%
%
%   Chapter 1 : Setting and main results
%
%

\chapter{Setting and main results}
\label{sec : Intro}

%\fontNeedsEdit{!}{keep question wording consistent throughout thesis---literally copy and paste}



\section{Guiding questions}

In this thesis we study nice varieties $X$ defined over a field $k$.% Begin footnote.
\footnote{A \fontDefWord{$k$-variety} is a separated scheme of finite type over $k$. A $k$-variety is \fontDefWord{nice} if it is smooth, projective, and geometrically integral.} % End footnote.
The proximate question is an inverse Galois problem.

\begin{question}[Inverse Galois problem for varieties]
\label{q : Inverse Galois Problem For Varieties}
Given a group $G$ and a field $k$, is there a $k$-variety $X$ (perhaps from a particular family) such that the image of the representation of the absolute Galois group of $k$ on the geometric Picard group of $X$ is isomorphic to $G$?
\end{question}

While this question may be of independent interest, it arises naturally from another, more ultimate question.

\begin{question}[Existence of $k$-points]
\label{q : Existence Of k-Points}
Given a nice $k$-variety $X$, is the set $X(k)$ of $k$-points of $X$ nonempty?
\end{question}

This chapter surveys our area of research. We first present a working view of Galois theory, the Picard group, and $k$-points on varieties. Then, starting from Question \ref{q : Existence Of k-Points} on the existence of $k$-points, we see how obstructions to $k$-points lead to the study of Brauer groups and the inverse Galois problem of Question \ref{q : Inverse Galois Problem For Varieties}. We introduce two families of varieties, del Pezzo surfaces and K3 surfaces, on which this thesis focuses. We close the chapter with a preview of our main results.



\section{{G}alois theory, {P}icard groups, and $k$-points}

This section presents a concrete, working view of concepts in Galois theory and algebraic geometry that feature in our work: elementary Galois theory of groups, the Picard group of a variety, group representations, and $k$-points of a variety. Readers acquainted with these concepts are invited to jump to Section \ref{sec : Intro : Existence Of k Points}, which starts our course of inquiry from Question \ref{q : Existence Of k-Points} on the existence of $k$-points; or to Section \ref{sec : Main Results}, which presents our main results.



\subsection{{G}alois theory and the inverse {G}alois problem}

The fundamental idea of Galois theory is to associate to an extension $K \fieldExtension k$ of fields a group $\automorphismGroup(K \fieldExtension k)$. A field extension $K \fieldExtension k$ is a map $k \rightarrow K$ that preserves the field structure---namely, addition, multiplication, and the multiplicative unit.% Begin footnote.
\footnote{More precisely, let $k$ and $K$ be fields. A \fontDefWord{field extension} (also called a \fontDefWord{field homomorphism}) from $k$ to $K$ is a map $\varphi : k \rightarrow K$ such that, for all $a_{1},a_{2} \in k$,
\begin{align*}
\varphi(a_{1} + a_{2})
&=
\varphi(a_{1}) + \varphi(a_{2})
&
\varphi(a_{1} a_{2})
&=
\varphi(a_{1}) \varphi(a_{2})
&
\varphi(1_{k})
&=
1_{K}
\end{align*}%
} % End footnote.
The group $\automorphismGroup(K \fieldExtension k)$ is the collection of bijective such maps from $K$ to itself---that is, automorphisms of $K$---that fix the base field $k$ point-wise.% Begin footnote.
\footnote{The operation of function composition gives this collection the structure of a group.} % End footnote.
Under suitable hypotheses,% Begin footnote.
\footnote{For the construction to yield the stated correspondence, the field extension must satisfy two properties: It must be normal and separable. In this case, the field extension is said to be Galois. In the case of infinite-degree Galois extensions, the correspondence requires that we impose additional structure, namely, the Krull topology on the Galois group, and that we restrict attention to the closed subgroups under this topology. The applications in this thesis deal exclusively with finite-degree, separable field extensions.} % End footnote.
this association gives an explicit, bijective correspondence between the intermediate fields of the field extension $K \fieldExtension k$ and the subgroups of $\automorphismGroup(K \fieldExtension k)$. In this case, the group $\automorphismGroup(K \fieldExtension k)$ is called the \fontDefWord{Galois group of $K \fieldExtension k$}, denoted $\galoisGroup(K \fieldExtension k)$.

\begin{remark}
Given a field $k$, we are frequently interested in a particular field extension of $k$, namely, a separable closure $\separableClosure{k}$ of $k$, which always exists. Formally, we fix an algebraic closure $\algebraicClosure{k}$ of $k$---another field extension of $k$ that always exists, unique up to isomorphism---then take the separable closure $\separableClosure{k}$ of $k$ in $\algebraicClosure{k}$. All elements of $\separableClosure{k}$, as elements of $\algebraicClosure{k}$, are algebraic over $k$, that is, they are roots of polynomials with coefficients in $k$. The additional property of being separable guarantees that for certain polynomials defining the elements of $\separableClosure{k}$, the number of distinct roots equals the degree of the polynomial. Our work in Chapter \ref{sec : dP1} illustrates these properties. For now, we can think of a separable closure $\separableClosure{k}$ of $k$ as a largest field containing $k$ such that Galois theory applies. The corresponding Galois group $\galoisGroup(\separableClosure{k} \fieldExtension k)$ is called the \fontDefWord{absolute Galois group of $k$} and features prominently in our work.
\end{remark}

Fix a base field $k$. We may view the Galois correspondence between field extensions and subgroups from two directions, depending on what objects we take as primitive, or given. If we take the field extension as the primitive, then we get the Galois problem:

\begin{question}[Galois problem for groups]
\label{q : Intro : Group Theory : Galois Problem}
Let $K \fieldExtension k$ be a Galois field extension. Find $\galoisGroup(K \fieldExtension k)$, up to isomorphism.
\end{question}

If we invert our view, instead taking the isomorphism class of the Galois group as the primitive, then we get the inverse Galois problem:

\begin{question}[Inverse Galois problem for groups]
\label{q : Intro : Group Theory : Inverse Galois Problem}
Let $G$ be a group. Find a Galois field extension $K \fieldExtension k$ such that $\galoisGroup(K \fieldExtension k)$ is isomorphic to $G$.
\end{question}

We emphasize: In Question \ref{q : Intro : Group Theory : Galois Problem}, we are given the field extension, and we seek the group. In Question \ref{q : Intro : Group Theory : Inverse Galois Problem}, we are given the group, and we seek the field extension. In both questions, the base field $k$ of the field extension is assumed to be fixed.

In Question \ref{q : Intro : Group Theory : Galois Problem}, the object sought (a group) always exists, even if it is difficult to determine. In contrast, in Question \ref{q : Intro : Group Theory : Inverse Galois Problem}, the object sought (a field extension of $k$) may not exist. For example, let $k$ be a finite field. Any Galois extension $K \fieldExtension k$ necessarily has a cyclic Galois group. Taking the ``target'' group $G$ to be any non-cyclic group thus gives an instance of Question \ref{q : Intro : Group Theory : Inverse Galois Problem} for which no solution exists.

