic. ∎

######  Lemma 4.14.

  1. (1)

Let $X$ be a disjoint union of trees of smooth projective rational curves,
with $r$ irreducible components and $s$ connected components. Then, $X$ is
piecewise isomorphic to the disjoint union of $r$ copies of ${\mathbb{A}}^{1}$
and $s$ points.

  2. (2)

Let $X$, $Y$ be two disjoint unions of trees of smooth projective rational
curves. Then, $X$ and $Y$ are piecewise isomorphic if and only if they have
the same number of irreducible components and the same number of connected
components.

###### Proof.

(1): If $T$ is a tree of smooth projective rational curves, with $r$
irreducible components, it is piecewise isomorphic to the disjoint union of
one point and $r$ copies of ${\mathbb{A}}^{1}$. Hence, if $X$ has $r$
irreducible components and $s$ connected components, then $X$ is piecewise
isomorphic to the disjoint union of $r$ copies of ${\mathbb{A}}^{1}$ and $s$
points, and thus satisfies $\chi(X)=r+s$ (Lemma 4.2 and Example 4.4)

(2): By Lemma 4.13, $X$ and $Y$ are piecewise isomorphic if and only if they
are birational and $\chi(X)=\chi(Y)$. The first assertion is equivalent to ask
that $X$ and $Y$ have the same number of irreducible components. The above
calculation shows that the two assertions are equivalent to ask that $X$ and
$Y$ have the same number of irreducible components and the same number of
connected components. ∎

######  Lemma 4.15.

Let $U\subseteq{\mathbb{P}}^{2}$ be a dense open subset. For $i=1,2$, let

$X_{i}\coloneqq U\amalg S_{i}\amalg R_{i}\,,$

where $S_{i}$ is a disjoint union of $s_{i}\geq 0$ points and $R_{i}$ is the
disjoint union of $r_{i}\geq 0$ irreducible curves. If $X_{1}$ and $X_{2}$ are
piecewise isomorphic, then:

  1. (1)

$R_{1}$ and $R_{2}$ are birational; in particular $r_{1}=r_{2}$.

  2. (2)

If moreover all irreducible components of $R_{1}$, $R_{2}$ are isomorphic,
then $s_{1}=s_{2}$.

###### Proof.

By Lemma 4.10 there exist open dense subsets $U_{1},U_{2}\subseteq U$ that are
isomorphic and such that $E_{1}\coloneqq X_{1}\setminus U_{1}$ and
$E_{2}\coloneqq X_{2}\setminus U_{2}$ are piecewise isomorphic.

Let $({\mathbb{P}}^{2}\setminus U_{i})^{\prime}$ and $(U\setminus
U_{i})^{\prime}$ be the union of one-dimensional irreducible components of
${\mathbb{P}}^{2}\setminus U_{i}$ and $U\setminus U_{i}$, respectively. The
isomorphism $U_{1}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}U_{2}$
gives a birational map ${\mathbb{P}}^{2}\dasharrow{\mathbb{P}}^{2}$ that
decomposes into $n$ blow-ups and $n$ blow-downs. Hence,
$({\mathbb{P}}^{2}\setminus U_{1})^{\prime}$ and $({\mathbb{P}}^{2}\setminus
U_{2})^{\prime}$ are birational and thus $(U\setminus U_{1})^{\prime}$ and
$(U\setminus U_{2})^{\prime}$ are birational as well. By Lemma 4.13, the
unions of one-dimensional irreducible components of $E_{1}$ and $E_{2}$ are
birational and since $E_{i}\coloneqq(U\setminus U_{i})\amalg S_{i}\amalg
R_{i}$ it follows that $R_{1}$ and $R_{2}$ are birational. If the irreducible
components of $R_{1}$ and $R_{2}$ are all isomorphic to $C$, we get
$\chi(E_{i})=\chi(U)-\chi(U_{i})+s_{i}+r_{i}\chi(C)$ and since
$\chi(E_{1})=\chi(E_{2})$, $\chi(U_{1})=\chi(U_{2})$ we get $s_{1}=s_{2}$. ∎

######  Example 4.16.

We show that Lemma 4.13 does not generalise to higher dimension.

The irreducible surfaces $X={\mathbb{A}}^{2}$ and
$Y={\mathbb{P}}^{2}\setminus\\{[1:0:0],[0:1:0]\\}$ are birational and satisfy
$\chi(X)=\chi(Y)=1$ (Lemma 4.2 and Example 4.4), but $X$ and $Y$ are not
piecewise isomorphic. Indeed, this last assertion follows from Lemma 4.15 and
the fact that $X$ is piecewise isomorphic to
${\mathbb{A}}^{2}\setminus\\{(0,0)\\}\amalg\\{pt\\}$ and $Y$ is piecewise
isomorphic to ${\mathbb{A}}^{2}\setminus\\{(0,0)\\}\amalg{\mathbb{A}}^{1}$.

######  Proposition 4.17.

Let $X$, $Y$ be two varieties, that are piecewise isomorphic and let
$E\subseteq X$, $F\subseteq Y$ be locally closed subvarieties of dimension at
most $1$. If $E$ and $F$ are piecewise isomorphic, then $X\setminus E$ and
$Y\setminus F$ are piecewise isomorphic.

###### Proof.

By definition, $X$ and $Y$ can be decomposed as disjoint unions of $n\geq 1$
locally closed subsets $X_{1},\ldots,X_{n}$ and $Y_{1},\ldots,Y_{n}$,
respectively, such that there is an isomorphism $\varphi_{i}\colon
X_{i}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}Y_{i}$ for each
$i\in\\{1,\ldots,n\\}$.

$\mathbf{(A)}$ We first consider the case where $E$ is one point (and thus $F$
is also one point). In this case, there is exactly one $i\in\\{1,\ldots,n\\}$
and one $j\in\\{1,\ldots,n\\}$ such that $E\in X_{i}$ and $F\subseteq Y_{j}$.
If $i=j$, the isomorphism $\varphi_{i}$ induces an isomorphism from
$X_{i}\setminus(E\cup\varphi_{i}^{-1}(F))$ to
$Y_{i}\setminus(F\cup\varphi_{i}(E))$, and we are done if $\varphi_{i}(E)=F$
or otherwise use an isomorphism from the point $\varphi_{i}^{-1}(F)$ to the
point $\varphi_{i}(E)$. If $i\neq j$, then $\varphi_{i}$ gives an isomorphisms
$X_{i}\setminus
E\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}Y_{i}\setminus\varphi_{i}(E)$
and $\varphi_{j}$ gives an isomorphism
$X_{j}\setminus\varphi_{j}^{-1}(F)\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}Y_{j}\setminus
F$. It then remains to use an isomorphism from the point $\varphi_{j}^{-1}(F)$
to the point $\varphi_{i}(E)$.

$\mathbf{(B)}$ If $E$ (and thus $F$) is finite, we proceed by induction on the
number of points, applying $\mathbf{(A)}$, and obtain the result.

$\mathbf{(C)}$ In the general case, we proceed by induction on the number of
irreducible components of $E$ of dimension $1$. When no such component exists,
we are done by $\mathbf{(B)}$. As $E$ and $F$ are piecewise isomorphic, there
are open subsets $U^{\prime}\subseteq E$ and $V^{\prime}\subseteq F$ such that
$E\setminus U^{\prime}$ and $F\setminus V^{\prime}$ are finite, with the same
number of points, and there is an isomorphism $\psi\colon
U^{\prime}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}V^{\prime}$ (Lemma
4.10). We take an open subset $U\subseteq U^{\prime}$, that is irreducible and
infinite, take $V=\psi(U)\subseteq V^{\prime}$, which is also open,
irreducible and infinite, and obtain that $E\setminus U$ and $F\setminus V$
are piecewise isomorphic. As $U,V$ are irreducible curves, there is exactly
one $i\in\\{1,\ldots,n\\}$ and one $j\in\\{1,\ldots,n\\}$ such that $U\cap
X_{i}$ and $V\cap Y_{j}$ are infinite. By removing finitely many points of $U$
and $V$ respectively, we may assume that $U\subseteq X_{i}$, $V\subseteq
Y_{j}$ and still assume that $U$, $V$ are isomorphic (we remove points outside
of $X_{i}$ or $Y_{j}$ and their images under $\psi$ or $\psi^{-1}$), and that
$E\setminus U$, $F\setminus V$ are piecewise isomorphic. We now prove that
$X\setminus U$ and $Y\setminus V$ are piecewise isomorphic, which will achieve
the proof, as $E\setminus U$ and $F\setminus V$ are piecewise isomorphic, with
one irreducible component of dimension $1$ less.

If $i=j$, the isomorphism $\varphi_{i}$ restricts to an isomorphism

$X_{i}\setminus(U\cup\varphi_{i}^{-1}(V))\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}Y_{i}\setminus(V\cup\varphi_{i}(U))\,,$

so we only need to see that
$\varphi_{i}^{-1}(V)\setminus(U\cap\varphi_{i}^{-1}(V))$ and
$\varphi_{i}(U)\setminus(V\cap\varphi_{i}(U))$ are piecewise isomorphic.
Applying $\varphi_{i}$ to the first set, we need to show that
$V\setminus(\varphi_{i}(U)\cap V)$ and
$\varphi_{i}(U)\setminus(V\cap\varphi_{i}(U))$ are piecewise isomorphic. If
$\varphi_{i}(U)\cap V$ is finite, this follows from $\mathbf{(B)}$. If
$\varphi_{i}(U)\cap V$ is infinite, then $V\setminus(\varphi_{i}(U)\cap V)$
and $\varphi_{i}(U)\setminus(V\cap\varphi_{i}(U))$ are both finite, with the
same number of points by using the topological Euler characteristic (Lemma
4.2).

