%
%
%   Chapter 4 : The inverse Galois problem for del Pezzo surfaces of degree 1
%
%

\chapter{The inverse {G}alois problem for del {P}ezzo surfaces of degree 1}
\label{sec : dP1}

In this chapter we investigate the inverse Galois problem for del Pezzo surfaces of degree $1$. We first recall key definitions and geometric features of del Pezzo surfaces. We then restrict the inverse Brauer problem to del Pezzo surfaces of degree $1$ whose Brauer group modulo constants contains $5$-torsion. Corn \cite{Corn2005} showed that the only such groups are $\integers/5 \integers$ and $(\integers/5 \integers)^{2}$. Using Dolgachev's classification of del Pezzo surfaces of degree $1$ by their automorphism groups \cite{Dolgachev2012}*{Table 8.14}, we focus on three families---Dolgachev's types 4, 8, and 13---and analyze a representative from each. For the representatives of types 4 and 8, we show that their Brauer groups modulo constants are trivial when considered as surfaces over $\rationals$, but become $\integers/5 \integers$ and $(\integers/5 \integers)^{2}$ when the base field is raised to certain Galois extensions of $\rationals$. We present partial results for the representative of type 13.



\section{Definitions and geometry}
\label{sec : dP1 : Definitions And Geometry}

In this section we present del Pezzo surfaces as blow-ups of the projective plane. After outlining the general framework of monoidal transformations, we review definitions and properties of del Pezzo surfaces relevant to our present work. For more extensive treatment, see \cite{Hartshorne1977}*{Sections I.4, II.7, V.3} for blow-ups and \cite{VarillyAlvarado2013}*{Section 1} for del Pezzo surfaces.



\subsection{Monoidal transformations}

Let $k$ be a field, and let $X,Y$ be $k$-varieties. Given a nonempty open subset $U \subseteq X$, we may consider all morphisms $\varphi_{U} : U \rightarrow Y$. We may define an equivalence relation on the collection of all such morphisms: For $i \in \{1,2\}$, let $U_{i}$ be a nonempty open subset of $X$, and let $\varphi_{U_{i}} : U_{i} \rightarrow Y$ be a morphism. We define
\begin{align}
\varphi_{U_{1}}
&\sim
\varphi_{U_{2}}
&
&\Leftrightarrow
&
\varphi_{U_{1}}|_{U_{1} \cap U_{2}}
&=
\varphi_{U_{2}}|_{U_{1} \cap U_{2}}%
\label{eq : Rational Map : Equivalence Relation}
\end{align}
A \fontDefWord{rational map} from $X$ to $Y$, denoted $\varphi : X \dashrightarrow Y$, is an equivalence class of morphisms $\varphi_{U}$ under the equivalence relation \eqref{eq : Rational Map : Equivalence Relation}. The rational map $\varphi$ is \fontDefWord{dominant} if there exists a representative $\varphi_{U}$ of $\varphi$ such that $\image \varphi_{U}$ is dense in $Y$.% Begin footnote.
\footnote{If such a representative $\varphi_{U}$ exists, then every representative of $\varphi$ satisfies this property.} % End footnote.
The rational map $\varphi$ is \fontDefWord{birational} if there exists a rational map $\psi : Y \dashrightarrow X$ such that $\psi \circ \varphi = \identityMap_{X}$ and $\varphi \circ \psi = \identityMap_{Y}$ as rational maps. In this case we say that $X$ and $Y$ are \fontDefWord{birationally equivalent} or simply \fontDefWord{birational}.

\begin{remark}
Birational maps are the isomorphisms in the category whose objects are $k$-varieties and whose morphisms are dominant rational maps.
\end{remark}

First we define the blow-up of $\affine_{k}^{n}$ at the origin $O = (0,\ldots,0)$. Consider the quasi-projective $k$-variety $\affine_{k}^{n} \times \projective_{k}^{n - 1}$. Denote the coordinates $((x_{1},\ldots,x_{n}),(y_{1} : \ldots : y_{n}))$. The \fontDefWord{blow-up of $\affine_{k}^{n}$ at $O$} is the closed subset
\begin{align*}
\blowUp_{O}(\affine_{k}^{n})
=
\zeroSet\left(\left\{x_{i} y_{j} - x_{j} y_{i} \st i,j \in \{1,\ldots,n\}\right\}\right)
\end{align*}
of $\affine_{k}^{n} \times \projective_{k}^{n - 1}$. Consider the natural morphism $\varphi : \blowUp_{O}(\affine_{k}^{n}) \rightarrow \affine_{k}^{n}$ given by restricting the projection map $\affine_{k}^{n} \times \projective_{k}^{n - 1} \rightarrow \affine_{k}^{n}$ (that is, projection onto the first factor) to $\blowUp_{O}(\affine_{k}^{n})$.
\begin{enumerate}
\item\label{itm : Blow Ups : Away From Center} For each point $P = (a_{1},\ldots,a_{n}) \in \affine_{k}^{n}$ such that $P \neq O$,
\begin{align*}
\varphi^{-1}(P)
=
\left\{((a_{1},\ldots,a_{n}),(a_{1} : \ldots : a_{n}))\right\}
\end{align*}
Thus
\begin{align*}
\blowUp_{O}(\affine_{k}^{n}) - \varphi^{-1}(O)
\isomorphism
\affine_{k}^{n} - \{O\}
\end{align*}

\item\label{itm : Blow Ups : At Center} For the point $O \in \affine_{k}^{n}$,
\begin{align*}
\varphi^{-1}(O)
=
\left\{((0,\ldots,0),(y_{1} : \ldots : y_{n})) \st y_{i} \in k,\exists i_{0} : y_{i_{0}} \neq 0\right\}
\end{align*}
Thus
\begin{align*}
\varphi^{-1}(O)
\isomorphism
\projective_{k}^{n - 1}
\end{align*}
One can view the projective coordinates $(y_{1} : \ldots : y_{n}) \in \projective_{k}^{n - 1}$ as representing all possible slopes of lines through a given point in $\affine_{k}^{n}$ (specifically, the chosen point $O$).

\item\label{itm : Blow Ups : Irreducible} $\blowUp_{O}(\affine_{k}^{n})$ is irreducible.
\end{enumerate}
$\varphi^{-1}(O)$ is called the \fontDefWord{exceptional curve} of the blow-up $\blowUp_{O}(\affine_{k}^{n})$.

Next we generalize this construction to closed subvarieties of $\affine_{k}^{n}$. Let $X \subseteq \affine_{k}^{n}$ be a closed subvariety containing the origin $O$, and define $\varphi : \blowUp_{O}(\affine_{k}^{n}) \rightarrow \affine_{k}^{n}$ as above. The \fontDefWord{blow-up of $X$ at $O$}, denoted $\blowUp_{O}(X)$, is
\begin{align*}
\blowUp_{O}(X)
=
\topologicalClosure{\varphi^{-1}(X - O)}
\end{align*}
where the overline denotes topological closure (using the Zariski topology). Let $P \in X$ be any point. The \fontDefWord{blow-up of $X$ at $P$}, denoted $\blowUp_{P}(X)$, is given by first making a linear change of coordinates that sends $P$ to $O$, then forming $\blowUp_{O}(X)$. As previously, $\varphi^{-1}(P) \subseteq \blowUp_{P}(\affine_{k}^{n})$ is called the \fontDefWord{exceptional curve} associated to the blow-up.

\begin{remark}
By property \ref{itm : Blow Ups : Away From Center} of blow ups, the projection morphism $\varphi$ induces an isomorphism
\begin{align*}
\blowUp_{P}(X) - \varphi^{-1}(P)
\isomorphism
X - \{P\}
\end{align*}
Because $\varphi^{-1}(P) \subseteq \blowUp_{P}(X)$ and $\{P\} \subseteq X$ are both closed subsets, it follows that $\blowUp_{P}(X)$ and $X$ are birationally equivalent.
\end{remark}

The procedure of blowing up a variety $X$ at a single point $P$ is called a \fontDefWord{monoidal transformation} of $X$ with center $P$. The procedure of blowing up can be generalized to blowing up a scheme at a closed subscheme (see \cite{Hartshorne1977}*{Section II.7}).