To go from the inverse Galois problem for groups to the inverse Galois problem for varieties, we introduce the Picard group.



\subsection{The {P}icard group}
\label{sec : Intro : Primer : The Picard Group}

Let $k$ be a field, and let $X$ be a nonsingular, geometrically integral $k$-variety.% Begin footnote.
\footnote{These hypotheses on $X$ are stronger than needed to obtain the isomorphism in \eqref{eq : Intro : Picard Group Weil Divisor Class Group Isomorphism}. More generally, we may let $X$ be a noetherian, integral, separated, locally factorial scheme. Our hypothesis that $X$ is nonsingular implies that it is regular in codimension one.} % End footnote.
The \fontDefWord{Picard group of $X$}, denoted $\picardGroup X$, is the group of isomorphism classes of invertible sheaves on $X$, equipped with the operation of tensor product.

Under our hypotheses on $X$, we can identify $\picardGroup X$ with a more concrete construction, the Weil divisor class group of $X$, denoted $\weilDivisorClassGroup X$. Loosely, we can think of $\weilDivisorClassGroup X$ as a collection of formal sums of certain subobjects on $X$ that are ``one dimension smaller'' than $X$. For example, if $X$ is a curve, then these subobjects are points on $X$; if $X$ is a surface, then they are curves on $X$; and so forth.

Formally, we construct $\weilDivisorClassGroup X$ as follows. A \fontDefWord{prime divisor} on $X$ is a closed, integral subscheme of codimension one. The Weil divisor group of $X$, denoted $\weilDivisorGroup X$, is the free abelian group generated by the prime divisors. That is, an element $D \in \weilDivisorGroup X$---a \fontDefWord{Weil divisor}---is a formal sum
\begin{align*}
D
=
\sum_{i} n_{i} Y_{i}
\end{align*}
where the sum is taken over all prime divisors $Y_{i}$ of $X$, and all but finitely many of the coefficients $n_{i}$ equal $0$. For each nonzero rational function $f$ in the function field $K(X)$ of $X$,% Begin footnote.
\footnote{Loosely, we can think of the function field $K(X)$ as the collection of $k$-valued rational functions defined on ``most of'' $X$, equipped with the operations of addition and multiplication defined pointwise. Formally, let $P \in X$. A function $f : X \rightarrow k$ is \fontDefWord{regular at $P$} if there exist an open subset $U_{P} \subseteq X$ with $P \in U_{P}$, and polynomials $g,h$, such that for all $Q \in U_{P}$, $h(Q) \neq 0$ and $f(Q) = \frac{g(Q)}{h(Q)}$. (If $X$ is a projective variety, then we further require that $g,h$ be homogeneous of the same degree.) $f$ is \fontDefWord{regular on X} if for all $P \in X$, $f$ is regular at $P$. The \fontDefWord{function field of $X$}, denoted $K(X)$, is the collection of equivalence classes of regular functions $f : U \rightarrow k$ on all nonempty open subsets $U \subseteq X$, with $f_{1} \sim f_{2}$ if $f_{1}|_{U_{1} \cap U_{2}} = f_{2}|_{U_{1} \cap U_{2}}$. A \fontDefWord{rational function on $X$} is an element of $K(X)$.} % End footnote.
we define the \fontDefWord{divisor of $f$}, denoted $(f)$, to be
\begin{align*}
(f)
=
\sum_{i} v_{Y_{i}}(f) Y_{i}
\end{align*}
where the sum is taken over all prime divisors $Y_{i}$ of $X$, and $v_{Y_{i}}$ denotes the discrete valuation associated to the local ring of $X$ at the generic point of $Y_{i}$. The hypotheses on $X$ (specifically, the fact that $X$ is noetherian) guarantee that all but finitely many of the coefficients $v_{Y_{i}}(f)$ equal $0$ in this construction, as required for a Weil divisor.

To illustrate these ideas, let $X$ be a curve. In this case, the prime divisors $Y_{i}$ are points on $X$. Given a nonzero rational function $f$ on $X$, the valuation $v_{Y_{i}}(f)$ equals the order of the zero or pole of $f$ evaluated at the point $Y_{i}$, positive if $f$ has a zero at $Y_{i}$, negative if $f$ has a pole at $Y_{i}$.

A \fontDefWord{principal divisor} on $X$ is a Weil divisor $D$ on $X$ such that there exists a rational function $f \in K(X)^{\times}$ such that $D = (f)$. The principal divisors form a subgroup $\principalDivisors X$ of $\weilDivisorGroup X$. The \fontDefWord{Weil divisor class group of $X$}, denoted $\weilDivisorClassGroup X$, is the quotient
\begin{align*}
\weilDivisorClassGroup X
=
\weilDivisorGroup X/\principalDivisors X
\end{align*}

We introduced the Weil divisor class group to make the Picard group more concrete. To this end, we note that a nice $k$-variety $X$ can be described as the zero set of a set $\ideal(X)$ of certain polynomials $f$:
\begin{align*}
X
=
\zeroSet(\ideal(X))
=
\left\{x \in \projective^{n}_{k} \st \forall f \in \ideal(X),f(x) = 0\right\}
\end{align*}
Similarly, a prime divisor $Y$ on $X$ can be described as the zero set of a set $\ideal(Y)$ of certain polynomials $g$. A Weil divisor $D$ is built from prime divisors, and an element of the Weil divisor class group can be represented by a Weil divisor. Thus, an element of the Weil divisor class group can be understood in terms of polynomials---in fact, in terms of a finite number of polynomials.

The following result allows us to think of the Picard group and the more concrete Weil divisor class group as essentially the same.

\begin{proposition}[\cite{Hartshorne1977}*{Corollary II.6.16}]
Let $X$ be an integral, locally factorial, noetherian, separated scheme. Then
\begin{align}
\picardGroup X
\isomorphism
\weilDivisorClassGroup X%
\label{eq : Intro : Picard Group Weil Divisor Class Group Isomorphism}
\end{align}
\end{proposition}

Let $K \fieldExtension k$ be a field extension. The map $k \rightarrow K$ allows us to view the polynomials $f$ that define a $k$-variety $X$, and the polynomials $g$ that define a prime divisor $Y$ of $X$, as polynomials with coefficients in $K$. By allowing coefficients in the larger field $K$, we may get new Weil divisors on $X$, and new divisor classes in the Picard group of $X$. Let $\picardGroup X_{K}$ denote the Picard group of $X$ obtained when we take polynomial coefficients in $K$. In particular, if we take $K$ to be a separable closure $\separableClosure{k}$ of $k$, then the corresponding group $\picardGroup X_{\separableClosure{k}}$ is called the \fontDefWord{geometric Picard group of $X$}.