If $i\neq j$, then $\varphi_{i}$ gives an isomorphisms $X_{i}\setminus
U\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}Y_{i}\setminus\varphi_{i}(U)$
and $\varphi_{j}$ gives an isomorphism
$X_{j}\setminus\varphi_{j}^{-1}(V)\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}Y_{j}\setminus
V$. It then remains to use an isomorphism from the curve $\varphi_{j}^{-1}(V)$
to the point $\varphi_{i}(U)$. ∎

######  Lemma 4.18.

Let $X$ be a smooth projective rational surface with Picard group
$\operatorname{Pic}(X)\simeq{\mathbb{Z}}^{n}$. Then, there is an open subset
$U\subseteq X$ isomorphic to ${\mathbb{A}}^{2}$, a point $p\in X$ and locally
closed curves $C_{1},\ldots,C_{n}\subseteq X$ such that each $C_{i}$ is
isomorphic to ${\mathbb{A}}^{1}$ and $X$ is the disjoint union

$X=U\amalg C_{1}\amalg\cdots\amalg C_{n}\amalg\\{p\\}\,.$

Moreover, for each finite set $\Delta\subseteq X$ and each closed curve
$E\subseteq X$, we may choose the above decomposition such that
$\Delta\subseteq U$, $\\{p\\}\cap(\Delta\cup E)=\varnothing$ and
$C_{i}\not\subseteq E$ for each $i\in\\{1,\ldots,n\\}$.

###### Proof.

If $X={\mathbb{P}}^{2}$, we choose a general line
$\ell\subseteq{\mathbb{P}}^{2}$ and a general point $p\in\ell$, and let
$U={\mathbb{P}}^{2}\setminus\ell$, $C_{1}=\ell\setminus\\{p\\}$. This gives
the statement in case $X={\mathbb{P}}^{2}$.

We may thus assume that $X$ is not isomorphic to ${\mathbb{P}}^{2}$, so there
is a birational morphism $X\to\mathbb{F}_{d}$ for some Hirzebruch surface
$\mathbb{F}_{d}$ where $d\geq 0$. We proceed by induction on the number of
blow-ups of $X\to{\mathbb{F}}_{d}$. If $X={\mathbb{F}}_{d}$, we choose a
general section $\ell_{1}$ of ${\mathbb{F}}_{d}\to{\mathbb{P}}^{1}$ of self-
intersection $d$ and a general fibre $\ell_{2}$ of
${\mathbb{F}}_{d}\to{\mathbb{P}}^{1}$ and let
$p\coloneqq\ell_{1}\cap\ell_{2}$,
$U\coloneqq{\mathbb{F}}_{d}\setminus(\ell_{1}\cup\ell_{2})$. Then,
$C_{i}=\ell_{i}\setminus\\{p\\}\simeq{\mathbb{A}}^{1}$ for each
$i\in\\{1,2\\}$.

We then do successive blow-ups, where we may always assume that the points
blown-up are in the subset $U\simeq{\mathbb{A}}^{2}$. Then, taking a general
line of ${\mathbb{A}}^{2}$ passing through the point, the strict transform of
the line is a closed curve $C_{i}\simeq{\mathbb{A}}^{1}$ of the blow-up of
${\mathbb{A}}^{2}$, with complement isomorphic to ${\mathbb{A}}^{2}$. ∎

######  Proposition 4.19.

Let $Y$ be a normal projective rational surface, that admits a
desingularisation $X\to Y$ with $\operatorname{Pic}(X)\simeq{\mathbb{Z}}^{n}$
and exceptional divisor being a finite disjoint union of trees of smooth
rational curves with $r\geq 0$ irreducible components $($this holds for
instance when $Y$ only has Du Val singularities$)$. Then, $Y$ is piecewise
isomorphic to the disjoint union of ${\mathbb{A}}^{2}$, one point, and $n-r$
copies of ${\mathbb{A}}^{1}$, with $n-r\geq 1$.

###### Proof.

If $Y$ is smooth, then the result, with $r=0$, directly follows from Lemma
4.18. We then assume that $Y$ is singular. Let $E\subseteq X$ be the
exceptional divisor of the desingularisation $X\to Y$, let $r$ be the number
of irreducible components of $E$ and let $s$ be the number of connected
components of $E$. Then $s$ is equal to the number of points of the image of
$E$ under $X\to Y$, since all fibres of $X\to Y$ are connected by the
normality of $Y$, see Zariski’s main theorem, [Har77, Ch. III, Corollary
11.4]. Moreover, $r,s\geq 1$, since otherwise every fibre of the proper
birational morphism $X\to Y$ onto the normal variety $Y$ would be a single
point, i.e. $X\to Y$ would be an isomorphism, see [Gro61, Corollaire 4.4.9].

We observe that $n-r\geq 1$. Indeed, the $r$ irreducible components
$E_{1},\ldots,E_{r}$ of $E$, together with the pull-back $P$ of an irreducible
curve of $Y$ not passing through the image of $E$ under $Y\to X$ are $r+1$
effective divisors on $X$, that are linearly independent: Assume $D\coloneqq
aP+\sum_{i=1}^{r}a_{i}E_{i}=0$ in $\operatorname{Pic}(X)$ for some
$a,a_{1},\ldots,a_{r}\in{\mathbb{Z}}$. Then $0=L\cdot D=aL^{2}$, which shows
that $a=0$. Since the intersection matrix $(E_{i}\cdot E_{j})$ is negative
definite (see [Art62, §2, (2.1)]), we get $a_{1}=\ldots=a_{r}=0$.

By Lemma 4.14(1), $E$ is piecewise isomorphic to the disjoint union of $r$
copies of ${\mathbb{A}}^{1}$ and $s$ points. By Lemma 4.18, $X$ is piecewise
isomorphic to the disjoint union of ${\mathbb{A}}^{2}$ with one point and $n$
copies of ${\mathbb{A}}^{1}$.

Hence, Proposition 4.17 shows that $X\setminus E$ is piecewise isomorphic to
the disjoint union of ${\mathbb{A}}^{2}\setminus\Delta$ with $n-r$ copies of
${\mathbb{A}}^{1}$, where $\Delta$ is a finite set of $s-1$ points.

Let $S$ be the image of $E$ under $X\to Y$. As $Y\setminus S$ is isomorphic to
$X\setminus E$ and $S$ contains $s$ points, this proves that $Y$ is piecewise
isomorphic to the disjoint union of ${\mathbb{A}}^{2}$ with one point and
$n-r$ copies of ${\mathbb{A}}^{1}$. ∎

######  Example 4.20.

Let $f,g\in{\mathbf{k}}[x,y,z]\setminus\\{0\\}$ be homogeneous polynomials of
degree $3$ and $4$, respectively, without common factor. Then,
$V_{{\mathbb{P}}^{2}}(f,g)$ consists of $r$ points, with $1\leq r\leq 12$. We
define $X\subseteq{\mathbb{P}}^{3}$ to be the quartic rational surface given
by

$X=\\{[w:x:y:z]\in{\mathbb{P}}^{3}\mid wf(x,y,z)=g(x,y,z)\\},$

and consider the open subset $X_{f}\subseteq X$ where $f$ is non-zero and its
complement $F\coloneqq X\setminus X_{f}=V_{{\mathbb{P}}^{3}}(f,g)$ The
hypersurface $X$ is normal, since the singular locus is finite, see e.g.
[Har77, Ch. II, Proposition 8.23]. We have an isomorphism

$X_{f}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{P}}^{2}_{f}\,,\quad[w:x:y:z]\mapsto[x:y:z]\,,$

and $F$ is the union of $r$ lines through $q=[1:0:0:0]$. Hence, $X$ is
piecewise isomorphic to the disjoint union of ${\mathbb{P}}^{2}_{f}$ with one
point and $r$ copies of ${\mathbb{A}}^{1}$. We now distinguish different cases
for the zero locus $\Gamma\coloneqq
V_{{\mathbb{P}}^{2}}(f)\subseteq{\mathbb{P}}^{2}$.

If $\Gamma$ consists of one, two or three lines meeting in one point, then $X$
is piecewise isomorphic to the disjoint union of ${\mathbb{A}}^{2}$ with one
point and $s$ copies of ${\mathbb{A}}^{1}$, for some $s\leq r$, similarly as
in Proposition 4.19. The same holds if $\Gamma$ is the union of a conic and a
line, meeting at only one point.

If $\Gamma$ consists of three general lines, then $X$ is piecewise isomorphic
to the disjoint union of ${\mathbb{A}}^{2}$ with _two_ points and $r-2$ copies
of ${\mathbb{A}}^{1}$. It is then _not_ piecewise isomorphic to a smooth
projective rational surface. Indeed, this follows from Lemma 4.15, since a
smooth projective rational surface is piecewise isomorphic to a disjoint union
of ${\mathbb{A}}^{2}$, _one_ point and some copies of ${\mathbb{A}}^{1}$, see
Lemma 4.18. A similar result holds if $\Gamma$ is the union of a conic and a
line meeting at two points.

If $\Gamma$ is a cuspidal cubic curve, it is piecewise isomorphic to a line,
so ${\mathbb{P}}^{2}_{f}$ is piecewise isomorphic to ${\mathbb{A}}^{2}$
(Proposition 4.17). Hence, this case is similar to the one of a line. If
$\Gamma$ is a nodal cubic, it is piecewise isomorphic to
${\mathbb{A}}^{1}\setminus\\{0\\}$ and thus $X$ is piecewise isomorphic to the
disjoint union of ${\mathbb{A}}^{2}$ with _two_ points and $r$ copies of
${\mathbb{A}}^{1}$, and thus $X$ is again not piecewise isomorphic to a smooth
rational projective surface.