\subsection{Blow-ups and {P}icard groups}
\label{sec : dP1 : Blow Ups And Picard Groups}

Let $k$ be a field, let $X$ be a nice $k$-surface, let $(-,-)_{X}$ denote the intersection pairing on $X$, and let $P \in X$. In this setting one can show that $\widetilde{X} = \blowUp_{P}(X)$ is again a nonsingular projective $k$-surface, and the exceptional curve $E = \varphi^{-1}(P)$ satisfies $E \isomorphism \projective_{k}^{1}$ and $(E,E)_{\widetilde{X}} = -1$. Conversely, let $\widetilde{X}$ be a projective $k$-surface with intersection pairing $(-,-)_{\widetilde{X}}$, and let $E \subseteq \widetilde{X}$ be a curve such that $E \isomorphism \projective_{k}^{1}$ and $(E,E)_{\widetilde{X}} = -1$. Then there exists a projective $k$-surface $X$ and a point $P \in X$ such that $\widetilde{X} \isomorphism \blowUp_{P}(X)$ and $E = \varphi^{-1}(P)$ is the corresponding exceptional curve.

Let $\widetilde{X} = \blowUp_{P} X$, let $\pi : \widetilde{X} \dashrightarrow X$ be the corresponding rational map, and let $E = \varphi^{-1}(P) \subseteq \widetilde{X}$ be the corresponding exceptional curve. The natural maps
\begin{align*}
\picardGroup \projective_{k}^{1}
\isomorphism
\integers
&\rightarrow
\picardGroup \widetilde{X}
&
\pi^{*}
:
\picardGroup X
&\rightarrow
\picardGroup \widetilde{X}
\\
1
&\mapsto
1 \cdot E
&
&
\end{align*}
yield an isomorphism
\begin{align}
\picardGroup \widetilde{X}
\isomorphism
\picardGroup X \directSum \integers%
\label{eq : Blow Ups : Isomorphism Of Picard Groups}
\end{align}
Moreover, the intersection pairing on $\widetilde{X}$ is determined by the following properties:
\begin{enumerate}
\item\label{itm : dP : Intersection Pairing : Up} Let $D_{1},D_{2} \in \picardGroup X$. Then
\begin{align*}
(\pi^{*} D_{1},\pi^{*} D_{2})_{\widetilde{X}}
=
(D_{1},D_{2})_{X}
\end{align*}

\item\label{itm : dP : Intersection Pairing : Up EC} Let $D \in \picardGroup X$. Then
\begin{align*}
(\pi^{*} D,E)_{\widetilde{X}}
=
0
\end{align*}

\item\label{itm : dP : Intersection Pairing : EC Self Intersection} The exceptional curve $E$ on $\widetilde{X}$ satisfies
\begin{align*}
(E,E)_{\widetilde{X}}
=
-1
\end{align*}

\item\label{itm : dP : Intersection Pairing : Up Down} Let $\pi_{*} : \picardGroup \widetilde{X} \rightarrow \picardGroup X$ denote projection onto the first factor in the isomorphism in \eqref{eq : Blow Ups : Isomorphism Of Picard Groups}, let $D \in \picardGroup X$, and let $\widetilde{D} \in \picardGroup \widetilde{X}$. Then
\begin{align*}
(\pi^{*} D,\widetilde{D})_{\widetilde{X}}
=
(D,\pi_{*} \widetilde{D})_{X}
\end{align*}
\end{enumerate}
Let $K_{X}$ denote the canonical divisor of $X$. The canonical divisor of $\widetilde{X}$ is
\begin{align*}
K_{\widetilde{X}}
=
\pi^{*} K_{X} + E
\end{align*}
Applying properties \ref{itm : dP : Intersection Pairing : Up}--\ref{itm : dP : Intersection Pairing : Up Down} above, we compute
\begin{align*}
(K_{\widetilde{X}},K_{\widetilde{X}})_{\widetilde{X}}
=
(K_{X},K_{X})_{X} - 1
\end{align*}

For certain embeddings of del Pezzo surfaces into projective space, the general shape of its exceptional curves is known. For example, a del Pezzo surface over $k$ of degree $1$ can be embedded as a smooth sextic hypersurface $X = \zeroSet(f)$ in the weighted projective space $\projective_{k,[1,1,2,3]}^{3}$. If $\characteristic k \neq 2,3$, then we may apply a change of variables to put the homogeneous degree-$6$ polynomial $f$ in the form
\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*}
where $f_{4},f_{6} \in k[x_{0},x_{1}]$ are homogeneous polynomials of degrees $4$ and $6$, respectively. In this setting, each exceptional curve $E$ on $X$ has the form
\begin{align}
E
=
\zeroSet(x_{2} - Q(x_{0},x_{1}),x_{3} - C(x_{0},x_{1}))%
\label{eq : dP1 : General Form Of EC On dP1}
\end{align}
where $Q,C \in \separableClosure{k}[x_{0},x_{1}]$ are homogeneous polynomials of degrees $2$ and $3$, respectively. More precisely, V\'{a}rilly-Alvarado \cite{VarillyAlvarado2009} proved the following result.

\begin{theorem}[\cite{VarillyAlvarado2009}*{Theorem 1.5.9}]
Let $X$ be a del Pezzo surface of degree $1$. If $E$ as defined in \eqref{eq : dP1 : General Form Of EC On dP1} is a divisor on $X$, then $E$ is an exceptional curve. Conversely, if $E$ is an exceptional curve on $X$, then $E$ has the form \eqref{eq : dP1 : General Form Of EC On dP1}.
\end{theorem}

Another, classical example in which the general shape of the exceptional curves is known is the embedding of a del Pezzo surface over $k$ of degree $3$ as a smooth cubic surface in $\projective_{k}^{3}$. In this case, the $27$ exceptional curves on $X$ are exactly the $27$ lines on the cubic surface.



\subsection{Del {P}ezzo surfaces as blow-ups}
\label{sec : dP1 : Del Pezzo Surfaces As Blow Ups}

Let $k$ be a field, and let $X$ be a del Pezzo surface of degree $d$ over $k$. Assume that $X \notIsomorphism \projective_{k}^{1} \times \projective_{k}^{1}$.% Begin footnote.
\footnote{Only in degree $d = 8$ does the isomorphism we exclude possibly hold for $X$.} % End footnote.
Then $X$ is isomorphic to a blow-up of $\projective_{k}^{2}$ at $r = 9 - d$ closed points $P_{1},\ldots,P_{r}$ of $\projective_{k}^{2}(k)$ 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.
Combining the fact that $\picardGroup \projective_{k}^{2} \isomorphism \integers$ with \eqref{eq : Blow Ups : Isomorphism Of Picard Groups}, we obtain
\begin{align*}
\picardGroup X
\isomorphism
\integers^{1 + r}
=
\integers^{10 - d}
\end{align*}
Let $\pi : X \dashrightarrow \projective_{k}^{2}$ be the composition of the $r$ rational maps, one associated to each successive blow-up.