\subsection{{G}alois representations on the {P}icard group}
\label{sec : Intro : Galois Representations On The Picard Group}

Let $X$ be a nice $k$-variety, and let $K \fieldExtension k$ be a Galois field extension. Form the Galois group $\galoisGroup(K \fieldExtension k)$ and the Picard group $\picardGroup X_{K}$. The elements of $\galoisGroup(K \fieldExtension k)$ act naturally on the coefficients of the polynomials $g$ defining the building blocks of the divisor classes in $\weilDivisorClassGroup X_{K}$---that is, the prime divisors of $X$---which by \eqref{eq : Intro : Picard Group Weil Divisor Class Group Isomorphism} we may identify with $\picardGroup X_{K}$. Specifically, an element $\sigma \in \galoisGroup(K \fieldExtension k)$ maps each divisor (class) on $X$ to another divisor (class) on $X$, in a manner that gives an automorphism of $\picardGroup X_{K}$. The data of these mappings, for all elements of the Galois group, can be encoded in a map
\begin{align}
\rho_{K}
:
\galoisGroup(K \fieldExtension k)
\rightarrow
\automorphismGroup(\picardGroup X_{K})%
\label{eq : Intro : Galois Representation On Picard Group}
\end{align}
called a \fontDefWord{(Galois) representation} of $\galoisGroup(K \fieldExtension k)$ on $\picardGroup X_{K}$. Note that $\rho_{K}$ is a group homomorphism, from the Galois group of $K \fieldExtension k$ to the automorphism group of $\picardGroup X_{K}$. Hence its image is a group.

We may view this action of $\galoisGroup(K \fieldExtension k)$ on $\picardGroup X_{K}$ in the setting of modules over rings.% Begin footnote.
\footnote{The two views are equivalent. See Section \ref{sec : GLnZ : Representation Theory}.} % End footnote.
Specifically, let the \fontDefWord{group ring} of $\galoisGroup(K \fieldExtension k)$ over $\integers$, denoted $\integers \galoisGroup(K \fieldExtension k)$, be the collection of formal sums
\begin{align*}
\sum_{i} a_{i} \sigma_{i}
\end{align*}
where for each $i$, $a_{i} \in \integers$ and $\sigma_{i} \in \galoisGroup(K \fieldExtension k)$; and all but finitely many of the coefficients $a_{i}$ equal zero. The ring operations of addition and multiplication are defined as expected.% Begin footnote.
\footnote{More precisely, the ring operations in $\integers \galoisGroup(K \fieldExtension k)$ are defined by
\begin{align*}
\sum_{i} a_{i} \sigma_{i} + \sum_{i} b_{i} \sigma_{i}
&=
\sum_{i} (a_{i} + b_{i}) \sigma_{i}
&
\left(\sum_{i} a_{i} \sigma_{i}\right) \left(\sum_{i} b_{i} \sigma_{i}\right)
&=
\sum_{i,j} a_{i} b_{j} \sigma_{i} \sigma_{j}
\end{align*}
where to multiply, we formally expand the product of the two sums, perform the resulting multiplications in $\integers$ and in $\galoisGroup(K \fieldExtension k)$, and simplify.} % End footnote.
Finally, we equip $\picardGroup X_{K}$ with the structure of a $\integers \galoisGroup(K \fieldExtension k)$-module by defining scalar multiplication in the module using the group action of $\galoisGroup(K \fieldExtension k)$ on $\picardGroup X_{K}$:
\begin{align*}
\left(\sum_{i} a_{i} \sigma_{i}\right) \cdot Y
=
\sum_{i} a_{i} \sigma_{i}(Y)
\end{align*}

Within either framework, we may construct a cochain complex, and thus a sequence of cohomology groups $\cohomology[i]{\galoisGroup(K \fieldExtension k),\picardGroup X_{K}}$. We will relate this first cohomology group to the Brauer group of $X$ in \eqref{eq : Intro : Br1/Br0 Galois Cohomology}.



\subsection{The inverse {G}alois problem for varieties}

We now translate the Galois problems for groups into Galois problems for varieties. Fix a base field $k$. Choose a separable closure $\separableClosure{k}$ of $k$, and form the absolute Galois group $\galoisGroup(\separableClosure{k} \fieldExtension k)$. Given a nice $k$-variety $X$, form its geometric Picard group $\picardGroup X_{\separableClosure{k}}$.

If we take the $k$-variety $X$ as the primitive, then we get a translation of Question \ref{q : Intro : Group Theory : Galois Problem}, the Galois problem for varieties:

\begin{question}[Galois problem for varieties]
Let $X$ be a nice $k$-variety. Find the image of the representation \eqref{eq : Intro : Galois Representation On Picard Group} of the absolute Galois group, up to isomorphism.
\end{question}

If we invert our view, taking instead the isomorphism class of the image as the primitive, then we get a translation of Question \ref{q : Intro : Group Theory : Inverse Galois Problem}, the inverse Galois problem for varieties:

\begin{question}[Inverse Galois problem for varieties]
\label{q : Intro : Inverse Galois Problem For Varieties}
Let $G$ be a group. Find a nice $k$-variety $X$ such that the image of the representation \eqref{eq : Intro : Galois Representation On Picard Group} of the absolute Galois group is isomorphic to $G$.
\end{question}



\subsection{The inverse Brauer problem for varieties}

Using \eqref{eq : Intro : Br1/Br0 Galois Cohomology} below, we may append an additional association, to shift the focus from Galois representations to Brauer group modulo constants. Let us refer to these variants as Brauer problems.

If we take the $k$-variety $X$ as the primitive, then we get a translation of Question \ref{q : Intro : Group Theory : Galois Problem}, the Brauer problem for varieties:

\begin{question}[Brauer problem for varieties]
Let $X$ be a nice $k$-variety. Find the algebraic Brauer group of $X$ modulo constants, up to isomorphism.
\end{question}

If we invert our view, taking instead the isomorphism class of the target group as the primitive, then we get a translation of Question \ref{q : Intro : Group Theory : Inverse Galois Problem}, the inverse Brauer problem for varieties:

\begin{question}[Inverse Brauer problem for varieties]
\label{q : Intro : Inverse Brauer Problem}
Let $G$ be a group. Find a nice $k$-variety $X$ such that the algebraic Brauer group of $X$ modulo constants is isomorphic to $G$.
\end{question}

These are the problems we tackle in this thesis.

\begin{remark}
Let $X$ be a nice $k$-variety such that $\picardGroup X_{\separableClosure{k}}$ is free and finitely generated. Then the algebraic Brauer group of $X$ modulo constants is a finite, commutative group. Thus in Question \ref{q : Intro : Inverse Brauer Problem} we should restrict $G$ to finite, commutative groups if we want a nontrivial answer.
\end{remark}



\subsection{$k$-points of a variety}

A nice $k$-variety $X$ can be defined by a set $\ideal(X)$ of homogeneous polynomials $f \in k[x_{0},\ldots,x_{n}]$. These polynomials embed $X$ into an ambient projective space. Under this embedding, we say a point $x = (x_{0},\ldots,x_{n}) \in k^{n + 1}$ \fontDefWord{lies on $X$} if for all $f \in \ideal(X)$, $f(x) = 0$. Such a point is called a \fontDefWord{$k$-point} of $X$. We denote the set of all $k$-points of $X$ by $X(k)$.