If $\Gamma$ is a smooth irreducible cubic, this one is not rational. Let
${\mathbb{P}}^{1}\subseteq{\mathbb{P}}^{2}$ be a line that intersects $\Gamma$
in exactly one point $p\in{\mathbb{P}}^{2}$ and let
$\Gamma_{0}\coloneqq\Gamma\setminus p$. As $X$ is piecewise isomorphic to
${\mathbb{A}}^{2}\setminus\Gamma_{0}\amalg\\{pt\\}\amalg\coprod_{i=1}^{r+1}{\mathbb{A}}^{1}$,
it follows that $X$ is not piecewise isomorphic to a smooth rational
projective surface $Y$, since such a $Y$ is piecewise isomorphic to
${\mathbb{A}}^{2}\setminus\Gamma_{0}\amalg\\{pt\\}\amalg\Gamma_{0}\amalg\coprod_{i=1}^{s}{\mathbb{A}}^{1}$
for some $s\geq 1$, see Lemma 4.15.

######  Example 4.21.

In Example 4.20, we may choose $f$ to obtain a singular normal rational
quartic $X\subseteq{\mathbb{P}}^{3}$ which is not piecewise isomorphic to a
smooth rational projective surface, and thus which does not admit a
desingularisation by trees of smooth rational curves (follows from Proposition
4.19).

We now choose $f=xy(x+y)$ and $g\in{\mathbf{k}}[x,y,z]_{4}$ general, and check
that this gives an example of a normal rational quartic

$X=\\{[w:x:y:z]\in{\mathbb{P}}^{3}\mid wf(x,y,z)=g(x,y,z)\\}\,,$

which admits a desingularisation by trees of smooth rational curves, but that
does not have Du Val singularities. For this, we consider the birational map

$\psi\colon{\mathbb{P}}^{2}\dasharrow
X\,,[x:y:z]\mapsto[g(x,y,z):xf(x,y,z):yf(x,y,z):zf(x,y,z)]\,,$

and observe that it has exactly twelve base-points, being
$V_{{\mathbb{P}}^{2}}(f,g)$. Denoting by $\eta\colon Y\to{\mathbb{P}}^{2}$ the
blow-up of the twelve points, we obtain a birational morphism
$\psi\circ\eta\colon Y\to X$, that contracts the strict transforms of the
three lines defined by $V_{{\mathbb{P}}^{2}}(f)$. These curves are smooth
rational curves on $Y$ with self-intersection $-3$, intersecting into a common
point. This tree of smooth rational curves is contracted to the singular point
of multiplicity three of $X$.

##  5\. Partial answers in low degree

In this chapter, we give for certain cases affirmative answers to Questions
(2) and (3). Moreover, over an algebraically closed field, we show for any
irreducible hypersurface $H\subseteq{\mathbb{P}}^{n}$ of degree $d$ that there
exists an element of $\mathrm{Aut}({\mathbb{P}}^{3}\setminus H)$ that does not
extend to an automorphism of ${\mathbb{P}}^{3}$ in case $d=2$ or $n=d=3$ and
$H$ is singular. This gives thus a negative answer to Question (1) in the
above mentioned cases.

###  5.1. Degree two

The goal of this subsection is to prove Theorem B, which gives an affirmative
answer to Question (2) (and thus (3)) for irreducible quadric hypersurfaces in
${\mathbb{P}}^{n}_{\mathbf{k}}$ over an algebraically closed field
$\mathbf{k}$. For doing this, we count the ${\mathbb{F}}_{q}$-rational points,
where ${\mathbb{F}}_{q}$ denotes the finite field with $q$ elements.

We start with a lemma that is certainly well-known to the specialists. For
lack of a reference, we insert a proof.

######  Lemma 5.1.

Let ${\mathbb{Z}}\subseteq A$, ${\mathbb{Z}}\subseteq B$ be finitely generated
ring extensions such that $A$, $B$ are integral domains. If there exists a
field $K$ and a $K$-isomorphism between $K\otimes_{{\mathbb{Z}}}A$ and
$K\otimes_{{\mathbb{Z}}}B$, then ${\mathbb{F}}_{q}\otimes_{{\mathbb{Z}}}A$ and
${\mathbb{F}}_{q}\otimes_{{\mathbb{Z}}}B$ are ${\mathbb{F}}_{q}$-isomorphic
for some prime power $q$.

###### Proof.

As ${\mathbb{Z}}$ is a principal ideal domain, it is a regular ring. Using
that $A$, $B$ are integral domains, we get that $A$, $B$ are flat
${\mathbb{Z}}$-modules; see [Har77, Proposition 9.7, Ch. III]. Hence, every
ring extension $R_{1}\subseteq R_{2}$ induces injections

$R_{1}\otimes_{{\mathbb{Z}}}A\subseteq
R_{2}\otimes_{{\mathbb{Z}}}A\quad\textrm{and}\quad
R_{1}\otimes_{{\mathbb{Z}}}B\subseteq R_{2}\otimes_{{\mathbb{Z}}}B\,.$

Let $a_{1},\ldots,a_{n}\in A$ with $A={\mathbb{Z}}[a_{1},\ldots,a_{n}]$ and
$b_{1},\ldots,b_{m}\in B$ with $B={\mathbb{Z}}[b_{1},\ldots,b_{m}]$. Denote by
$\varphi\colon
K\otimes_{{\mathbb{Z}}}A\xlongrightarrow{\sim}K\otimes_{{\mathbb{Z}}}B$ a
$K$-isomorphism. There exist finitely many $c_{1},\ldots,c_{r}\in K$ such that
the ${\mathbb{Z}}$-algebra $R$ spanned by $c_{1},\ldots,c_{r}$ inside $K$
satisfies $\varphi(a_{i})\in R\otimes_{{\mathbb{Z}}}B$ for all $1\leq i\leq n$
and $\varphi^{-1}(b_{j})\in R\otimes_{{\mathbb{Z}}}A$ for all $1\leq j\leq m$.
In particular, $\varphi$ restricts to an $R$-isomorphism
$R\otimes_{{\mathbb{Z}}}A\simeq R\otimes_{{\mathbb{Z}}}B$. We may assume that
$R$ is non-zero and thus contains a maximal ideal ${\mathfrak{m}}$. Then we
get an $R/{\mathfrak{m}}$-isomorphism

$(R/{\mathfrak{m}})\otimes_{{\mathbb{Z}}}A\xlongrightarrow{\sim}(R/{\mathfrak{m}})\otimes_{{\mathbb{Z}}}B\,.$

Denote by $F$ the prime field of $R/{\mathfrak{m}}$, i.e. the quotient field
of the image of the unique ring homomorphism ${\mathbb{Z}}\to
R/{\mathfrak{m}}$. As $R$ is finitely generated as a ${\mathbb{Z}}$-algebra,
$R/{\mathfrak{m}}$ is finitely generated as an $F$-algebra and by Noether’s
normalization theorem, the field extension $F\subseteq R/{\mathfrak{m}}$ is
finite. By the Artin-Tate lemma (see [AM69, Proposition 7.8]) it follows that
$F$ is a finitely generated ${\mathbb{Z}}$-algebra. Hence, $F$ cannot be
isomorphic to ${\mathbb{Q}}$ and thus $F={\mathbb{F}}_{p}$ where
$p=\operatorname{char}(R/{\mathfrak{m}})$. As the field extension $F\subseteq
R/{\mathfrak{m}}$ is finite, $R/{\mathfrak{m}}={\mathbb{F}}_{p^{r}}$ for some
$r\geq 1$. ∎

######  Corollary 5.2.

Let $X,Y$ be affine integral schemes of finite type over ${\mathbb{Z}}$ such
that $X\to\mathrm{Spec}({\mathbb{Z}})$, $Y\to\mathrm{Spec}({\mathbb{Z}})$ are
dominant. If the pull-backs $X_{K}$, $Y_{K}$ are $K$-isomorphic for some field
$K$, then there exists a prime power $q$ and a bijection between the
${\mathbb{F}}_{q}$-rational points $X({\mathbb{F}}_{q})\to
Y({\mathbb{F}}_{q})$ that maps the regular ${\mathbb{F}}_{q}$-rational points
of $X$ bijectively onto the regular ${\mathbb{F}}_{q}$-rational points of $Y$.

###### Proof.

This is a direct consequence of Lemma 5.1. ∎

In the following lemmas, we denote by $Z({\mathbb{F}}_{q})$
($Z({\mathbb{F}}_{q})_{\operatorname{reg}},Z({\mathbb{F}}_{q})_{\operatorname{sing}}$)
the set of (regular, singular) ${\mathbb{F}}_{q}$-rational points of a scheme
$Z$ that is defined over ${\mathbb{Z}}$. Note that
$Z({\mathbb{F}}_{q})_{\operatorname{reg}}=Z({\mathbb{F}}_{q})\setminus
Z({\mathbb{F}}_{q})_{\operatorname{sing}}$.

In case $Z$ is a hypersurface in ${\mathbb{A}}^{n+1}_{{\mathbb{Z}}}$ given by
a polynomial $f\in{\mathbb{Z}}[x_{0},\ldots,x_{n}]$, the set
$Z({\mathbb{F}}_{q})$ consists of those points in
$(a_{0},\ldots,a_{n})\in{\mathbb{F}}_{q}^{n+1}$ that satisfy
$f(a_{0},$…$,a_{n})=0$ and $(a_{0},\ldots,a_{n})\in Z({\mathbb{F}}_{q})$ is
regular, if not all the partial derivatives $\frac{\partial f}{\partial
x_{0}},\ldots,\frac{\partial f}{\partial
x_{n}}\in{\mathbb{Z}}[x_{0},\ldots,x_{n}]$ vanish at $(a_{0},\ldots,a_{n})$.