The exceptional curves $E$ on $X$ are characterized by the conditions $(E,E) = -1$ and $(E,K_{X_{\separableClosure{k}}}) = -1$. For del Pezzo surfaces $X$ of degree $d \leq 7$, $\picardGroup X_{\separableClosure{k}}$ is generated by the exceptional curves. We list the number of exceptional curves on $X$ in Table \ref{tab : dP1 : Number Of Exceptional Curves On Del Pezzo}.
\begin{table}[t]
\centering
\begin{tabular}{l *{7}{c}}
\hline\hline
Degree of del Pezzo surface	&	1	&	2	&	3	&	4	&	5	&	6	&	7	\\
Number of exceptional curves	&	240	&	56	&	27	&	16	&	10	&	6	&	3	\\
\hline
\end{tabular}
\caption{The number of exceptional curves on a del Pezzo surface of degree $1,\ldots,7$.}
\label{tab : dP1 : Number Of Exceptional Curves On Del Pezzo}
\end{table}
We can deduce these numbers as follows. For each $i \in \{1,\ldots,r\}$, let $E_{i} = \pi^{-1}(P_{i})$ denote the exceptional curve associated to the point $P_{i} \in \projective_{k}^{2}(k)$, let $\ell \subseteq \projective_{k}^{2}$ be a line not passing through any of the points $P_{i}$, and let $L = \pi^{-1}(\ell)$. Then $L,E_{1},\ldots,E_{r}$ is a basis of $\picardGroup X_{\separableClosure{k}}$. Using properties \ref{itm : dP : Intersection Pairing : Up}--\ref{itm : dP : Intersection Pairing : Up Down} of the intersection pairing on $X$, we compute
\begin{align}
(E_{i},E_{j})_{X}
&=
\begin{dcases*}
-1	&	if $i = j$	\\
0	&	if $i \neq j$
\end{dcases*}
&
(E_{i},L)_{X}
&=
0
&
(L,L)_{X}
=
1%
\label{eq : dP : Intersections For Picard Group Basis}
\end{align}
The canonical class $K_{X}$ of $X$ is given by
\begin{align*}
K_{X}
=
-3 L + \sum_{i = 1}^{r} E_{i}
\end{align*}
Let $E \in \picardGroup X_{\separableClosure{k}}$. Because $L,E_{1},\ldots,E_{r}$ is a basis of $\picardGroup X_{\separableClosure{k}}$, there exist $a_{i} \in \integers$ such that
\begin{align}
E
=
a_{0} L - \sum_{i = 1}^{r} a_{i} E_{i}%
\label{eq : dP : General EC In Terms Of Basis}
\end{align}
Using \eqref{eq : dP : General EC In Terms Of Basis}, bilinearity of the intersection pairing, and the intersection numbers in \eqref{eq : dP : Intersections For Picard Group Basis}, we compute
\begin{align*}
-1
&=
(E,E)
=
a_{0}^{2} - \sum_{i = 1}^{r} a_{i}^{2}
\\
-1
&=
(E,K_{X})
=
-3 a_{0} + \sum_{i = 1}^{r} a_{i}
\end{align*}
The number of integer solutions to this system of equations depends on $d = 9 - r$ as shown in Table \ref{tab : dP1 : Number Of Exceptional Curves On Del Pezzo}.

\begin{remark}
Because the Galois action on the Picard group preserves the intersection pairing, it follows that the Galois action sends exceptional curves to exceptional curves. (See \eqref{eq : Intro : Galois Action Preserves Intersection Pairing}, taking $D_{1} = D_{2} = E$ for any exceptional curve $E$.)
\end{remark}

\begin{remark}
Let $d \leq 7$, so that the exceptional curves on $X$ generate $\picardGroup X_{\separableClosure{k}}$, and let $r = 9 - d$ as before. The intersection numbers in \eqref{eq : dP : Intersections For Picard Group Basis} imply that the intersection matrix for the basis $L,E_{1},\ldots,E_{r}$ is
\begin{align*}
\begin{bmatrix}
1	&		&			&		\\
	&	-1	&			&		\\
	&		&	\ddots	&		\\
	&		&			&	-1	\\
\end{bmatrix}
\end{align*}
Because the exceptional curves generate $\picardGroup X_{\separableClosure{k}}$, there exists an exceptional curve $E_{r + 1}$ such that $E_{1},\ldots,E_{r + 1}$ is a basis for $\picardGroup X_{\separableClosure{k}}$. The intersection matrix for this basis is
\begin{align}
\begin{bmatrix}
-1	&			&		&		&		\\
	&	\ddots	&		&		&		\\
	&			&	-1	&		&	1	\\
	&			&		&	-1	&	1	\\
	&			&	1	&	1	&	-1	\\
\end{bmatrix}%
\label{eq : dP1 : Intersection Matrix For Basis Of ECs}
\end{align}
We use the matrix in \eqref{eq : dP1 : Intersection Matrix For Basis Of ECs} in our Magma implementation.
\end{remark}

When $d \leq 6$, the automorphism group $\automorphismGroup(\picardGroup X_{\separableClosure{k}})$ is isomorphic to the Weyl group $\weylGroup R_{d}$ of the root system $R_{d}$, as given in Table \ref{tab : dP1 : Root Systems}.
\begin{table}[t]
\centering
\begin{tabular}{l *{6}{c}}
\hline\hline
Degree of del Pezzo surface	&	1		&	2		&	3		&	4		&	5		&	6	\\
Associated root system		&	$E_{8}$	&	$E_{7}$	&	$E_{6}$	&	$D_{5}$	&	$A_{4}$	&	$A_{1} \times A_{2}$	\\
\hline
\end{tabular}
\caption{The root system $R_{d}$, for a del Pezzo surface $X$ of degree $d \in \{1,\ldots,6\}$ over a field $k$, such that the automorphism group of the geometric Picard group of the del Pezzo surface is isomorphic to the Weyl group of $R_{d}$.}
\label{tab : dP1 : Root Systems}
\end{table}
The image of the Galois representation $\rho_{\separableClosure{k}} : \galoisGroup(\separableClosure{k} \fieldExtension k) \rightarrow \automorphismGroup(\picardGroup X_{\separableClosure{k}})$ is a subgroup of $\weylGroup R_{d}$.