Similarly, under this embedding, each prime divisor $Y$ of $X$ can be defined by a set $\ideal(Y)$ of homogeneous polynomials $g \in k[x_{0},\ldots,x_{n}]$. As with the polynomials defining $X$, we may ask whether these polynomials defining $Y$ evaluate to $0$ at various points. This underlies our analysis of del Pezzo surfaces in Chapter \ref{sec : dP1}.

Let $K \fieldExtension k$ be a field extension of $k$. As before, the map $k \rightarrow K$ allows us to work with polynomial coefficients in $K$. We say a point $x \in K^{n + 1}$ lies on $X$ if for all $f \in \ideal(X)$, $f(x) = 0$. We denote the set of all $K$-points of $X$ by $X(K)$.



\section{Existence of $k$-points}
\label{sec : Intro : Existence Of k Points}

Let $k$ be a \fontDefWord{global field}, that is, a finite extension of either the field $\rationals$ of rational numbers or the field $\fontField{F}_{q}(t)$ of rational functions in one variable over a finite field $\fontField{F}_{q}$. Let $X$ be a nice $k$-variety. Given these primitives, we may ask Question \ref{q : Existence Of k-Points}: Is the set $X(k)$ of $k$-points of $X$ nonempty?

If $X$ has a $k$-point, then for each valuation $v$ of $k$, this global $k$-point specializes to a local $k_{v}$-point, where $k_{v}$ denotes the completion of $k$ at $v$. This gives a necessary condition for $X$ to have a $k$-point:
\begin{align}
\text{for all completions $k_{v}$ of $k$},
X(k_{v})
\neq
\emptyset%
\label{eq : Intro : Locally Soluble}
\end{align}

We may bundle these local data in the product $\prod_{v \in \Omega} X(k_{v})$, taken over the set $\Omega$ of all valuations $v$ of $k$. Because $X$ is projective over $k$, it is proper over $k$, so this product of local points equals the set of adelic points of $X$:% Begin footnote.
\footnote{The \fontDefWord{adele ring} $\adeles_{k}$ of $k$ is the restricted product
\begin{align*}
\adeles_{k}
=
\prod_{v \in \Omega}{}' k_{v}
=
\left\{(a_{v})_{v \in \Omega} \st \text{for all but finitely many $v$},a_{v} \in \valuationRing_{v} \right\}
\end{align*}
where $\valuationRing_{v}$ denotes the valuation ring of $k_{v}$.}% End footnote.
\begin{align*}
\prod_{v \in \Omega} X(k_{v})
=
X(\adeles_{k})
\end{align*}
Thus we may write the necessary condition for $X$ to have a $k$-point as
\begin{align*}
X(k)
\neq
\emptyset
\qquad
\text{implies}
\qquad
X(\adeles_{k})
\neq
\emptyset
\end{align*}

A nice $k$-variety $X$ is \fontDefWord{locally soluble} if it satisfies \eqref{eq : Intro : Locally Soluble}. The \fontDefWord{Hasse principle}, also called the \fontDefWord{local-to-global principle}, states that if $X$ is locally soluble, then $X$ has a (global) $k$-point. Several families of varieties satisfy the local-to-global principle, including quadratic forms (the varieties originally studied by Hasse \cite{Hasse1924}), Severi--Brauer varieties (see \cite{Gille2006}*{Chapter 5}(, and del Pezzo surfaces of degree at least five (see \cite{VarillyAlvarado2013}*{Theorem 2.1}).

In general, this necessary condition is not sufficient, as examples have shown since at least the 1940s.% Begin footnote.
\footnote{In a series of articles from 1874--1882, P\'{e}pin \cites{Pepin1879,Pepin1880} includes examples of genus-$1$ curves that are everywhere locally solvable, and he asserts the nonexistence of rational points on these curves. See Lemmermeyer \cite{Lemmermeyer1999} for discussion and proofs.} % End footnote.
Lind \cite{Lind1940} and Reichardt \cite{Reichardt1942} independently showed that the genus-$1$ curve $X = \zeroSet(f) \subseteq \projective_{\rationals,[1,2,1]}^{2}$ defined by the vanishing of
\begin{align*}
f
=
x_{0}^{4} - 2 x_{1}^{2} - 17 x_{2}^{4}
\end{align*}
is locally soluble, yet $X(\rationals) = \emptyset$.% Begin footnote.
\footnote{For a concise discussion of Lind's and Reichardt's example, see \cite{ColliotThelene2021}*{pages 331--332}.} % End footnote.
Such a variety is called a \fontDefWord{counterexample to the Hasse principle}.

Using the cohomological Brauer group $\brauerGroup X = \cohomologyEtale[2]{X,\groupSchemeMultiplicative}_{\torsion}$, Manin \cite{Manin1971} constructed a candidate obstruction to the existence of $k$-points, the Brauer--Manin set $X(\adeles_{k})^{\brauerGroup X}$. We can view this obstruction set as lying between the (global) $k$-points of $X$ and the (bundled local) adelic points of $X$:
\begin{align*}
X(k)
\subseteq
X(\adeles_{k})^{\brauerGroup X}
\subseteq
X(\adeles_{k})
\end{align*}
In particular, if the intermediate Brauer--Manin set $X(\adeles_{k})^{\brauerGroup X}$ is empty, then so is $X(k)$, even if $X(\adeles_{k})$ is not.

In this way, the study of $k$-points leads naturally to the study of Brauer groups.



\section{{B}rauer groups}
\label{sec : Intro : Brauer Groups}

Let $k$ be a field, and let $X$ be a $k$-variety. Fix an algebraic closure $\algebraicClosure{k}$ of $k$, and consider the separable closure $\separableClosure{k}$ of $k$ in $\algebraicClosure{k}$. Form the absolute Galois group $\galoisGroup(\separableClosure{k} \fieldExtension k)$ and the fiber product $X_{\separableClosure{k}}$ of $X$ and $\Spec \separableClosure{k}$ over $\Spec k$:
\begin{align*}
\begin{tikzpicture}
% Nodes:
\node (Xksep) {$X_{\separableClosure{k}}$};
\node (X) [right = of Xksep] {$X$};
\node (ksep) [below = of Xksep] {$\Spec \separableClosure{k}$};
\node (k) [below = of X] {$\Spec k$};
% Arrows:
\draw[-stealth] (Xksep) -- node [] {} (ksep);
\draw[-stealth] (X) -- node [] {} (k);
\draw[-stealth] (Xksep) -- node [] {} (X);
\draw[-stealth] (ksep) -- node [] {} (k);
\end{tikzpicture}
\end{align*}
To this diagram, we may apply the contravariant functor $\brauerGroup(-) = \cohomologyEtale[2]{-,\groupSchemeMultiplicative}_{\torsion}$ that defines the cohomological Brauer group, then use the isomorphism $\brauerGroup(\Spec k) \isomorphism \brauerGroup k$. This gives a sequence
\begin{align*}
\brauerGroup k \rightarrow \brauerGroup X \rightarrow \brauerGroup X_{\separableClosure{k}}
\end{align*}
of commutative groups,% Begin footnote.
\footnote{In fact, $\image(\brauerGroup k \rightarrow \brauerGroup X) \subseteq \ker(\brauerGroup X \rightarrow \brauerGroup X_{\separableClosure{k}})$, so this sequence is a (co)chain complex.} % End footnote.
which filters the Brauer group $\brauerGroup X$ of $X$ into three parts:
\begin{enumerate}
\item Elements in $\brauerGroupConstant X \defeq \image(\brauerGroup k \rightarrow \brauerGroup X)$ are called \fontDefWord{constant}.
\item Elements in $\brauerGroupAlgebraic X \defeq \ker(\brauerGroup X \rightarrow \brauerGroup X_{\separableClosure{k}})$ are called \fontDefWord{algebraic}.
\item Elements in $\brauerGroup X \setminus \brauerGroupAlgebraic X$ are called \fontDefWord{transcendental}.
\end{enumerate}