######  Lemma 5.3.

For all integers $m,n$ with $1\leq 2m-1\leq n$, let $X_{m,n}$ be the
hypersurface in ${\mathbb{A}}^{n+1}_{{\mathbb{Z}}}$ given by

$\sum_{i=0}^{m-1}x_{2i}x_{2i+1}=1.$

Then for any prime power $q$ we get

$\\#X_{m,n}({\mathbb{F}}_{q})=\\#X_{m,n}({\mathbb{F}}_{q})_{\operatorname{reg}}=q^{n-m}(q^{m}-1)\,.$

###### Proof.

As $X_{m,n}$ is smooth over $\mathrm{Spec}({\mathbb{Z}})$,
$X_{m,n}({\mathbb{F}}_{q})_{\operatorname{reg}}=X_{m,n}({\mathbb{F}}_{q})$. We
proceed now by induction on $m\geq 1$. For any $n\geq 1$, we get
$\\#X_{1,n}({\mathbb{F}}_{q})=q^{n-1}(q-1)$, as $X_{1,n}$ and
$({\mathbb{A}}_{{\mathbb{Z}}}^{1}\setminus\\{0\\})\times{\mathbb{A}}_{{\mathbb{Z}}}^{n-1}$
are isomorphic over ${\mathbb{Z}}$. Moreover, for $m>1$:

$X_{m,n}({\mathbb{F}}_{q})=V_{X_{m,n}}(x_{2m-1})({\mathbb{F}}_{q})\coprod(X_{m,n})_{x_{2m-1}}({\mathbb{F}}_{q})\,.$

As $(X_{m,n})_{x_{2m-1}}$ and
$({\mathbb{A}}_{{\mathbb{Z}}}^{1}\setminus\\{0\\})\times{\mathbb{A}}_{{\mathbb{Z}}}^{n-1}$
are isomorphic over ${\mathbb{Z}}$, it follows that
$\\#(X_{m,n})_{x_{2m-1}}({\mathbb{F}}_{q})=q^{n-1}(q-1)$. As
$V_{X_{m,n}}(x_{2m-1})$ and $X_{m-1,n-1}$ are isomorphic over ${\mathbb{Z}}$,
we get by induction

$\displaystyle\\#X_{m,n}({\mathbb{F}}_{q})$ $\displaystyle=$
$\displaystyle\\#X_{m-1,n-1}({\mathbb{F}}_{q})+q^{n-1}(q-1)$ $\displaystyle=$
$\displaystyle q^{n-m}(q^{m-1}-1)+q^{n-1}(q-1)=q^{n-m}(q^{m}-1)\,.\qed$

######  Lemma 5.4.

For all integers with $0\leq 2m\leq n$, let $Y_{m,n}$ be the hypersurface in
${\mathbb{A}}^{n+1}_{{\mathbb{Z}}}$ that is given by

$\left(\sum_{i=0}^{m-1}x_{2i}x_{2i+1}\right)+x_{2m}^{2}=1.$

Then for every _odd_ prime power $q$ we get

$\\#Y_{m,n}({\mathbb{F}}_{q})=\\#Y_{m,n}({\mathbb{F}}_{q})_{\operatorname{reg}}=q^{n-m}(q^{m}+1)\,,$

and for every _even_ prime power $q$, we have

$\\#(Y_{m,n})({\mathbb{F}}_{q})=q^{n}\quad\textrm{and}\quad\\#(Y_{m,n})({\mathbb{F}}_{q})_{\operatorname{reg}}=q^{n-2m}(q^{2m}-1)\,.$

###### Proof.

Suppose first that $q$ is odd. Here, $Y_{m,n}$ is smooth over
$\mathrm{Spec}({\mathbb{Z}})\setminus\\{(2)\\}$, so
$Y_{m,n}({\mathbb{F}}_{q})_{\operatorname{reg}}=Y_{m,n}({\mathbb{F}}_{q})$. We
proceed by induction on $m\geq 0$. For any $n\geq 1$, we get
$\\#Y_{0,n}=2q^{n}$, as $Y_{0,n}$ is given by $x_{0}^{2}-1$ in
${\mathbb{A}}^{n+1}_{{\mathbb{Z}}}$ over ${\mathbb{Z}}$. Assume $m>0$. As in
Lemma 5.3, we count the points where $x_{2m-1}=0$ and the points where
$x_{2m-1}\neq 0$, and obtain:

$\displaystyle\\#Y_{m,n}({\mathbb{F}}_{q})$ $\displaystyle=$
$\displaystyle\\#Y_{m-1,n-1}({\mathbb{F}}_{q})+q^{n-1}(q-1)$ $\displaystyle=$
$\displaystyle q^{n-m}(q^{m-1}+1)+q^{n-1}(q-1)=q^{n-m}(q^{m}+1)\,.$

Suppose now that $q$ is a power of $2$. As
${\mathbb{F}}_{q}\to{\mathbb{F}}_{q},t\to t^{2}$ is bijective, the projection
$Y_{m,n}\to{\mathbb{A}}_{{\mathbb{Z}}}^{n}$,
$(x_{0},\ldots,x_{n})\mapsto(x_{0},\ldots,x_{2m-1},x_{2m+1},\ldots,x_{n})$
induces a bijection
$Y_{m,n}({\mathbb{F}}_{q})\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{A}}_{{\mathbb{Z}}}^{n}({\mathbb{F}}_{q})$,
which yields $\\#(Y_{m,n})({\mathbb{F}}_{q})=q^{n}$. The set of singular
${\mathbb{F}}_{q}$-rational points of $Y_{m,n}$ is equal to

$(Y_{m,n})({\mathbb{F}}_{q})_{\operatorname{sing}}=\\{\,(0,\ldots,0,1,x_{2m+1},\ldots,x_{n})\
|\ x_{2m+1},\ldots,x_{n}\in{\mathbb{F}}_{q}\,\\}$

and hence $(Y_{m,n})({\mathbb{F}}_{q})_{\operatorname{sing}}=q^{n-2m}$. This
gives the result. ∎

######  Proposition 5.5.

If ${\mathbf{k}}$ is algebraically closed, every irreducible quadric
hypersurface in ${\mathbb{P}}^{n}$ is given, under a suitable change of
coordinates, by

$\sum_{i=0}^{m-1}x_{2i}x_{2i+1}\quad\textrm{or}\quad\left(\sum_{i=0}^{m-1}x_{2i}x_{2i+1}\right)+x_{2m}^{2}\,,$

where $1<2m-1\leq n$ in the first case and $0<2m\leq n$ in the second case.

###### Proof.

This follows from [Pfi95, Theorem 1.8] if
$\operatorname{char}({\mathbf{k}})\neq 2$ and from [Pfi95, Theorem 4.3] if
$\operatorname{char}(\mathbf{k})=2$. ∎

###### Proof of Theorem B.

Let

$\mathcal{M}=\mathcal{M}_{X}\coprod\mathcal{M}_{Y}$

where

$\mathcal{M}_{X}\coloneqq\\{\,X_{m,n}\ |\ 1<2m-1\leq
n\,\\}\quad\textrm{and}\quad\mathcal{M}_{Y}\coloneqq\\{\,Y_{m,n}\ |\ 0<2m\leq
n\,\\}$

and $X_{m,n}$, $Y_{m,n}$ are defined in Lemmas 5.3 and 5.4, respectively.
Using Proposition 2.1, Remark 2.2 and Proposition 5.5, it is enough to show
that distinct elements of $\mathcal{M}$ are non-isomorphic over
${\mathbf{k}}$. By Corollary 5.2, an isomorphism over $\mathbf{k}$ between two
elements from $\mathcal{M}$ would give a bijection between the
${\mathbb{F}}_{q}$-rational points and between the regular
${\mathbb{F}}_{q}$-rational points, for some prime power $q$.

We consider for every prime power $q$ the map

$\Phi_{q}\colon\mathcal{M}\to{\mathbb{N}}_{0}\,,\quad
Z\mapsto\\#Z({\mathbb{F}}_{q})_{\operatorname{reg}}\,.$

By Lemmas 5.3-5.4, we get

$\Phi_{q}(X_{m,n})=q^{n-m}(q^{m}-1)\quad\textrm{for all prime powers $q$}$

and

$\Phi_{q}(Y_{m,n})=\left\\{\begin{array}[]{rl}q^{n-m}(q^{m}+1)&\textrm{if $q$
is an odd prime power}\\\ q^{n-2m}(q^{2m}-1)&\textrm{if $q$ is a power of
$2$}\end{array}\right.\,.$

Hence, for odd prime powers $q$, the map $\Phi_{q}$ is injective. If $q=2^{r}$
for some $r\geq 1$, then the restrictions $\Phi_{2^{r}}|_{\mathcal{M}_{X}}$
and $\Phi_{2^{r}}|_{\mathcal{M}_{Y}}$ are still injective. By the Lemmas
5.3-5.4, all ${\mathbb{F}}_{2^{r}}$-rational points of $X_{m,n}$ are regular,
whereas $Y_{m,n}$ has singular ${\mathbb{F}}_{2^{r}}$-rational points. This
achieves to prove that distinct elements from $\mathcal{M}$ are non-isomorphic
over ${\mathbf{k}}$. ∎

The following example shows that Question (1) has a negative answer for degree
$2$ and any $n\geq 2$:

######  Example 5.6.

Let $n\geq 2$. For each $f\in{\mathbf{k}}[x_{3},\ldots,x_{n}]$, homogeneous of
degree $2$ (we choose $f=0$ if $n=2$), the following morphism defines a
${\mathbb{G}}_{a}$-action

$\begin{array}[]{rcl}\rho\colon{\mathbb{G}}_{a}\times{\mathbb{P}}^{n}_{x_{0}x_{1}+x_{2}^{2}+f}&\to&{\mathbb{P}}^{n}_{x_{0}x_{1}+x_{2}^{2}+f}\\\
(t,[x_{0}:x_{1}:\cdots:x_{n}])&\mapsto&[x_{0}:x_{1}-2tx_{2}-t^{2}x_{0}:x_{2}+tx_{0}:x_{3}:\cdots:x_{n}].\end{array}$

Since $\frac{x_{0}^{2}}{x_{0}x_{1}+x_{2}^{2}+f}$ is a
${\mathbb{G}}_{a}$-invariant function on
${\mathbb{P}}^{n}_{x_{0}x_{1}+x_{2}^{2}+f}$, it follows that for every non-
constant univariate polynomial $q$ the following map

$[x_{0}:x_{1}:\cdots:x_{n}]\mapsto\rho\left(q\left(\frac{x_{0}^{2}}{x_{0}x_{1}+x_{2}^{2}+f}\right),[x_{0}:x_{1}:\cdots:x_{n}]\right)$

defines an automorphism of ${\mathbb{P}}^{n}_{x_{0}x_{1}+x_{2}^{2}+f}$ that
doesn’t extend to an element of $\mathrm{Aut}({\mathbb{P}}^{n})$.

Similarly, for each $n\geq 3$ and each polynomial
$g\in{\mathbf{k}}[x_{4},\ldots,x_{n}]$ (again $g=0$ if $n=3$), the following
morphism defines a ${\mathbb{G}}_{a}$-action

$\begin{array}[]{rcl}\rho\colon{\mathbb{G}}_{a}\times{\mathbb{P}}^{n}_{x_{0}x_{1}+x_{2}x_{3}+g}&\to&{\mathbb{P}}^{n}_{x_{0}x_{1}+x_{2}x_{3}+g}\\\
(t,[x_{0}:x_{1}:\cdots:x_{n}])&\mapsto&[x_{0}:x_{1}-tx_{3}:x_{2}+tx_{0}:x_{3}:\cdots:x_{n}].\end{array}$

Since $\frac{x_{0}^{2}}{x_{0}x_{1}+x_{2}x_{3}+g}$ is a
${\mathbb{G}}_{a}$-invariant function on
${\mathbb{P}}^{n}_{x_{0}x_{1}+x_{2}x_{3}+g}$, it follows that for every non-
constant univariate polynomial $q$ the following map

$[x_{0}:x_{1}:\cdots:x_{n}]\mapsto\rho\left(q\left(\frac{x_{0}^{2}}{x_{0}x_{1}+x_{2}x_{3}+g}\right),[x_{0}:x_{1}:\cdots:x_{n}]\right)$

defines an automorphism of ${\mathbb{P}}^{n}_{x_{0}x_{1}+x_{2}x_{3}+g}$ that
doesn’t extend to an element of $\mathrm{Aut}({\mathbb{P}}^{n})$.

If ${\mathbf{k}}$ is algebraically closed, it follows from Proposition 5.5
that every irreducible quadric is of one of the above form, up to change of
coordinates.

###  5.2. Degree three and beyond in dimension three

Throughout this subsection we assume always that $\mathbf{k}$ is algebraically
closed.

Recall that an irreducible hypersurface in ${\mathbb{P}}^{3}$ is normal if and
only if its singular locus is finite [Har77, Ch. II, Proposition 8.23].

######  Remark 5.7.

Let $X\subseteq{\mathbb{P}}^{2}$ be a normal rational irreducible cubic
hypersurface. Then there exist six (possibly infinitely near) points
$p_{1},\ldots,p_{6}$ in ${\mathbb{P}}^{2}$ such that the blow-up
$\pi\colon\tilde{X}\to{\mathbb{P}}^{2}$ of them satisfies: There exists a
birational morphism $\eta\colon\tilde{X}\to X$ and it is the minimal
desingularisation of $X$. In case $X$ is singular this follows from the proof
of Lemma 5.9 and in case $X$ is smooth this follows from [Har77, Ch. V,
Corollary 4.7, Remark 4.7.1]. Moreover in the smooth case $p_{1},\ldots,p_{6}$
belong to ${\mathbb{P}}^{2}$ (and are in general position).

######  Lemma 5.8.

An irreducible cubic hypersurface in ${\mathbb{P}}^{3}$ is non-rational if and
only if it is the cone over a smooth cubic curve in ${\mathbb{P}}^{2}$.

###### Proof.

Let $X\subseteq{\mathbb{P}}^{3}$ be a non-rational irreducible cubic
hypersurface. Then $X$ has a singularity (as otherwise it would be the blow-up
of ${\mathbb{P}}^{2}$ at six points, see Remark 5.7), say at
$P=[1:0:0:0]\in{\mathbb{P}}^{3}$ and it is given by a homogeneous polynomial
of the form $f_{2}(x,y,z)w+f_{3}(x,y,z)$, where $[w:x:y:z]$ denote the
homogeneous coordinates of ${\mathbb{P}}^{3}$ and $f_{i}\in\mathbf{k}[x,y,z]$
is homogeneous of degree $i$. As $X$ is non-rational, we get $f_{2}=0$
(otherwise the projection to $x,y,z$ gives a birational map to
${\mathbb{P}}^{2}$) and $X$ is the cone in ${\mathbb{P}}^{3}$ over a cubic
curve $\Gamma\subset{\mathbb{P}}^{2}$ given by $f_{3}=0$. As $X$ is
irreducible, so is $\Gamma$. Then, $X$ is birational to
$\Gamma\times{\mathbb{P}}^{1}$, and thus $\Gamma$ is non-rational as $X$ is.
This implies that $\Gamma$ is smooth, and that $X$ is the cone in
${\mathbb{P}}^{3}$ over a smooth cubic curve in ${\mathbb{P}}^{2}$.

Assume now that $X$ is a cone over a smooth cubic curve $C$. Blowing up the
vertex yields a projective smooth ruled surface $S$ over $C$. In particular
$H^{1}(S,\mathcal{O}_{S})$ doesn’t vanish (see [Har77, Ch. V, Corollary 2.5])
and hence $X$ is not rational (see [KSC04, Theorem 3.2]). ∎

The next lemma is certainly known to the specialists. For lack of a proof in
any characteristic we provide one:

######  Lemma 5.9.

Let $X\subseteq{\mathbb{P}}^{3}$ be a normal rational cubic hypersurface. Then
$X$ has only Du Val singularities and is piecewise isomorphic to the disjoint
union of ${\mathbb{A}}^{2}$, one point, and $n$ copies of ${\mathbb{A}}^{1}$,
with $1\leq n\leq 7$. Moreover, $3\leq\chi(X)=2+n\leq 9$ and $\chi(X)=9$ if
and only if $X$ is smooth.

###### Proof.

If $X$ is smooth, then $X$ is the blow-up of $6$ points of ${\mathbb{P}}^{2}$
(Remark 5.7) and is thus piecewise isomorphic to the disjoint union of
${\mathbb{A}}^{2}$, one point and seven copies of ${\mathbb{A}}^{1}$ by Lemma
4.18. In particular, $\chi(X)=9$.

Thus we may assume that $X$ is singular and after some coordinate change we
may assume that $X$ is given by $f_{2}(x,y,z)w+f_{3}(x,y,z)$, where
$f_{2},f_{3}\in\mathbf{k}[x,y,z]$ are homogeneous polynomials of degree $2,3$,
respectively, and $[w:x:y:z]$ denote the homogeneous coordinates in
${\mathbb{P}}^{3}$. If $f_{2}=0$, then, $X$ is a cone over
$V_{{\mathbb{P}}^{2}}(f_{3})$. As $X$ is a rational cubic, $f_{3}$ is
irreducible and $V_{{\mathbb{P}}^{2}}(f_{3})$ has to be singular (Lemma 5.8),
contradicting the fact that $X$ is normal. Hence, $f_{2}\neq 0$.

The projection $X\setminus\\{[1:0:0:0]\\}\to{\mathbb{P}}^{2}$,
$[w:x:y:z]\mapsto[x:y:z]$ gives a birational morphism whose inverse
$\Phi\colon{\mathbb{P}}^{2}\dashrightarrow X$ is given by

$[x:y:z]\mapsto[f_{3}(x,y,z):xf_{2}(x,y,z):yf_{2}(x,y,z):zf_{2}(x,y,z)]\,.$

By [Sak10, Proposition 1.5], there exists a smooth projective surface
$\tilde{X}$ which is the blow-up $\pi\colon\tilde{X}\to{\mathbb{P}}^{2}$ of
six (possibly infinitely near) points in the zero locus $S=V(f_{2},f_{3})$
such that $\eta\coloneqq\Phi\circ\pi\colon\tilde{X}\to X$ is a morphism and
$\eta$ is given by the complete linear system $|-K_{\tilde{X}}|$, where
$K_{\tilde{X}}$ denotes the canonical divisor in $\tilde{X}$. Let
$X_{\operatorname{reg}}$ be the open subvariety of regular points in $X$ and
let $X_{\operatorname{sing}}$ be its complement in $X$.