Given a choice of $\separableClosure{k}$, the \fontDefWord{splitting field of $X$} is the smallest intermediate field $K$ in $\separableClosure{k} \fieldExtension k$ such that all exceptional curves of $X$ are defined over $K$. Thus $\picardGroup X_{K} \isomorphism \picardGroup X_{\separableClosure{k}}$. The Galois action of $\galoisGroup(\separableClosure{k} \fieldExtension k)$ on $\picardGroup X_{\separableClosure{k}}$ factors through the quotient $\galoisGroup(K \fieldExtension k)$, yielding a representation
\begin{align*}
\rho_{K}
:
\galoisGroup(K \fieldExtension k)
\rightarrow
\automorphismGroup(\picardGroup X_{K})
\end{align*}
Choosing a basis of $\picardGroup X_{K} \isomorphism \picardGroup X_{\separableClosure{k}}$ gives isomorphisms
\begin{align*}
\picardGroup X_{K}
&\isomorphism
\integers^{n}
&
&\text{and}
&
\automorphismGroup(\picardGroup X_{K})
&\isomorphism
\generalLinear(n,\integers)
\end{align*}
where $n = 10 - d$, allowing us to extend the representation $\rho_{K}$ to a matrix representation
\begin{align*}
\rho_{K}
:
\galoisGroup(K \fieldExtension k)
\rightarrow
\generalLinear(n,\integers)
\end{align*}



\subsection{Equivalent definition of del Pezzo surface}

Let $k$ be a field, and let $X$ be a nice $k$-variety. Let $\omega_{X}$ denote the canonical sheaf of $X$, and $\omega_{X}^{\tensor -1}$ the \fontDefWord{anticanonical sheaf of $X$}. Let $K_{X}$ denote the class of $\omega_{X}$ in the Picard group $\picardGroup X$. Under the hypotheses on $X$, the Picard group of $X$ is isomorphic to the Weil divisor class group of $X$, as discussed in Section \ref{sec : Intro : Primer : The Picard Group}.

\begin{definition}
A nice $k$-surface $X$ is a \fontDefWord{del Pezzo surface} if its anticanonical sheaf $\omega_{X}^{\tensor -1}$ is ample. The \fontDefWord{degree} of $X$ is the self-intersection number $(K_{X},K_{X})$.
\end{definition}



\subsection{{B}rauer groups of del {P}ezzo surfaces}

Let $k$ be a field, and let $X$ be a del Pezzo surface over $k$. Then $X$ has no transcendental Brauer elements. That is, the algebraic Brauer group is the full Brauer group:% Begin footnote.
\footnote{In particular, all Brauer elements become trivial upon base change to a separable closure of the base field.}% End footnote.
\begin{align*}
\brauerGroupAlgebraic X
=
\brauerGroup X
\end{align*}
For this reason, in this chapter, we will write $\brauerGroup X/\brauerGroupConstant X$ rather than $\brauerGroupAlgebraic X/\brauerGroupConstant X$.





\section{Families of del {P}ezzo surfaces studied}

Why focus on Brauer groups with $5$-torsion? For del Pezzo surfaces of degree $1$, such Brauer groups are theoretically possible. However, to date, no examples of such surfaces defined over $\rationals$ have been found. Corn \cite{Corn2005}*{Theorem 1.4.1} presents the isomorphism classes of candidate Brauer groups modulo constants of del Pezzo surfaces of all degrees. The only degree in which $5$-torsion appears is degree $1$, in the candidate groups $\integers/5 \integers$ and $(\integers/5 \integers)^{2}$. Group cohomology theory gives a necessary (but not sufficient) condition for $\cohomology[1]{\galoisGroup(\separableClosure{k} \fieldExtension k),\picardGroup X_{\separableClosure{k}}}$ to have $5$-torsion: The image of $\galoisGroup(\separableClosure{k} \fieldExtension k)$ under the representation $\rho_{\separableClosure{k}}$ on $\picardGroup X_{\separableClosure{k}}$ must have $5$-torsion.

Table \ref{tab : dP1 Types} presents the three types of del Pezzo surfaces of degree $1$ we selected for study.
\begin{table}[t]
\centering
\begin{tabular}{c c c c c c}
\hline\hline
Type	&	$f_{4}$		&	$f_{6}$						&	$\automorphismGroup X$	&	$\order \automorphismGroup X$	&	$\order \galoisGroup(K \fieldExtension \rationals)$	\\
\hline
4	&	$0$			&	$x_{0} (x_{0}^{5} + x_{1}^{5})$		&	$C_{30}$				&	$30$							&	$240$						\\
8	&	$x_{0}^{4}$	&	$x_{0} x_{1}^{5}$				&	$C_{20}$				&	$20$							&	$160$						\\
13	&	$x_{0}^{4}$	&	$x_{0} (a x_{0}^{5} + x_{1}^{5})$	&	$C_{10}$				&	$10$							&	$2400$						\\
\hline
\end{tabular}
\caption{Defining polynomials and automorphism groups of three families of del Pezzo surfaces of degree $1$. For each family (type), the del Pezzo surface is defined in weighted projective space $\projective^{3}_{\rationals,[1,1,2,3]}$ by the zero set of the polynomial $x_{3}^{2} - x_{2}^{3} - f_{4}(x_{0},x_{1}) x_{2} - f_{6}(x_{0},x_{1})$. The polynomials $f_{4}$ and $f_{6}$ come from Dolgachev's classification \cite{Dolgachev2012}*{Table 8.14}. For Type 13, we set the parameter $a = 1$.}
\label{tab : dP1 Types}
\end{table}
For each type, the automorphism group has order divisible by $5$. We conjectured that the order of the the image of $\rho_{\separableClosure{\rationals}}$ might share this property. This was indeed the case for the representatives we studied: The values of $\order \galoisGroup(K \fieldExtension \rationals)$ in the final column of Table \ref{tab : dP1 Types} equal the corresponding values of $\order \image(\rho_{\separableClosure{\rationals}})$.



\begin{remark}
The exact relationship between the automorphism group and the image of the Galois representation on the geometric Picard group remains unknown to me. For the surfaces we studied, the order of the image of the Galois representation is one or two orders of magnitude larger than the order of the corresponding automorphism group. Moreover, the relative orders of the images of the Galois representations does not correspond to the relative orders of the full automorphism groups. See Table \ref{tab : dP1 Types}.
\end{remark}

\begin{remark}
\label{rmk : dP1 : Orders Of Subgroups Weyl E8 Yielding 5 Torsion}
As discussed in Section \ref{sec : dP1 : Del Pezzo Surfaces As Blow Ups}, for a del Pezzo surface of degree $1$, the image of the Galois representation on the geometric Picard group is isomorphic to a subgroup of $\weylGroup E_{8}$. We took a matrix representation of $\weylGroup E_{8}$ and computed the first cohomology group for each $\integers$-conjugacy class of subgroup, noting those that produced $5$-torsion. The orders of these subgroups were
\begin{align*}
5,10,20,25,50,60,100,120,200,300,400,600,1200,2400,3600,7200,14400
\end{align*}
\end{remark}

\begin{remark}
Carter \cite{Carter2006}*{Section 4.3} characterizes nontrivial $5$-torsion in $\brauerGroup X/\brauerGroupConstant X$ in terms of how the 240 exceptional curves of $X$ behave under the action of $\galoisGroup(\separableClosure{k} \fieldExtension k)$. In particular, see Theorems 20 and 22.
\end{remark}





\section{Computing {G}alois representations}
\label{sec : dP1 : Computing Galois Representations}

This section describes our approach to computing a representation of the absolute Galois group of the base field on the geometric Picard group of a del Pezzo surface of degree $1$. The approach returns a subgroup of $\generalLinear(9,\integers)$ isomorphic to the image of this representation. Computing the first cohomology group of this subgroup gives the Brauer group modulo constants of the del Pezzo surface.