This filtration allows us to connect the Brauer group to Galois cohomology, as follows. The Leray spectral sequence
\begin{align*}
E_{2}^{p,q}
\defeq
\cohomology[p]{\galoisGroup(\separableClosure{k} | k),\cohomologyEtale[q]{X_{\separableClosure{k}},\groupSchemeMultiplicative}}
\Rightarrow
\cohomologyEtale[p + q]{X,\groupSchemeMultiplicative}
\end{align*}
for the structure morphism $X \rightarrow \Spec k$ gives rise to an exact sequence of low-degree terms.% Begin footnote.
\footnote{See, for example, Colliot-Th\'{e}l\`{e}ne and Skorobogatov \cite{ColliotThelene2021}*{Section 5.4.1}.} % End footnote.
If $k$ is a number field, and if $X$ satisfies the condition $\separableClosure{k}[X]^{\times} \isomorphism (\separableClosure{k})^{\times}$, then this exact sequence implies
\begin{align}
\brauerGroupAlgebraic X/\brauerGroupConstant X
\xrightarrow{\isomorphism}
\cohomology[1]{\galoisGroup(\separableClosure{k} \fieldExtension k),\picardGroup X_{\separableClosure{k}}}%
\label{eq : Intro : Br1/Br0 Galois Cohomology}
\end{align}
That is, up to isomorphism, we can compute the algebraic Brauer group of $X$ modulo constants from the action of the absolute Galois group $\galoisGroup(\separableClosure{k} \fieldExtension k)$ on the geometric Picard group $\picardGroup X_{\separableClosure{k}}$ of $X$.

For any subset $B$ of the full Brauer group $\brauerGroup X$ of $X$, Manin's construction yields an obstruction set $X(\adeles_{k})^{B}$:
\begin{align*}
X(k)
\subseteq
X(\adeles_{k})^{\brauerGroup X}
\subseteq
X(\adeles_{k})^{B}
\subseteq
X(\adeles_{k})
\end{align*}
In particular, we may perform this construction for $B = \brauerGroupAlgebraic X$, the subgroup of all algebraic Brauer elements. Constant Brauer elements never obstruct the existence of $k$-points. Thus the obstruction set $X(\adeles_{k})^{\brauerGroupAlgebraic X}$ gives useful information only if $\brauerGroupAlgebraic X \neq \brauerGroupConstant X$, that is, only if $\brauerGroupAlgebraic X/\brauerGroupConstant X$ is nontrivial. In this case, $X(\adeles_{k})^{\brauerGroupAlgebraic X}$ may be empty, and thus explain the nonexistence of $k$-points on $X$.

The isomorphism in \eqref{eq : Intro : Br1/Br0 Galois Cohomology} lets us investigate $\brauerGroupAlgebraic X/\brauerGroupConstant X$ indirectly. It requires that we understand the action of the absolute Galois group of $k$ on the geometric Picard group of $X$. This is the focus of our present study.



\section{Del {P}ezzo surfaces and {K}3 surfaces}

A \fontDefWord{surface} is a variety of dimension $2$. In this thesis, we study two classes of surfaces: del Pezzo surfaces of degree $1$, and algebraic K3 surfaces.



\subsection{General features of surfaces}
\label{sec : Intro : General Features Of Surfaces}

One feature of surfaces that facilitates their study is the intersection pairing. Let $X$ be a smooth, projective surface. Then we may define a pairing
\begin{align*}
(-,-)
:
\weilDivisorGroup X \times \weilDivisorGroup X
\rightarrow
\integers
\end{align*}
on the group $\weilDivisorGroup X$ of Weil divisors of $X$. This pairing is bilinear and symmetric, and it captures the intuition of how many times two curves on $X$ intersect, in the following sense: If $C_{1}$ and $C_{2}$ are smooth curves on $X$ whose intersections are transverse,% Begin footnote.
\footnote{More precisely, let $C_{1}$ and $C_{2}$ be curves on $X$, let $P \in C_{1} \cap C_{2}$ be a point of intersection of $C_{1}$ and $C_{2}$, and let $f_{1}$ and $f_{2}$ be local equations for $C_{1}$ and $C_{2}$, respectively, at $P$. The curves $C_{1}$ and $C_{2}$ \fontDefWord{meet transversally at $P$} if $f_{1}$ and $f_{2}$ generate the maximal ideal $\fontIdeal{m}_{P}$ of the local ring $\fontSheaf{O}_{X,P}$ on $X$ at $P$.} % End footnote.
then $(C_{1},C_{2})$ equals the number of points of intersection. Moreover, this pairing depends only on the linear equivalence class of the curves. This last property allows us to descend the pairing to the Weil divisor class group $\weilDivisorClassGroup X = \weilDivisorGroup X/\principalDivisors X$, the quotient of $\weilDivisorGroup X$ in which linearly equivalent divisors are identified. Using the isomorphism $\picardGroup X \isomorphism \weilDivisorClassGroup X$ from \eqref{eq : Intro : Picard Group Weil Divisor Class Group Isomorphism}, we may view this pairing as defined on the Picard group $\picardGroup X$ of $X$, and we call it the \fontDefWord{intersection pairing} on $\picardGroup X$.

The natural Galois action on the Picard group preserves this intersection pairing. More precisely, let $X$ be a $k$-surface, and let $K \fieldExtension k$ be any Galois extension of $k$. Then for all $\sigma \in \galoisGroup(K \fieldExtension k)$, and for all $D_{1},D_{2} \in \picardGroup X$,
\begin{align}
(\sigma D_{1},\sigma D_{2})
=
(D_{1},D_{2})%
\label{eq : Intro : Galois Action Preserves Intersection Pairing}
\end{align}