By applying the adjunction formula to the smooth closed hypersurface
$X_{\operatorname{reg}}$ in ${\mathbb{P}}^{3}\setminus
X_{\operatorname{sing}}$, we get that the canonical divisor $K_{X}$ of $X$ is
equal to $-H$, where $H$ is a hyperplane section of
$X\subseteq{\mathbb{P}}^{3}$. This gives $\eta^{\ast}K_{X}=K_{\tilde{X}}$. By
[Art62, Theorem 2.7] it follows that the exception locus of $\eta$ is a finite
union of trees and that in fact $X$ has only Du Val singularities.

By Proposition 4.19, $X$ is piecewise isomorphic to a disjoint union of
${\mathbb{A}}^{2}$, one point and $7-r$ copies of ${\mathbb{A}}^{1}$, with
$1\leq r\leq 6$. In particular, $3\leq\chi(X)\leq 8$. ∎

We will use the following classification of non-normal irreducible cubic
surfaces in ${\mathbb{P}}^{3}$:

######  Proposition 5.10 (see [LPS11, Theorem 3.1]).

Let $X\subseteq{\mathbb{P}}^{3}$ be a non-normal irreducible cubic surface.
Denote by $[w:x:y:z]$ the homogeneous coordinates in ${\mathbb{P}}^{3}$.

  * $\bullet$

If $X$ is a cone over a curve in ${\mathbb{P}}^{2}$, then up to a coordinate
change of ${\mathbb{P}}^{3}$, the hypersurface $X$ is given by

$x^{2}w+y^{3}\quad\textrm{or}\quad x^{2}w+y^{3}+xy^{2}\quad\textrm{or}\quad
xyw+x^{3}+y^{3}\,.$

Moreover, if $\operatorname{char}(\mathbf{k})\neq 3$, then $x^{2}w+y^{3}$ and
$x^{2}w+y^{3}+xy^{2}$ are the same up to a linear coordinate change in
$w,x,y$.

  * $\bullet$

If $X$ is not a cone over a curve in ${\mathbb{P}}^{2}$, then up to a
coordinate change of ${\mathbb{P}}^{3}$, the hypersurface $X$ is given by

$x^{2}w+y^{2}z\quad\textrm{or}\quad xyw+y^{2}z+x^{3}\quad\textrm{or}\quad
xyw+(x^{2}+y^{2})z\,.$

Moreover, if $\operatorname{char}(\mathbf{k})\neq 2$, then $x^{2}w+y^{2}z$ and
$xyw+(x^{2}+y^{2})z$ are the same up to a linear coordinate change in
$w,x,y,z$.

In particular, $X$ always contains a line with multiplicity $2$ $($the line
given by $x=y=0$ in the above equations$)$.

In the following lemma, we give a similar decompositions into locally closed
subsets as in Lemma 5.9 for irreducible non-normal cubic hypersurfaces in
${\mathbb{P}}^{3}$:

######  Lemma 5.11.

Let

$\displaystyle f_{1}$ $\displaystyle=xyw+x^{3}+y^{3}\,,$ $\displaystyle f_{2}$
$\displaystyle=x^{2}w+y^{3}\,,$ $\displaystyle f_{3}$
$\displaystyle=x^{2}w+y^{3}+xy^{2}\,,$ $\displaystyle f_{4}$
$\displaystyle=xyw+y^{2}z+x^{3}\,,$ $\displaystyle f_{5}$
$\displaystyle=x^{2}w+y^{2}z\,,$ $\displaystyle f_{6}$
$\displaystyle=xyw+(x^{2}+y^{2})z\,.$

Then, $X_{i}=V_{{\mathbb{P}}^{3}}(f_{i})\subseteq{\mathbb{P}}^{3}$ is
piecewise isomorphic to the disjoint union of ${\mathbb{A}}^{2}$, one point,
and $n_{i}$ copies of ${\mathbb{A}}^{1}$, with $n_{1}=0$,
$n_{2}=n_{3}=n_{4}=1$ and $n_{5}=n_{6}=2$. In particular,
$\chi(X_{i})=2+n_{i}\in\\{2,3,4\\}$.

###### Proof.

We can decompose $X_{i}$ into $X_{i}\cap{\mathbb{P}}^{3}_{x}$ and $X_{i}\cap
V_{{\mathbb{P}}^{3}}(x)$. Then we get the following table

Equation | $X_{i}\cap{\mathbb{P}}^{3}_{x}$ | $X_{i}\cap V_{{\mathbb{P}}^{3}}(x)$
---|---|---
$xyw+x^{3}+y^{3}$ | ${\mathbb{A}}^{1}\times({\mathbb{A}}^{1}\setminus\\{0\\})$ | ${\mathbb{P}}^{1}$
$x^{2}w+y^{3}$ | ${\mathbb{A}}^{2}$ | ${\mathbb{P}}^{1}$
$x^{2}w+y^{3}+xy^{2}$ | ${\mathbb{A}}^{2}$ | ${\mathbb{P}}^{1}$
$xyw+y^{2}z+x^{3}$ | ${\mathbb{A}}^{1}\times({\mathbb{A}}^{1}\setminus\\{0\\})$ | ${\mathbb{P}}^{1}\vee{\mathbb{P}}^{1}$
$x^{2}w+y^{2}z$ | ${\mathbb{A}}^{2}$ | ${\mathbb{P}}^{1}\vee{\mathbb{P}}^{1}$
$xyw+(x^{2}+y^{2})z$ | ${\mathbb{A}}^{2}$ | ${\mathbb{P}}^{1}\vee{\mathbb{P}}^{1}$

where ${\mathbb{P}}^{1}\vee{\mathbb{P}}^{1}$ denotes two copies of
${\mathbb{P}}^{1}$ in ${\mathbb{P}}^{2}$ (that intersect transversally in
exactly one point). This implies the statement. ∎

######  Corollary 5.12.

Let $X$ be a rational cubic hypersurface of ${\mathbb{P}}^{3}$. Then, $X$ is
piecewise isomorphic to the disjoint union of ${\mathbb{A}}^{2}$, one point,
and $n$ copies of ${\mathbb{A}}^{1}$, with $0\leq n\leq 7$. Moreover,
$2\leq\chi(X)=2+n\leq 9$.

###### Proof.

If $X$ is normal, this follows from Lemma 5.9. Otherwise, this follows from
Lemma 5.11 in combination with Proposition 5.10. ∎

######  Lemma 5.13.

Let $X\subseteq{\mathbb{P}}^{3}$ be an irreducible cubic hypersurface. If $X$
is rational, then $\chi(X)\geq 2$. If $X$ is not rational, then $\chi(X)=1$.

###### Proof.

If $X$ is rational, then Corollary 5.12 gives $\chi(X)\geq 2$.

Assume now that $X$ is non-rational. By Lemma 5.8, $X$ is the cone over a
smooth cubic curve $C$ in ${\mathbb{P}}^{2}$. Let $p\in X$ be the unique
singularity. Then $X\setminus\\{p\\}$ is a locally trivial
${\mathbb{A}}^{1}$-bundle over $C$ (with respect to the Zariski topology).
Hence $\chi(X)=1+\chi([X\setminus p])=1+\chi(C)=1$, where the second and third
equality follow from Example 4.5 and Example 4.3, respectively. ∎

######  Proposition 5.14.

Let $X_{1},X_{2}$ be normal projective rational surfaces having only
singularities that can be solved by trees of smooth rational curves $($e.g. Du
Val singularities$)$. Then the following statements are equivalent

  1. (1)

$X_{1}$ and $X_{2}$ are piecewise isomorphic;

  2. (2)

$[X_{1}]=[X_{2}]$ inside $K_{0}(\operatorname{Var}_{\mathbf{k}})$;

  3. (3)

$\chi(X_{1})=\chi(X_{2})$.

###### Proof.

The implications “(1) $\Rightarrow$ (2)” and “(2) $\Rightarrow$ (3)” both
follow from Lemma 4.10. Proposition 4.19 says that $X_{i}$ is piecewise
isomorphic to the disjoint union of ${\mathbb{A}}^{2}$, one point and
$s_{i}\geq 0$ copies of ${\mathbb{A}}^{1}$. If $\chi(X_{1})=\chi(X_{2})$, then
we get $s_{1}=s_{2}$, i.e. $X_{1}$, $X_{2}$ are piecewise isomorphic. This
gives the implication “(3) $\Rightarrow$ (1)”. ∎

######  Corollary 5.15.

Let $f,g\in{\mathbf{k}}[x,y,z,w]$ be irreducible homogeneous polynomials such
that ${\mathbb{P}}^{3}_{f}\simeq{\mathbb{P}}^{3}_{g}$. Assume moreover that
the zero loci $X\coloneqq V_{{\mathbb{P}}^{3}}(f)$, $Y\coloneqq
V_{{\mathbb{P}}^{3}}(g)$

  1. a)

are both normal, rational and each admits a desingularisation by trees of
smooth rational curves or

  2. b)

are both of degree $3$ and rational.

Then $X$, $Y$ are piecewise isomorphic.

###### Proof.

Since ${\mathbb{P}}^{3}_{f}\simeq{\mathbb{P}}^{3}_{g}$ we get $[X]=[Y]$ inside
$K_{0}(\operatorname{Var}_{\mathbf{k}})$ and thus $\chi(X)=\chi(Y)$, see Lemma
4.10. In case a), the statement follows from Proposition 5.14. So assume we
are in case b). Then the statement follows from Proposition 4.19 in
combination with Lemma 5.9 if $X$ and $Y$ are normal and from Proposition 5.10
in combination with Lemma 5.11 in the other case. ∎

We now focus on cones over smooth cubic curves, i.e. on non-rational cubic
hypersurfaces of ${\mathbb{P}}^{3}$. This consists the last unproven part of
Theorem C. To prove that for such hypersurfaces, Question (2) has an
affirmative answer, we will need the following result:

######  Lemma 5.16.

Let $f\in{\mathbf{k}}[x,y,z]$ be an irreducible homogeneous polynomial of
degree $3$ such that $V_{{\mathbb{P}}^{2}}(f)$ is a smooth cubic curve. Then,
the surface

$X=V_{{\mathbb{A}}^{3}}(f-1)$

admits no non-trivial ${\mathbb{G}}_{a}$-action.