We list here the steps in our approach. We discuss each step below.
\begin{enumerate}
\item\label{itm : dP1 : Step Construct Splitting Field} Construct a candidate splitting field for the surface.
\item\label{itm : dP1 : Step Compute ECs} Compute equations for all exceptional curves.
\item\label{itm : dP1 : Step Finite Field} Find a finite field over which all exceptional curves specialize.
\item\label{itm : dP1 : Step Specialize ECs} Specialize the exceptional curves to this finite field.
\item\label{itm : dP1 : Step Basis Pic} Specify a basis of the geometric Picard group.
\item\label{itm : dP1 : Step Galois Gens} Construct elements of the Galois group.
\item\label{itm : dP1 : Step Matrix Reps} Compute matrix representations of these Galois elements.
\item\label{itm : dP1 : Step Subgroup Matrix Rep} Compute the subgroup generated by these matrix representations.
\item\label{itm : dP1 : Step Brauer Group} Compute the Brauer group modulo constants of the del Pezzo surface.
\end{enumerate}



\subsection{Step \ref{itm : dP1 : Step Construct Splitting Field}: Construct a candidate splitting field for the surface}
\label{sec : dP1 : Step Construct Splitting Field}

Let $K_{0}$ be a field of characteristic not $2$ or $3$. We start with an explicit equation
\begin{align}
f
=
x_{3}^{2} - x_{2}^{3} - f_{4}(x_{0},x_{1}) x_{2} - f_{6}(x_{0},x_{1})%
\label{eq : Defining Equation dP1 In P11234}
\end{align}
that defines a del Pezzo surface $X = \zeroSet(f)$ of degree $1$ in weighted projective space $\projective_{K_{0},[1,1,2,3]}^{3}$.% Begin footnote.
\footnote{In particular, $f$ must be specified so that $X = \zeroSet(f)$ is smooth as a subset of $\projective_{K_{0},[1,1,2,3]}^{3}$.} % End footnote.
The variables $x_{0},x_{1},x_{2},x_{3}$ have weights $1,1,2,3$, respectively. The polynomials $f_{4}$ and $f_{6}$ in $K_{0}[x_{0},x_{1}]$ are homogeneous of degrees $4$ and $6$, respectively.

As we discuss in Sections \ref{sec : dP1 : Blow Ups And Picard Groups}--\ref{sec : dP1 : Del Pezzo Surfaces As Blow Ups}, the geometric Picard group of $X$ is generated by the exceptional curves on $X$. When $\characteristic K_{0} \neq 2$ or $3$, a divisor $E$ on $X_{\separableClosure{K_{0}}}$ is an exceptional curve of $X$ if and only if there exist homogeneous polynomials $Q(x_{0},x_{1})$ and $C(x_{0},x_{1})$ of degrees $2$ and $3$, respectively, such that
\begin{align}
E
=
\zeroSet(x_{2} - Q(x_{0},x_{1}),x_{3} - C(x_{0},x_{1}))%
\label{eq : Divisor On dP1 In P1123}
\end{align}
Define generic polynomials
\begin{align*}
C(x_{0},x_{1})
&=
c_{3,0} x_{0}^{3} + c_{2,1} x_{0}^{2} x_{1} + c_{1,2} x_{0} x_{1}^{2} + c_{0,3} x_{1}^{3}
\\
Q(x_{0},x_{1})
&=
q_{2,0} x_{0}^{2} + q_{1,1} x_{0} x_{1} + q_{0,2} x_{1}^{2}
\end{align*}
treating the coefficients $c_{i,j}$ and $q_{i,j}$ as indeterminates. Given a field extension $K$ of $K_{0}$, by \eqref{eq : Divisor On dP1 In P1123} a sequence $(c_{3,0},\ldots,q_{0,2}) \in K^{7}$ of particular values for these indeterminates corresponds to a curve in $\projective_{K,[1,1,2,3]}^{2}$. To be a divisor on $X$, this curve must satisfy the equation \eqref{eq : Defining Equation dP1 In P11234} defining $X$. That is, the function $f$ evaluated at $x_{2} = Q(x_{0},x_{1})$ and $x_{3} = C(x_{0},x_{1})$ must be identically zero, as a polynomial in $K[x_{0},x_{1}]$. This gives
\begin{align*}
0
&=
f(x_{0},x_{1},Q(x_{0},x_{1}),C(x_{0},x_{1}))
\\
&=
C(x_{0},x_{1})^{2} - Q(x_{0},x_{1})^{3} - f_{4}(x_{0},x_{1}) Q(x_{0},x_{1}) - f_{6}(x_{0},x_{1})
\end{align*}
Each term on the right side is a homogeneous polynomial of degree $6$ in the variables $x_{0},x_{1}$. Expanding the products and collecting terms by their monomials $x_{0}^{i} x_{1}^{j}$, we get
\begin{align}
0
=
\sum_{\ell = 0}^{6} a_{\ell}(c_{i,j},q_{i',j'}) x_{0}^{\ell} x_{1}^{6 - \ell}%
\label{eq : Divisor On dP1 As Sum Of Seven Polynomials}
\end{align}
where the $a_{\ell}$ are polynomials in the seven indeterminates $c_{3,0},\ldots,q_{0,2}$ with coefficients in $K_{0}$.

Equation \eqref{eq : Divisor On dP1 As Sum Of Seven Polynomials} holds if and only if all seven polynomials $a_{\ell}$ are identically zero. A sequence $(c_{3,0},\ldots,q_{0,2}) \in K^{7}$ of seven particular values for which this is true is precisely a point in the zero set of the ideal
\begin{align*}
\fontIdeal{a}
=
(a_{0},\ldots,a_{6})
\end{align*}
generated by the seven polynomials $a_{\ell}$. Note that each polynomial $a_{\ell}$ is defined with coefficients in $K_{0}$, so for any field extension $K$ of $K_{0}$ there is a natural way to view each $a_{\ell}$, and hence the ideal $\fontIdeal{a}$, in $K[c_{3,0},\ldots,q_{0,2}]$.

We have thus established the following correspondence:
\begin{center}
\begin{tabular}{*{5}{c}}
Exceptional curve
&
\multirow{2}{*}{\hspace{5mm}$\leftrightarrow$\hspace{5mm}}
&
Sequence
&
\multirow{2}{*}{\hspace{5mm}$\leftrightarrow$\hspace{5mm}}
&
Ideal
\\
$E \in \picardGroup X_{\separableClosure{K_{0}}}$
&
&
$(c_{3,0},\ldots,q_{0,2})$
&
&
$P \in \zeroSet(\fontIdeal{a})$
\end{tabular}
\end{center}
This allows us to translate our search for exceptional curves on $X$ into a search for points in the zero set of an ideal. In the latter setting, we can use tools like elimination ideals and Gr\"{o}bner bases to facilitate our search.

Over $\separableClosure{K_{0}}$---indeed, over any intermediate field between the splitting field of $X$ in $\separableClosure{K_{0}}$ and $\separableClosure{K_{0}}$---there are 240 exceptional curves on $X$, corresponding to 240 distinct points in $\zeroSet(\fontIdeal{a})$. In the setting of ideals, our goal becomes to find a Galois field extension $K$ of $K_{0}$ such that $\order \zeroSet(\fontIdeal{a}) = 240$ when we view $\fontIdeal{a}$ as an ideal in $K[c_{3,0},\ldots,q_{0,2}]$.

All subsequent steps except Step \ref{itm : dP1 : Step Brauer Group} require only that all 240 exceptional curves on $X$ be defined over $K$. However, to ensure the correctness of our computation of $\brauerGroup X/\brauerGroupConstant X$ in Step \ref{itm : dP1 : Step Brauer Group}, we will ask that the extension $K \fieldExtension K_{0}$ be minimal, in the sense that $K$ is a splitting field of $X$. See Remark \ref{rmk : dP1 : Galois Group And Field Extension}.