If $X$ is projective, then it has a hyperplane class $H \in \picardGroup X$ defined over the base field $k$. This class is fixed by all elements of the Galois group. That is, for all $\sigma \in \galoisGroup(K \fieldExtension k)$, the automorphism $\rho_{K}(\sigma)$ of $\picardGroup X_{K}$, given by the representation \eqref{eq : Intro : Galois Representation On Picard Group}, has a nontrivial eigenspace associated to the eigenvalue $1$.% Begin footnote.
\footnote{The Picard group $\picardGroup X_{K}$ is a commutative group, or equivalently, a $\integers$-module. If $\picardGroup X_{K}$ is free and finitely generated, then $\picardGroup X_{K} \isomorphism \integers^{n}$ for some $n \in \integersPositive$. In this case, $\picardGroup X_{K} \tensor[\integers] \rationals$ is a $\rationals$-vector space of dimension $n$, and we may think of $\rho_{K}(\sigma)$ as a linear transformation on $\rationals^{n}$. Hence our use of the terms ``eigenspace'' and ``eigenvalue'' here.} % End footnote.
Denote this eigenspace by $E(\rho_{K}(\sigma),1)$. Thus, for a projective $k$-surface, and for any Galois extension $K \fieldExtension k$,
\begin{align*}
\bigcap_{\sigma \in \galoisGroup(K \fieldExtension k)} E(\rho_{K}(\sigma),1)
\neq
\{0\}
\end{align*}

A second feature of surfaces that facilitates their study is the classification of projective surfaces by their Kodaira dimension, an integer $\kodairaDimension(X)$ in $\{-1,0,1,2\}$.% Begin footnote.
\footnote{More generally, let $X$ be a projective variety of dimension $d$. Then $\kodairaDimension(X) \in \{-1,\ldots,d\}$. Moreover, every integer in this set is the Kodaira dimension of some variety of dimension $d$. For further discussion, see \cite{Hartshorne1977}*{Section V.6}.} % End footnote.
Del Pezzo surfaces have Kodaira dimension $-1$. Algebraic K3 surfaces have Kodaira dimension $0$.



\subsection{Specific constructions: Algebraic K3 surfaces}

A nice $k$-surface $X$ in $\projective_{k}^{n}$ is called a \fontDefWord{smooth complete intersection} if there exist $n - 2$ homogeneous polynomials $f_{i}$ such that $X = \cap_{i = 1}^{n - 2} \zeroSet(f_{i})$. Algebraic K3 surfaces include three such constructions (in each case, the intersection defining $X$ must be smooth):
\begin{enumerate}
\item Let $f \in k[x_{0},\ldots,x_{3}]$ have degree $4$. Then $X = \zeroSet(f) \subseteq \projective^{3}_{k}$ is an algebraic K3 surface, of degree $4$.
\item Let $f_{1},f_{2} \in k[x_{0},\ldots,x_{4}]$ have degrees $2$ and $3$, respectively. Then $X = \zeroSet(f_{1}) \cap \zeroSet(f_{2}) \subseteq \projective^{4}_{k}$ is an algebraic K3 surface, of degree $6$.
\item Let $f_{1},f_{2},f_{3} \in k[x_{0},\ldots,x_{5}]$ each have degree $2$. Then $X = \cap_{i = 1}^{3} \zeroSet(f_{i}) \subseteq \projective^{5}_{k}$ is an algebraic K3 surface, of degree $8$.
\end{enumerate}

Another accessible construction is a double cover of $\projective^{2}_{k}$ branched along a smooth curve $C = \zeroSet(f_{6}) \subseteq \projective^{2}_{k}$, where $f_{6} \in k[x_{0},\ldots,x_{2}]$ is a nonsingular, homogeneous polynomial of degree $6$. Then $X = \zeroSet(x_{3}^{2} - f_{6})$ in the weighted projective space $\projective^{3}_{k,[1,1,1,3]}$ is an algebraic K3 surface, of degree $2$.



\subsection{Specific constructions: Del Pezzo surfaces}

Del Pezzo surfaces are also accessible, via the procedure of blowing up a variety at a point. If $X$ is a del Pezzo surface over $k$ of degree $d$, then either $d = 8$ and $X$ is isomorphic to $\projective^{1}_{k} \times \projective_{k}^{1}$, or $X$ is isomorphic to the blow-up of $\projective_{k}^{2}$ at $9 - d$ points in general position.% Begin footnote.
\footnote{A set of points in $\projective_{k}^{2}(k)$ is in \fontDefWord{general position} if no three of the points lie on a line, no six of the points lie on a conic, and no eight of the points lie on singular cubic with one of the points at the singularity.} % End footnote.
In the latter case, the theory of monoidal transformations---that is, the operation of blowing up at a single point---implies that $\picardGroup X_{\separableClosure{k}} \isomorphism \integers^{10 - d}$. If $d \leq 7$, then the geometric Picard group $\picardGroup X_{\separableClosure{k}}$ is generated by the classes of the exceptional curves on $X$. The number of exceptional curves in each degree is known. Moreover, for certain embeddings of del Pezzo surfaces into projective space, the general shape of these exceptional curves is also known. In particular, given a del Pezzo surface of degree $1$ over a field $k$ of characteristic $\characteristic k \neq 2,3$, we can specify such an embedding (see Section \ref{sec : dP1 : Blow Ups And Picard Groups}).



\section{Main results}
\label{sec : Main Results}

In Chapter \ref{sec : GLnZ} we study finite subgroups of $\generalLinear(n,\integers)$, the group of $n \times n$ invertible matrices with integer entries, and their cohomology. Our ultimate goal is to use these results to study the geometry of varieties, such as del Pezzo surfaces and K3 surfaces, whose geometric Picard groups are free and of finite rank. Let $X$ be such a variety, defined over $k$, of geometric Picard rank $n$. Choosing a basis for the geometric Picard group establishes isomorphisms $\picardGroup X_{\separableClosure{k}} \isomorphism \integers^{n}$ and $\automorphismGroup(\picardGroup X_{\separableClosure{k}}) \isomorphism \generalLinear(n,\integers)$. Using the latter isomorphism, we may extend a representation \eqref{eq : Intro : Galois Representation On Picard Group} of the absolute Galois group of $k$ on the geometric Picard group of $X$ to a matrix representation
\begin{align*}
\rho_{\separableClosure{k}}
:
\galoisGroup(\separableClosure{k} \fieldExtension k)
\rightarrow
\generalLinear(n,\integers)
\end{align*}
which identifies the absolute Galois group with a subgroup $G = \image \rho_{\separableClosure{k}}$ of $\generalLinear(n,\integers)$. Thus a choice of basis of $\picardGroup X_{\separableClosure{k}}$ allows us to make an identification
\begin{align*}
\cohomology[1]{\galoisGroup(\separableClosure{k} \fieldExtension k),\picardGroup X_{\separableClosure{k}}}
\isomorphism
\cohomology[1]{G,\integers^{n}}
\end{align*}
In particular, by \eqref{eq : Intro : Br1/Br0 Galois Cohomology}, the algebraic Brauer group of $X$ modulo constants is isomorphic to $\cohomology[1]{G,\integers^{n}}$ for some subgroup $G \subgroupeq \generalLinear(n,\integers)$. In fact, this subgroup $G$ is finite.% Begin footnote.
\footnote{If $\picardGroup X_{\separableClosure{k}}$ is finitely generated---implied by our hypotheses that $\picardGroup X_{\separableClosure{k}}$ is free and of finite rank---then there exists a finite Galois extension $K \fieldExtension k$ such that all classes of $\picardGroup X_{\separableClosure{k}}$ are defined over $K$, and hence $\image \rho_{K} = \image \rho_{\separableClosure{k}}$. Moreover, the representation $\rho_{K} : \galoisGroup(K \fieldExtension k) \rightarrow \automorphismGroup(\picardGroup X_{K}) \isomorphism \generalLinear(n,\integers)$ is an isomorphism onto its image. Thus $G = \image \rho_{\separableClosure{k}} = \image \rho_{K} \isomorphism \galoisGroup(K \fieldExtension k)$ is finite.}% End footnote.