###### Proof.

Suppose for contradiction that $X$ admits a non-trivial
${\mathbb{G}}_{a}$-action. By [BFT23, Proposition 2.5.1], there exists a
${\mathbb{G}}_{a}$-invariant affine dense open subset $X^{\prime}\subseteq X$,
that is a ${\mathbb{G}}_{a}$-cylinder, i.e. $X^{\prime}$ is
${\mathbb{G}}_{a}$-isomorphic to ${\mathbb{G}}_{a}\times U$, where $U$ is a
smooth affine curve and where the ${\mathbb{G}}_{a}$-action on
${\mathbb{G}}_{a}\times U$ is given $s\cdot(t,u)\coloneqq(s+t,u)$ for
$s,t\in{\mathbb{G}}_{a}$, $u\in U$.

Using the canonical embedding
${\mathbb{A}}^{3}\hookrightarrow{\mathbb{P}}^{3},(x,y,z)\mapsto[1:x:y:z]$ we
can view $X$ as an open subset of the irreducible surface
$Y=V_{{\mathbb{P}}^{3}}(f-w^{3})$, where $[w:x:y:z]$ denote the homogeneous
coordinates on ${\mathbb{P}}^{3}$. Here, $X=Y_{w}$ is the complement in $Y$ of
a smooth curve $\Gamma$ and $\Gamma\simeq V_{{\mathbb{P}}^{2}}(f)$.

Suppose first that $Y$ is a cone over a smooth cubic curve $C$. Writing $p\in
Y$ the singular point, the projection away from $p$ gives to
$Y\setminus\\{p\\}$ an ${\mathbb{A}}^{1}$-bundle structure over $C$. Hence,
every closed rational curve on $Y$ is one of the lines through $p$. As
$\Gamma$ is a hyperplane section of $Y$ and as $\Gamma$ is not rational, it
intersects a general such line into at least one point outside of $p$. The
open subset $X^{\prime}\subseteq Y\setminus\Gamma$ being contained in the
smooth locus of $Y$, it contains only finitely many curves isomorphic to
${\mathbb{A}}^{1}$. This contradicts the fact that $X^{\prime}$ is a
${\mathbb{G}}_{a}$-cylinder.

In the remaining case, $Y$ is a rational cubic surface (Lemma 5.8). Hence, $U$
is a smooth affine rational curve, and is thus isomorphic to
${\mathbb{A}}^{1}\setminus\Delta$ for some finite set $\Delta$ of $r\geq 0$
points. Hence, $X^{\prime}$ is isomorphic to the open subset
$V^{\prime}={\mathbb{A}}^{1}\times({\mathbb{A}}^{1}\setminus\Delta)\subseteq{\mathbb{A}}^{2}$.
By Corollary 5.12, $Y$ is piecewise isomorphic to the disjoint union of
${\mathbb{A}}^{2}$, one point, and $n$ copies of ${\mathbb{A}}^{1}$, with
$0\leq n\leq 7$, and thus to the disjoint union of $X^{\prime}\simeq
V^{\prime}$ with one point and $n+r$ copies of ${\mathbb{A}}^{1}$. As the one-
dimensional variety $Y\setminus X^{\prime}$ contains $\Gamma$, that is not
rational, we get a contradiction, by Lemma 4.15. ∎

We are now able to prove part (3) of Theorem C, that is Proposition 5.17. If
${\mathbf{k}}={\mathbb{C}}$, then Proposition 5.17 follows essentially from
[LS12].

######  Proposition 5.17.

Let $f,g\in\mathbf{k}[x,y,z,w]$ be irreducible homogeneous polynomials of
degree three such that $V_{{\mathbb{P}}^{3}}(f)$, $V_{{\mathbb{P}}^{3}}(g)$
are non-rational. If ${\mathbb{P}}^{3}_{f}\simeq{\mathbb{P}}^{3}_{g}$, then
there exists $\varphi\in\mathrm{Aut}({\mathbb{P}}^{3})$ with
$\varphi(V_{{\mathbb{P}}^{3}}(f))=V_{{\mathbb{P}}^{3}}(g)$.

###### Proof.

By Lemma 5.8 we may assume that $f,g\in\mathbf{k}[x,y,z]$ and that the zero
loci $V_{{\mathbb{P}}^{2}}(f)$, $V_{{\mathbb{P}}^{2}}(g)$ are smooth cubic
curves.

By Lemma 5.16, every ${\mathbb{G}}_{a}$-action on the affine surfaces
$X_{0}\coloneqq V_{{\mathbb{A}}^{3}}(f-1)$ and $Y_{0}\coloneqq
V_{{\mathbb{A}}^{3}}(g-1)$ is trivial.

Let $\theta\colon{\mathbb{P}}_{f}^{3}\to{\mathbb{P}}^{3}_{g}$ be an
isomorphism. By Proposition 2.1 and Remark 2.2, we get an isomorphism

$\varphi\colon{\mathbb{A}}^{1}\times X_{0}=V_{{\mathbb{A}}^{4}}(f-1)\to
V_{{\mathbb{A}}^{4}}(g-1)={\mathbb{A}}^{1}\times Y_{0}$

such that $\pi_{g,1}\circ\varphi=\theta\circ\pi_{f,1}$, where the morphisms
$\pi_{f,1}\colon V_{{\mathbb{A}}^{4}}(f-1)\to{\mathbb{P}}^{3}_{f}$ and
$\pi_{g,1}\colon V_{{\mathbb{A}}^{4}}(g-1)\to{\mathbb{P}}^{3}_{g}$ denote the
canonical projections.

Using that $X_{0}$ admit no non-trivial ${\mathbb{G}}_{a}$-action, by [Cra04,
Proposition 4.7] it follows that the intersection of the invariant subalgebras
$\mathbf{k}[{\mathbb{A}}^{1}\times X_{0}]^{{\mathbb{G}}_{a}}$ inside
$\mathbf{k}[{\mathbb{A}}^{1}\times X_{0}]$ over all ${\mathbb{G}}_{a}$-actions
on ${\mathbb{A}}^{1}\times X_{0}$ is equal to $\mathbf{k}[X_{0}]$. Using that
a similar statement holds for $Y_{0}\times{\mathbb{A}}^{1}$, we get that
$\varphi$ maps the fibres of ${\mathbb{A}}^{1}\times X_{0}\to X_{0}$ onto the
fibres of ${\mathbb{A}}^{1}\times Y_{0}\to Y_{0}$. Using the commutative
diagrams

$\textstyle{{\mathbb{A}}^{1}\times
X_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(t,x_{0})\mapsto
x_{0}}$$\scriptstyle{\pi_{f,1}}$$\textstyle{{\mathbb{P}}^{3}_{f}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\begin{subarray}{c}\textrm{projection}\\\
\textrm{from
$[1:0:0:0]$}\end{subarray}}$$\textstyle{X_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathbb{P}}^{2}_{f}}$
$\textstyle{{\mathbb{A}}^{1}\times
Y_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(t,y_{0})\mapsto
y_{0}}$$\scriptstyle{\pi_{g,1}}$$\textstyle{{\mathbb{P}}^{3}_{g}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\begin{subarray}{c}\textrm{projection}\\\
\textrm{from
$[1:0:0:0]$}\end{subarray}}$$\textstyle{Y_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathbb{P}}^{2}_{g}}$

we get that $\theta\colon{\mathbb{P}}_{f}^{3}\to{\mathbb{P}}^{3}_{g}$ maps the
fibres of the locally trivial ${\mathbb{A}}^{1}$-bundle
${\mathbb{P}}^{3}_{f}\to{\mathbb{P}}^{2}_{f}$ (with respect to the Zariski
topology) onto those of ${\mathbb{P}}^{3}_{g}\to{\mathbb{P}}^{2}_{g}$. Hence,
we get an isomorphism ${\mathbb{P}}^{2}_{f}\simeq{\mathbb{P}}^{2}_{g}$. As
$V_{{\mathbb{P}}^{2}}(f)$ and $V_{{\mathbb{P}}^{2}}(g)$ are non-rational this
isomorphism extends to an automorphism of ${\mathbb{P}}^{2}$. ∎

Now we are able to give the proof of Theorem C:

###### Proof of Theorem C.

(1): This follows from Corollary 5.15.

(3): If $H$ is a non-rational cubic surface, then as before, $H^{\prime}$ is a
cubic as well. Since $\chi(H)=\chi(H^{\prime})$, Lemma 5.13 implies that
$H^{\prime}$ is non-rational. Thus (3) follows from Proposition 5.17.