To this end, compute elimination ideals of $\fontIdeal{a}$, one for each of the seven indeterminates $c_{i,j}$ and $q_{i',j'}$. This computation can be done using Gr\"{o}bner bases. More precisely, for each indeterminate $b$, compute the elimination ideal
\begin{align*}
\fontIdeal{a}_{b}
=
\fontIdeal{a} \cap K_{0}[b]
\end{align*}
The elimination ideal $\fontIdeal{a}_{b}$ captures all elements of $\fontIdeal{a}$ that involve the indeterminate $b$ only. Note that $\fontIdeal{a}_{b}$ is an ideal in $K_{0}[b]$, a polynomial ring in one indeterminate over a field, and hence a principal ideal domain. Thus the ideal $\fontIdeal{a}_{b}$ can be generated by a single polynomial $g_{b} \in K_{0}[b]$:
\begin{align*}
\fontIdeal{a}_{b}
=
(g_{b})
\end{align*}

As explained in Remark \ref{rmk : dP1 : Galois Group And Field Extension}, our approach to finding a splitting field for $X$ is to find a minimal field extension $K$ of $K_{0}$ over which all seven generators $g_{b}$ factor into linear factors, then certify the existence of a subgroup $G \subgroupeq \image \rho_{K}$ such that $\order(G) = [K : K_{0}]$. To this end, it is helpful to view the seven generators $g_{b}$ as elements of a common polynomial ring, say $K_{0}[t]$. (Simply replace each indeterminate $b$ with the indeterminate $t$.) Then apply the following algorithm.

\begin{algorithm}
\label{alg : dP1 : Splitting Field}
Set $i = 0$ and $K_{i} = K_{0}$.
\begin{enumerate}
\item\label{itm : dP1 : Splitting Field Algorithm : Step 1} Check whether all seven generators $g_{b}$ factor into linear factors over $K_{i}$. If yes, then stop. If no, then proceed to Step \ref{itm : dP1 : Splitting Field Algorithm : Step 2}.
\item\label{itm : dP1 : Splitting Field Algorithm : Step 2} Compute the irreducible factors $p$ of each $g_{b}$ in $K_{i}[t]$.
\begin{enumerate}
\item Select a nonlinear irreducible factor, call it $p_{i}$.
\item Adjoin to $K_{i}$ a (symbolic) root $\theta_{i}$ of $p_{i}$. That is, form the field $K_{i + 1} \defeq K_{i}(\theta_{i})$, or more precisely, $K_{i + 1} \defeq K_{i}[t]/(p_{i})$. Increment $i$ and return to Step \ref{itm : dP1 : Splitting Field Algorithm : Step 1}.
\end{enumerate}
\end{enumerate}
At each iteration, use the inclusion $K_{i} \hookrightarrow K_{i + 1}$ to view the seven generators $g_{b}$ as elements of $K_{i + 1}[t]$.
\end{algorithm}

Algorithm \ref{alg : dP1 : Splitting Field} is guaranteed to terminate: We have finitely many (seven) polynomials $g_{b}$, each with finitely many irreducible factors (at each iteration of the algorithm), each of finite degree. After each iteration, the degree of at least one irreducible factor has decreased. The degrees of the irreducible factors are bounded below by $1$. Thus the algorithm terminates in at most $\prod_{b} \degree(g_{b})$ steps.

Let $K$ denote the final field produced by Algorithm \ref{alg : dP1 : Splitting Field}. By construction, $\order \zeroSet(\fontIdeal{a}) = 240$. That is, all $240$ exceptional curves of $X$ are defined over $K$. The extension $K \fieldExtension K_{0}$ is normal, because by construction it is a splitting field for the set $\{g_{c_{3,0}},\ldots,g_{q_{0,2}}\}$. If $\characteristic K_{0} = 0$,% Begin footnote.
\footnote{This hypothesis is sufficient but not necessary for the following.} % End footnote.
then $K \fieldExtension K_{0}$ is also separable, and hence Galois. It remains to verify that there exists a subgroup $G \subgroupeq \image \rho_{K}$ such that $\order(G) = [K : K_{0}]$, in which case $K$ is indeed a splitting field of $X$. Steps \ref{itm : dP1 : Step Compute ECs}--\ref{itm : dP1 : Step Subgroup Matrix Rep} address this.

\begin{remark}
In practice, during some iterations of Algorithm \ref{alg : dP1 : Splitting Field}, there exist multiple irreducible factors such that adjoining a root of one factor did not affect irreducibility of the others. In this case, we adjoined all such roots at the same step of the algorithm. This seemed to facilitate computation.
\end{remark}

\begin{remark}
In practice, we did not analyze all seven generators $g_{b}$ at each step of Algorithm \ref{alg : dP1 : Splitting Field}. Instead, in the first iteration, we chose a indeterminate $b$ whose generator $g_{b}$ had relatively simple irreducible factors, and we iterated Algorithm \ref{alg : dP1 : Splitting Field} while monitoring a Gr\"{o}bner basis for $\fontIdeal{a}$ with respect to a lexicographical order with the chosen indeterminate as the smallest monomial.
\end{remark}

\begin{remark}
\label{rmk : dP1 : Galois Group And Field Extension}
Intuitively, a splitting field $K$ of $X$ is ``large enough'' so that the representation $\rho_{K} : \galoisGroup(K \fieldExtension K_{0}) \rightarrow \automorphismGroup(\picardGroup X_{K})$ is ``exhaustive'', in the sense that $\image \rho_{K} \isomorphism \image \rho_{\separableClosure{K_{0}}}$; and ``small enough'' so that $\rho_{K}$ is faithful, and therefore $\image \rho_{K} \isomorphism \galoisGroup(K \fieldExtension K_{0})$. It follows that if $K$ is a splitting field of $X$, then
\begin{align*}
\order(\image \rho_{\separableClosure{K_{0}}})
=
\order(\image \rho_{K})
=
\order \galoisGroup(K \fieldExtension K_{0})
=
[K : K_{0}]
\end{align*}

In steps \ref{itm : dP1 : Step Construct Splitting Field}--\ref{itm : dP1 : Step Subgroup Matrix Rep} we compute (i) a Galois field extension $K \fieldExtension K_{0}$ over which all 240 exceptional curves of $X$ are defined, and (ii) a subgroup $\rho(G) \subgroupeq \image \rho_{\separableClosure{K_{0}}}$. A priori, if
\begin{align}
\order(\rho(G))
<
[K : K_{0}]%
\label{eq : dP1 : Order Matrix Subgroup Less Than Degree Extension}
\end{align}
then we don't know whether this is because the subgroup $\rho(G)$ is too small (because we have failed to define a generating set in Step \ref{itm : dP1 : Step Galois Gens}) or because $K$, and hence $[K : K_{0}]$, is too big. If $K$ is a splitting field of $X$, then $[K : K_{0}]$ is the ``correct'' value $\order(\image \rho_{\separableClosure{K_{0}}})$, which is the smallest possible degree of a field extension of $K_{0}$ for which all 240 exceptional curves of $X$ are defined. In this case the inequality \eqref{eq : dP1 : Order Matrix Subgroup Less Than Degree Extension} indicates that $\rho(G)$ is too small, so we must define additional generators.