We implement an inductive algorithm presented by Brown et al. \cite{Brown1972i} to construct a representative from each $\integers$-conjugacy class of \emph{reducible} finite subgroup of $\generalLinear(n,\integers)$ for ranks $n \in \{1,\ldots,6\}$. We combine this with an analysis of the subgroups of maximal \emph{irreducible} finite subgroups of $\generalLinear(n,\integers)$ to enumerate a list of representatives from each $\integers$-conjugacy class of finite subgroup of $\generalLinear(n,\integers)$. For each class, we compute the associated first cohomology group.

The isomorphism in \eqref{eq : Intro : Br1/Br0 Galois Cohomology} relates the first cohomology group to the algebraic Brauer group modulo constants for varieties whose geometric Picard group is free of rank $n$. Recall (see Section \ref{sec : Intro : General Features Of Surfaces}) that the geometric Picard group of a projective variety has a nontrivial $\galoisGroup(\separableClosure{k} \fieldExtension k)$-subspace associated to the eigenvalue $1$. We identify the representative subgroups of $\generalLinear(n,\integers)$ that satisfy this condition, and we enumerate their first cohomology groups.% These are presented in Table \ref{tab : K3 : Candidate Algebraic Brauer Groups} on page \pageref{tab : K3 : Candidate Algebraic Brauer Groups}.

We summarize these results in the following theorem.
\begin{table}[t]
\centering
\begin{tabular}{c l *{9}{r}}
\hline\hline
Rank	&	\multicolumn{9}{c}{Invariant factors* of new first-cohomology groups}	\\
	&	\multicolumn{9}{c}{$[d_{1}]\cdots[d_{z}] \leftrightsquigarrow (\integers/d_{1} \integers) \oplus \ldots \oplus (\integers/d_{z} \integers)$}	\\
\hline
$0$	&	$[\;]$	\\
$1$	&	$[2]$	\\
$2$	&	$[2]^{2}$	&	$[3]$	\\
$3$	&	$[2]^{3}$	&	$[2][3]$		&	$[4]$	\\
$4$	&	$[2]^{4}$	&	$[2]^{2}[3]$	&	$[2][4]$		&	$[3]^{2}$		&	$[5]$	\\
$5$	&	$[2]^{5}$	&	$[2]^{3}[3]$	&	$[2]^{2}[4]$	&	$[2][3]^{2}$	&	$[2][5]$	&	$[3][4]$	\\
$6$	&	$[2]^{6}$	&	$[2]^{4}[3]$	&	$[2]^{3}[4]$	&	$[2]^{2}[3]^{2}$	&	$[2]^{2}[5]$	&	$[2][3][4]$	&	$[3]^{3}$	&	$[3][5]$	&	$[4]^{2}$	&	$[7]$	\\
\hline
\end{tabular}
\caption{First cohomology groups $\cohomology[1]{G,\integers^{n}}$ arising from all finite subgroups $G \subgroupeq \generalLinear(n,\integers)$ of rank $n \in \{1,\ldots,6\}$. (* The invariant factors may be deduced from the $d_{i}$. For example, $[2]^{2} [3]$ corresponds to the invariant factors $2 \divides 6$, and so forth.)}
\label{tab : Intro : First Cohomology Groups}
\end{table}



\begin{theorem}
Let $n \in \{1,\ldots,6\}$, let $G$ be a finite subgroup of $\generalLinear(n,\integers)$, and view $\integers^{n}$ as a left $G$-module with the natural action of $G$. Then $\cohomology[1]{G,\integers^{n}}$ is isomorphic to one of the groups in Table \ref{tab : Intro : First Cohomology Groups}, in the row corresponding to rank $n$ or smaller. If we impose the condition that $G$ must have a nontrivial eigenspace associated to the eigenvalue $1$, then $\cohomology[1]{G,\integers^{n}}$ is isomorphic to one of the groups in the same table, but in the row corresponding to rank $n - 1$ or smaller.
\end{theorem}

The isomorphism in \eqref{eq : Intro : Br1/Br0 Galois Cohomology} gives an immediate corollary.

\begin{corollary}
Let $k$ be a field, and let $X$ be a nice $k$-variety whose geometric Picard group is free and of finite rank $n \in \{1,\ldots,6\}$. Then the algebraic Brauer group of $X$ modulo constants is isomorphic to one of the groups in Table \ref{tab : Intro : First Cohomology Groups}, in the row corresponding to rank $n - 1$ or smaller.
\end{corollary}

In Chapter \ref{sec : K3} we study the inverse Galois problem for algebraic K3 surfaces over $\rationals$, using the candidate algebraic Brauer groups modulo constants for projective varieties and corresponding representative finite subgroups in $\generalLinear(n,\integers)$. We construct an algebraic K3 surface $X$ with geometric Picard rank $2$ such that $\brauerGroupAlgebraic X/\brauerGroupConstant X \isomorphism \integers/2 \integers$. Note that in geometric Picard rank $2$, $\integers/2 \integers$ is the only nontrivial algebraic Brauer group modulo constants. We present these results in the following theorem.

\begin{theorem}
Let $g_{1},g_{2},q_{1},q_{2} \in \rationals[x_{0},\ldots,x_{3}]$ be the homogeneous polynomials
\begin{align*}
g_{1}
&=
-x_{0}^{2} + x_{1} x_{3} + x_{2}^{2} + x_{2} x_{3} + x_{3}^{2}
\\
g_{2}
&=
x_{1}^{3} x_{3} + x_{1}^{2} x_{2}^{2} + x_{1}^{2} x_{2} x_{3} + x_{1}^{2} x_{3}^{2} + x_{2}^{2} x_{3}^{2}
\\
q_{1}
&=
2 x_{0}^{2} + 12 x_{0} x_{1} + 4 x_{0} x_{2} + 10 x_{0} x_{3} + 7 x_{1}^{2} + 19 x_{1} x_{2} + 7 x_{1} x_{3} + 3 x_{2} x_{3} + 12 x_{3}^{2}
\\
q_{2}
&=
8
\end{align*}
and let
\begin{align*}
X
=
\zeroSet(q_{1} g_{1} + q_{2} g_{2})
\end{align*}
Then $X$ is an algebraic K3 surface over $\rationals$ with geometric Picard rank $2$ such that
\begin{align*}
\brauerGroupAlgebraic X/\brauerGroupConstant X
\isomorphism
\integers/2 \integers.
\end{align*}
\end{theorem}