(2): If $H$ or $H^{\prime}$ is a cubic surface, then both are cubics (by using
the order of the Picard groups of the complements). If either $H$ or
$H^{\prime}$ is non-rational, then (3) implies that $H$ and $H^{\prime}$ are
isomorphic, and thus also piecewise isomorphic. We may thus assume that both
$H$ and $H^{\prime}$ are rational. By Corollary 5.12 it follows that $H$
(respectively $H^{\prime}$) is piecewise isomorphic to a disjoint union of
${\mathbb{A}}^{2}$, one point and $n$ (respectively $n^{\prime}$) copies of
${\mathbb{A}}^{1}$. As ${\mathbb{P}}^{3}\setminus H$ and
${\mathbb{P}}^{3}\setminus H^{\prime}$ are isomorphic, we get
$2+n=\chi(H)=\chi(H^{\prime})=2+n^{\prime}$ and hence $H$, $H^{\prime}$ are
piecewise isomorphic. If moreover $H$ is smooth, then $\chi(H)=9$ by Lemma
5.9. If $H^{\prime}$ would be non-normal, then Lemma 5.11 and Proposition 5.10
would imply that $\chi(H^{\prime})\leq 4$, which contradicts
$\chi(H)=\chi(H^{\prime})$. Thus Lemma 5.9 implies that $H^{\prime}$ is
smooth. This finishes the proof of (2). ∎

We finish this subsection with the following concrete question motivated by
Proposition 3.4:

######  Question 5.18.

Let $f,g\in{\mathbf{k}}[x,y,z,w]$ be given by

$f=x^{2}y+z^{3}\quad\textrm{and}\quad g=x^{2}y+z^{3}+xw^{2}\,.$

Are the varieties ${\mathbb{P}}^{3}_{f}$ and ${\mathbb{P}}^{3}_{g}$
isomorphic?

An affirmative answer would give a negative answer to Question (3) in degree
three and dimension three, since the zero locus of $f$ in ${\mathbb{P}}^{3}$
is non-normal, whereas the zero locus of $g$ in ${\mathbb{P}}^{3}$ is normal.
Moreover, $\chi(V_{{\mathbb{P}}^{3}}(f))=\chi(V_{{\mathbb{P}}^{3}}(g))$, since
${\mathbb{P}}^{4}_{f}\simeq{\mathbb{P}}^{4}_{g}$ by Proposition 3.4 and since
${\mathbb{P}}^{4}_{f}\to{\mathbb{P}}^{3}_{f}$,
${\mathbb{P}}^{4}_{g}\to{\mathbb{P}}^{3}_{g}$ are locally trivial
${\mathbb{A}}^{1}$-bundles with respect to the Zariski topology, see Example
4.5.

###  5.3. Automorphisms of complements of singular cubic surfaces

Throughout this subsection we always assume that $\mathbf{k}$ is algebraically
closed.

We finish this text by giving some examples, that give a negative answer to
Question (1), for singular irreducible cubic hypersurfaces of
${\mathbb{P}}^{3}$. As explained in the introduction, this question is wide
open for smooth cubics and known to have a negative answer for singular cubic
surfaces with Du Val singularities in characteristic zero [CDP18, Theorem C
and Theorem 4.3]. We now extend this to other singular cubics. Each
irreducible cubic hypersurface of ${\mathbb{P}}^{3}$ that does not have Du Val
singularities is either the cone over a smooth cubic curve in
${\mathbb{P}}^{2}$ or is rational and non-normal (this follows from Lemmas 5.8
and 5.9). In this latter case, it always contains a line with multiplicity $2$
(Proposition 5.10). These two cases are done in the next two simple lemmas,
that work in any dimension.

######  Lemma 5.19.

Let $d\geq 1$, $n\geq 3$ and let $X\subseteq{\mathbb{P}}^{n}$ be an
irreducible hypersurface of degree $d$, having a point of multiplicity $d$
$($i.e. being a cone$)$. Then, there exists an element
$\varphi\in\mathrm{Aut}({\mathbb{P}}^{n}\setminus X)$ that does not extend to
an element of $\mathrm{Aut}({\mathbb{P}}^{n})$.

###### Proof.

Changing coordinates, we may assume that $[0:\cdots:0:1]$ is a point of $X$ of
multiplicity $d$. Hence, $X=V_{{\mathbb{P}}^{n}}(f)$, where
$f\in{\mathbf{k}}[x_{0},\ldots,x_{n-1}]$ is irreducible and homogeneous of
degree $d$. We define $\varphi\in\mathrm{Aut}({\mathbb{P}}^{n}_{f})$ to be the
involution given by

$[x_{0}:\cdots:x_{n}]\mapsto\left[x_{0}:\cdots:x_{n-1}:-x_{n}+\frac{x_{0}^{d+1}}{f(x_{0},\ldots,x_{n-1})}\right]\,.$

∎

######  Lemma 5.20.

Let $d\geq 1$, $n\geq 3$ and let $X\subseteq{\mathbb{P}}^{n}$ be an
irreducible hypersurface of degree $d$, being of multiplicity $d-1$ along a
linear subspace $L\subseteq{\mathbb{P}}^{n}$ of dimension
$r\in\\{1,\ldots,n-2\\}$. Then, there exists an element
$\varphi\in\mathrm{Aut}({\mathbb{P}}^{n}\setminus X)$ that does not extend to
an element of $\mathrm{Aut}({\mathbb{P}}^{n})$.

###### Proof.

Changing coordinates, we may assume that
$L=V_{{\mathbb{P}}^{n}}(x_{r+1},\ldots,x_{n})$. Hence,
$X=V_{{\mathbb{P}}^{n}}(f)$, where $f\in{\mathbf{k}}[x_{0},\ldots,x_{n}]$ is
irreducible, homogeneous of degree $d$ and of the form

$f=x_{0}a_{0}+\cdots+x_{r}a_{r}+b\,,$

where $a_{0},\ldots,a_{r},b\in{\mathbf{k}}[x_{r+1},\ldots,x_{n}]$. We define a
$\mathbb{G}_{a}$-action on ${\mathbb{P}}^{n}_{f}$ by

$\begin{array}[]{ccc}\mathbb{G}_{a}\times{\mathbb{P}}^{n}_{f}&\to&{\mathbb{P}}^{n}_{f}\\\
(t,[x_{0}:\cdots:x_{n}])&\mapsto&\left[x_{0}+t\frac{a_{1}x_{n}^{2}}{f(x_{0},\ldots,x_{n})}:x_{1}-t\frac{a_{0}x_{n}^{2}}{f(x_{0},\ldots,x_{n})}:x_{2}:\cdots:x_{n}\right]\end{array}$

and choose $\varphi\in\mathrm{Aut}({\mathbb{P}}^{n}_{f})$ to be any non-
trivial element of
$\mathbb{G}_{a}\subseteq\mathrm{Aut}({\mathbb{P}}^{n}_{f})$. ∎

The following lemma follows from [CDP18, Theorem C and Theorem 4.3] in
characteristic zero. We insert a proof (in any characteristic), as the
argument is simple.

######  Lemma 5.21.

Let $X\subseteq{\mathbb{P}}^{3}$ be a normal rational singular cubic
hypersurface. Then ${\mathbb{P}}^{3}\setminus X$ contains an open affine
subset $U$ that is an ${\mathbb{A}}^{1}$-cylinder, i.e.
$U\simeq{\mathbb{A}}^{1}\times U^{\prime}$ for some affine variety
$U^{\prime}$.

###### Proof.

Since $X$ is singular, we may assume (after a coordinate change) that $X$ is
given by $f\coloneqq f_{2}(x,y,z)w+f_{3}(x,y,z)$ where
$f_{i}\in\mathbf{k}[x,y,z]$ is homogeneous of degree $i$ and $[w:x:y:z]$
denote the homogeneous coordinates of ${\mathbb{P}}^{3}$. As $X$ is normal and
rational, $f_{2}\neq 0$, see Lemma 5.8. Let $C\coloneqq
V_{{\mathbb{P}}^{2}}(f_{2})$. We may choose the coordinates $(x,y,z)$ in such
a way, that $V_{{\mathbb{P}}^{2}}(z)$ is contained in $C$ if $f_{2}$ is
reducible, and that $V_{{\mathbb{P}}^{2}}(z)$ is tangent to $C$ if $f_{2}$ is
irreducible. Then
${\mathbb{P}}^{2}_{zf_{2}}\simeq{\mathbb{A}}^{2}_{f_{2}(x,y,1)}\simeq
D\times{\mathbb{A}}^{1}$, where
$D\in\\{{\mathbb{A}}^{1},{\mathbb{A}}^{1}\setminus\\{0\\}\\}$, and we get an
isomorphism

$\begin{array}[]{cccc}{\mathbb{P}}^{3}_{zff_{2}}\simeq&{\mathbb{A}}^{3}_{f(w,x,y,1)f_{2}(x,y,1)}&\xrightarrow{\simeq}&({\mathbb{A}}^{1}\setminus\\{0\\})\times{\mathbb{A}}^{2}_{f_{2}(x,y,1)}\\\
&(w,x,y)&\mapsto&(f_{2}(x,y,1)w+f_{3}(x,y,1),x,y)\\\
&\left(\frac{u-f_{3}(x,y,1)}{f_{2}(x,y,1)},x,y\right)&\mapsfrom&(u,x,y)\,.\end{array}$

∎

######  Proposition 5.22.

For each singular irreducible cubic hypersurface $X\subseteq{\mathbb{P}}^{3}$,
there exists an element $\varphi\in\mathrm{Aut}({\mathbb{P}}^{3}\setminus X)$
that does not extend to an element of $\mathrm{Aut}({\mathbb{P}}^{3})$.

###### Proof.

If $X$ is a cone, this follows from Lemma 5.19. If $X$ is not normal, it
contains a line with multiplicity $2$ (Proposition 5.10), and the result
follows from Lemma 5.20. In the remaining case, $X$ is rational (Lemma 5.8)
and normal. As $\operatorname{Pic}({\mathbb{P}}^{3}\setminus
X)\simeq{\mathbb{Z}}/3{\mathbb{Z}}$ is finite, Lemma 5.21 implies that
${\mathbb{P}}^{3}\setminus X$ admits a non-trivial ${\mathbb{G}}_{a}$-action,
see e.g. [DK15, Propostion 2]. This implies the result. ∎

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