In practice, we may obtain a splitting field $K$ of $X$ as follows: After we compute equations for all exceptional curves of $X$ in Step \ref{itm : dP1 : Step Compute ECs}, form the subset $\Theta$ consisting of those $\theta_{i}$ that appear in the expressions for at least one of the exceptional curves. Form the set $H$ consisting of the irreducible factors $h$ (over $K_{0}$) of the $g_{b}$ in Step \ref{itm : dP1 : Step Construct Splitting Field} such that $h$ is the minimal polynomial (over $K_{0}$) for some $\theta \in \Theta$. Define $K$ to be the splitting field of the set $H$.

For the surfaces D4 and D8, the field $K$ obtained using this latter method coincides with the field $K$ obtained using the procedure in Algorithm \ref{alg : dP1 : Splitting Field}.
\end{remark}



\subsection{Step \ref{itm : dP1 : Step Compute ECs}: Compute equations for all exceptional curves}

Once we have a field $K$ over which all exceptional curves of the del Pezzo surface $X$ are defined, base change $X$ to $\Spec K$ and compute explicit equations for its primary components, which are its exceptional curves. (We need explicit equations of the exceptional curves to determine the Galois representation on $\picardGroup X_{\separableClosure{K_{0}}}$.) In our case, we used the computational algebra system Magma, which has a function called \fontComputer{PrimaryComponents}.



\subsection{Step \ref{itm : dP1 : Step Finite Field}: Find a finite field over which all exceptional curves specialize}

\begin{remark}
As we will see, this step does not depend on Step \ref{itm : dP1 : Step Compute ECs}. We perform this step to facilitate the computation of matrix representations of Galois elements, which we do using intersection numbers. Computing intersection numbers is much faster over finite fields than over number fields.
\end{remark}

Let $p_{0},\ldots,p_{z}$ denote the irreducible polynomials used to construct the field extension $K \fieldExtension K_{0}$ in Step \ref{itm : dP1 : Step Construct Splitting Field}.% Begin footnote.
\footnote{The index $i$ of each polynomial corresponds to the iteration of the algorithm in which the polynomial was used. In an iteration in which multiple roots were adjoined simultaneously, order those roots arbitrarily.} % End footnote.
Conduct the following search for a finite field over which specializations of all corresponding roots $\theta_{i}$ of the $p_{i}$ are defined. In Remark \ref{rmk : Finite Field Search} below we comment on the parameter space of pairs $(p,r)$ for the search.

\begin{algorithm}
\label{alg : dP1 : Finite Field Search}
Fix a prime power $q = p^{r}$. Initialize \fontTextVariable{all\_split} to \fontComputer{true}.
\begin{enumerate}
\item\label{itm : dP1 : Find Finite Field : Is Nonsingular} Check whether the polynomial $f$ that defines the surface over $\rationals$, when specialized to $\fontField{F}_{q}$, yields a nonsingular surface.
\begin{enumerate}
\item If no, then set \fontTextVariable{all\_split} to \fontComputer{false}. This choice of $q$ has failed.
\item If yes, then proceed.
\end{enumerate}
\end{enumerate}
Initialize $i = 0$.
\begin{enumerate}
\setcounter{enumi}{1}
\item\label{itm : dP1 : Find Finite Field : Factor Polynomials} While \fontTextVariable{all\_split} is \fontComputer{true} and $i \leq z$, check whether the specialization $\overline{p}_{i} \in \fontField{F}_{q}[t]$ of the polynomial $p_{i} \in K[t]$ splits into linear factors.
\begin{enumerate}
\item If no, then set \fontTextVariable{all\_split} to \fontComputer{false}. This choice of $q$ has failed.
\item If yes, then let $\overline{\theta}_{i}$ be any root of $\overline{p}_{i}(t)$.
\end{enumerate}
Increment $i$ and return to Step \ref{itm : dP1 : Find Finite Field : Factor Polynomials}.
\end{enumerate}
$\fontField{F}_{q}$ is a finite field over which all exceptional curves specialize if and only if, when this algorithm terminates, the value of \fontTextVariable{all\_split} is \fontComputer{true}.
\end{algorithm}

Algorithm \ref{alg : dP1 : Finite Field Search} is guaranteed to terminate: Each iteration of Step \ref{itm : dP1 : Find Finite Field : Factor Polynomials} either ends the process (because \fontComputer{all\_split} is \fontComputer{false}) or increments $i$. If \fontComputer{all\_split} is never reset to \fontComputer{false}, then eventually $i > z$, and Step \ref{itm : dP1 : Find Finite Field : Factor Polynomials} terminates.

\begin{remark}
\label{rmk : Finite Field Search}
In Table \ref{tab : dP1 : Finite Field Size} we present, for the representative of each type of del Pezzo surface we study in this thesis, the smallest finite fields $\fontField{F}_{q}$ (where $q = p^{r}$) over which all exceptional curves are defined.
\begin{table}[t]
\centering
\begin{tabular}{c r r r r}
\hline\hline
	&	\multicolumn{4}{c}{Smallest $p$ for given $r$}	\\
Type	&	$r = 1$	&	$2$		&	$3$	&	$4$	\\
\hline
4	&	6271		&	\textbf{29}	&	631	&	29	\\
8	&	3121		&	\textbf{19}	&	3121	&	13	\\
13	&	30391	&	89		&	571	&	\textbf{7}	\\
\hline
\end{tabular}
\caption{Smallest finite fields $\fontField{F}_{p^{r}}$, over which all exceptional curves of the del Pezzo surface of given type specialize. For each surface, the entry corresponding to the smallest such finite field, for any power $r$, is indicated in bold.}
\label{tab : dP1 : Finite Field Size}
\end{table}
More precisely, for each surface, we embedded Algorithm \ref{alg : dP1 : Finite Field Search} inside nested iterations over the power $r$ and primes $p$ (up to some upper bound). If desired, we may break the iteration over $p$ after the first suitable (that is, smallest valid) $p$ is found for a given $r$.

We exclude the primes $p = 2$ and $3$ from this analysis. For each surface, the polynomials $p_{i}$ used to construct a splitting field $K$ may have fractional coefficients. This requires% Begin footnote.
\footnote{This assumes the coefficients of each polynomial $p_{i}$ are in reduced form. If this is not the case, additional primes may be excluded unnecessarily.} % End footnote.
that we exclude all primes factors $p$ of the denominators of these coefficients from this analysis. For each surface, the set of such primes is as follows:
\begin{itemize}
\item Type 4: $\{2,3,5\}$
\item Type 8: $\{5\}$
\item Type 13: $\{2,5,31\}$
\end{itemize}

Performing the analysis for each surface, we find the smallest finite field over which all exceptional curves specialize to be as follows:
\begin{itemize}
\item Type 4: $q = p^{r} = 29^{2} = 841$
\item Type 8: $q = p^{r} = 19^{2} = 361$
\item Type 13: $q = p^{r} = 7^{4} = 2401$
\end{itemize}
Note that all these fields have size $7^{4}$ or smaller. Because $7$ is the smallest valid prime for each of these surfaces, analyzing prime powers larger than $r = 4$ will necessarily yield finite fields larger than those listed here. We conclude that the listed fields are in fact the smallest possible finite fields over which all exceptional curves of the given del Pezzo surface specialize.
\end{remark}



\subsection{Step \ref{itm : dP1 : Step Specialize ECs}: Specialize the exceptional curves to this finite field}

Take the equations for the exceptional curves computed in Step \ref{itm : dP1 : Step Compute ECs} and replace each appearance of $\theta_{i}$ with its corresponding specialization $\overline{\theta}_{i}$ chosen in Step \ref{itm : dP1 : Step Finite Field}.