In Chapter \ref{sec : dP1} we study the inverse Galois problem for del Pezzo surfaces of degree $1$. In particular, we search for a del Pezzo surface over $\rationals$ of degree $1$ such that $\brauerGroup X/\brauerGroupConstant X$ has $5$-torsion.% Begin footnote.
\footnote{For a del Pezzo surface $X$, its algebraic Brauer group is the full Brauer group, and therefore $\brauerGroupAlgebraic X/\brauerGroupConstant X = \brauerGroup X/\brauerGroupConstant X$.} % End footnote.
Work of Corn \cite{Corn2005}*{Theorem 1.4.1} implies that this is equivalent to the inverse Galois problem for del Pezzo surfaces of degree $1$ with $k = \rationals$ and the target group $G$ isomorphic to $\integers/5 \integers$ or $(\integers/5 \integers)^{2}$. We analyze representative surfaces from three types of del Pezzo surfaces of degree $1$, as classified by Dolgachev \cite{Dolgachev2012}. For each surface, we construct an explicit splitting field for the geometric Picard group. For two of the three families, we compute the image of the representation of the absolute Galois group on the geometric Picard group, and the corresponding Brauer group modulo constants. We present the results in the following theorem.
\begin{table}[t]
\centering
\begin{tabular}{*{5}{c}|*{2}{c}}
\hline\hline
	&				&							&					&										&	\multicolumn{2}{c}{$\exists L : \brauerGroup X_{L}/\brauerGroupConstant X_{L} \isomorphism G$?}	\\
Type	&	$f_{4}$		&	$f_{6}$					&	$[K : \rationals]$	&	$\brauerGroup X/\brauerGroupConstant X$	&	$G = \integers/5 \integers$	&	$G = (\integers/5 \integers)^{2}$	\\
\hline
4	&	$0$			&	$x_{0} (x_{0}^{5} + x_{1}^{5})$	&	$240$			&	$0$	&	Yes	&	Yes	\\
8	&	$x_{0}^{4}$	&	$x_{0} x_{1}^{5}$			&	$160$			&	$0$	&	Yes	&	Yes	\\
13	&	$x_{0}^{4}$	&	$x_{0} (x_{0}^{5} + x_{1}^{5})$	&	$2400$			&		&		&		\\
\hline
\end{tabular}
\caption{Attributes of representative del Pezzo surfaces $X$ of degree $1$ from Dolgachev's classification, types 4, 8, and 13. The column $[K : \rationals]$ presents the degree of the splitting field $K$ of $X$ as an extension of $\rationals$. The final two columns answer the question, Does there exist an intermediate field $L$ in the extension $K \fieldExtension \rationals$ such that the del Pezzo surface $X$, when viewed as defined over $L$, has Brauer group modulo constants isomorphic to the given group $G$?}
\label{tab : Intro : dP1 Results}
\end{table}

\begin{theorem}
Let $f_{4},f_{6} \in \rationals[x_{0},x_{1}]$ be homogeneous polynomials of degree $4$ and $6$, respectively; let $f \in \rationals[x_{0},\ldots,x_{3}]$ be given by
\begin{align*}
f
=
x_{3}^{2} - x_{2}^{3} - f_{4}(x_{0},x_{1}) x_{2} - f_{6}(x_{0},x_{1})
\end{align*}
and let $X = \zeroSet(f) \subseteq \projective^{3}_{\rationals,[1,1,2,3]}$ be the associated del Pezzo surface of degree $1$. Then for the choices of $f_{4}$ and $f_{6}$ shown in Table \ref{tab : Intro : dP1 Results}, the corresponding del Pezzo surface $X$ has the attributes summarized in the same table. Specifically,
\begin{itemize}
\item The type-4 degree-$1$ del Pezzo surface $X = \zeroSet(f)$, where
\begin{align*}
f
=
x_{3}^{2} - x_{2}^{2} - x_{0} (x_{0}^{5} + x_{1}^{5})
\end{align*}
has splitting field $K = \rationals(\theta_{0})\cdots(\theta_{5})$ of degree $240$ over $\rationals$, where for each $i$, $\theta_{i}$ is a root of the polynomial $p_{i}$, given below:
\begin{align*}
p_{0}
&=
t^{2} - 5
\\
p_{1}
&=
t^{2} - (78 - 174 \theta_{0}^{-1})
\\
p_{2}
&=
t^{2} - (3 \theta_{1} - (18 + 9 \theta_{0}))
\\
p_{3}
&=
t^{3} - (3 \theta_{1} - (19 + 9 \theta_{0}))
\\
p_{4}
&=
t^{2} + \left(\frac{1}{2} (39 \theta_{0} - 87) \theta_{1} - 9 \theta_{0} + 18\right)
\\
p_{5}
&=
t^{5} + \left(\frac{1}{36} (-23 \theta_{0} + 44) \theta_{1} + \frac{1}{60} (49 \theta_{0} + 45) \theta_{2}\right)
\end{align*}
If we view $X$ as a surface over $\rationals$, then $\brauerGroup X/\brauerGroupConstant X \isomorphism 0$. There exist intermediate fields $L_{1},L_{2}$ such that, if we view $X$ as a surface over $L_{i}$, then $\brauerGroup X_{L_{i}}/\brauerGroupConstant X_{L_{i}} \isomorphism (\integers/5 \integers)^{i}$.

\item The type-8 degree-$1$ del Pezzo surface $X = \zeroSet(f)$, where
\begin{align*}
f
=
x_{3}^{2} - x_{2}^{3} - x_{0}^{4} x_{2} - x_{0} x_{1}^{5}
\end{align*}
has splitting field $K = \rationals(\theta_{0})\cdots(\theta_{3})$ of degree $160$ over $\rationals$, where for each $i$, $\theta_{i}$ is a root of the polynomial $p_{i}$, given below:
\begin{align*}
p_{0}
&=
t^{2} + 1
\\
p_{1}
&=
t^{4} - (-1 - 2 \theta_{0})
\\
p_{2}
&=
t^{4} - (-1 + 2 \theta_{0})
\\
p_{3}
&=
t^{5} + \left(\frac{1}{125} (48 \theta_{0} - 14) \theta_{1}^{2} \theta_{2}^{2} + \frac{1}{5} (4 \theta_{0} - 2)\right)
\end{align*}
If we view $X$ as a surface over $\rationals$, then $\brauerGroup X/\brauerGroupConstant X \isomorphism 0$. There exist intermediate fields $L_{1},L_{2}$ such that, if we view $X$ as a surface over $L_{i}$, then $\brauerGroup X_{L_{i}}/\brauerGroupConstant X_{L_{i}} \isomorphism (\integers/5 \integers)^{i}$.

\item The type-13 degree-$1$ del Pezzo surface $X = \zeroSet(f)$, where
\begin{align*}
f
=
x_{3}^{2} - x_{2}^{3} - x_{0}^{4} x_{2} - x_{0} (x_{0}^{5} + x_{1}^{5})
\end{align*}
has splitting field $K = \rationals(\theta_{0})\cdots(\theta_{4})$ of degree $2400$ over $\rationals$.
\end{itemize}
\end{theorem}

We remark that there exist subgroups of the Weyl group of $E_{8}$ of order $2400$ whose first cohomology group has $5$-torsion.