\subsection{Step \ref{itm : dP1 : Step Basis Pic}: Specify a basis of the geometric Picard group}

As discussed in Section \ref{sec : dP1 : Definitions And Geometry} above, for a del Pezzo surface of degree $1$ defined over $K_{0}$, the geometric Picard group $\picardGroup X_{\separableClosure{K_{0}}}$ is isomorphic to $\integers^{9}$. Moreover, there exists a basis for $\picardGroup X_{\separableClosure{K_{0}}}$ consisting of nine exceptional curves whose intersection matrix is given by \eqref{eq : dP1 : Intersection Matrix For Basis Of ECs}. Because Fano schemes are flat, for any divisors $D_{i},D_{j} \in \picardGroup X_{\separableClosure{K_{0}}}$, their intersection number over $\separableClosure{K_{0}}$ equals the intersection number of their specializations $\overline{D}_{i},\overline{D}_{j} \in \picardGroup X_{\fontField{F}_{q}}$:
\begin{align}
(D_{i},D_{j})_{X_{\separableClosure{K_{0}}}}
=
(\overline{D}_{i},\overline{D}_{j})_{X_{\fontField{F}_{q}}}%
\label{eq : dP1 : Flat Scheme Intersection Numbers}
\end{align}
It is to facilitate these (and subsequent) computations of intersection numbers that we include Steps \ref{itm : dP1 : Step Finite Field} and \ref{itm : dP1 : Step Specialize ECs} above. Let \fontTextVariable{ECListFq} denote the list of specialized exceptional curves from Step \ref{itm : dP1 : Step Specialize ECs}, in an arbitrary but fixed order.

To find a basis of $\picardGroup X_{\separableClosure{K_{0}}}$, we use the following algorithm.

\begin{algorithm}
\label{alg : dP1 : Find Basis Of Picard Group Over Finite Field}
Initialize $\text{\fontTextVariable{basis}} = \emptyset$ and $\text{\fontTextVariable{i}} = 1$.
\begin{quote}
\hspace{0mm}Loop over the exceptional curves $\overline{E}$ in \fontTextVariable{ECListFq}.

\hspace{10mm}Set \fontTextVariable{noIntersection} to \fontTextVariable{true}.

\hspace{10mm}Loop over the exceptional curves $\overline{B}$ in \fontTextVariable{basis}.

\hspace{20mm}If $(\overline{E},\overline{B})_{X_{\fontField{F}_{q}}} \neq 0$, then

\hspace{30mm}Set \fontTextVariable{noInstersection} to \fontTextVariable{false}.

\hspace{30mm}Break from the loop over \fontTextVariable{basis}.

\hspace{10mm}If \fontTextVariable{noIntersection} is \fontTextVariable{true}, then

\hspace{20mm}Append $\overline{E}$ to \fontTextVariable{basis}.

\hspace{10mm}If \fontTextVariable{basis} has 8 elements, then

\hspace{20mm}Break from the loop over \fontTextVariable{ECListFq}.

\hspace{0mm}If \fontTextVariable{basis} has only 7 elements, then delete the 7th element and return to the preceding loop over \fontTextVariable{ECListFq}, starting at the index following the index corresponding to the deleted 7th element. See Remark \ref{rmk : Generator 7 Subtlety} below. After this step is complete, \fontTextVariable{basis} will have exactly 8 elements.

\hspace{0mm}Loop over the exceptional curves $\overline{E}$ in \fontTextVariable{ECListFq}, starting at the index immediately after the index of the 6th element in \fontTextVariable{basis}.

\hspace{10mm}Set \fontTextVariable{noIntersection} to \fontTextVariable{true}.

\hspace{10mm}Loop over the first 6 elements $\overline{B}$ of \fontTextVariable{basis}.

\hspace{20mm}If $(\overline{E},\overline{B})_{X_{\fontField{F}_{q}}} \neq 0$, then

\hspace{30mm}Set \fontTextVariable{noIntersection} to \fontTextVariable{false}.

\hspace{30mm}Break from the loop over the subset of \fontTextVariable{basis}.

\hspace{10mm}If \fontTextVariable{noIntersection} is \fontComputer{true}, $(\overline{E},\overline{B}_{7}) = 1$, and $(\overline{E},\overline{B}_{8}) = 1$, then

\hspace{20mm}Append $\overline{E}$ to \fontTextVariable{basis}.

\hspace{20mm}Break (we are done).
\end{quote}
\end{algorithm}

\begin{remark}
\label{rmk : Matching Bases Picard Group}
To determine a representation of the action of the absolute Galois group on the geometric Picard group (see Step \ref{itm : dP1 : Step Matrix Reps}), we must translate the basis of specialized exceptional curves on $X_{\fontField{F}_{q}}$ returned by Algorithm \ref{alg : dP1 : Find Basis Of Picard Group Over Finite Field} to a basis of exceptional curves on $X$. In theory, we may simply match each specialized exceptional curve $\overline{E}$ chosen for the basis of $\picardGroup X_{\fontField{F}_{q}}$ to its originator $E \in \picardGroup X_{K}$. In practice, our approach obscures the matching: In passing to the specialization, we chose a particular value in $\fontField{F}_{q}$ for each $\overline{\theta}_{i}$, and the computational algebra system reduces the $7$-tuple describing each exceptional curve of $X$ over $K$ ($7$ algebraic expressions involving elements of $K_{0}$ and the $\theta_{i}$) to $7$ numbers in $\fontField{F}_{q}$. Once a basis for $\picardGroup X_{\fontField{F}_{q}}$ was constructed, for each reduced basis element $\overline{E}$ we can no longer simply ``remove the bars'' from each $\overline{\theta}_{i}$ to recover $\theta_{i}$ and the corresponding exceptional curve $E$ in $\picardGroup X_{K}$.

To address this, we constructed two lists of all exceptional curves, one list defined over the splitting field $K$, the other list defined over the finite field $\fontField{F}_{q}$, such that the index in the first list of each curve over $K$ equaled the index in the second list of its specialization over $\fontField{F}_{q}$. The expressions for the list over $\fontField{F}_{q}$ were identical to those in the list over $K$, except that each $\theta_{i}$ was replaced by the symbol $\overline{\theta}_{i}$. Once a basis for $\picardGroup X_{\fontField{F}_{q}}$ was constructed, we could then take each $7$-tuple of numbers in $\fontField{F}_{q}$ and identify the index in the second list whose $7$ expressions equaled these $7$ numbers. That index, in the first list of exceptional curves over $K$, gave the corresponding basis element for $\picardGroup X_{K}$.
\end{remark}

\begin{remark}
The ``if \fontTextVariable{basis} has 8 elements'' check in Algorithm \ref{alg : dP1 : Find Basis Of Picard Group Over Finite Field} is optional. Every exceptional curve has nonzero intersection with at least one of the first eight elements of \fontTextVariable{basis}. In particular, this is true for all exceptional curves appearing after the 8th element of \fontTextVariable{basis} in \fontTextVariable{ECListFq}, so all iterations of the first loop over \fontTextVariable{ECListFq} after \fontComputer{basis} has eight elements will not extend \fontComputer{basis} further. In practice, the eighth element appended to \fontTextVariable{basis} tends to occur toward the end of \fontTextVariable{ECListFq}, so the computation saved is minor.
\end{remark}

\begin{remark}
\label{rmk : Generator 7 Subtlety}
There is a subtlety in choosing the seventh generator for the basis. From the geometry of del Pezzo surfaces (see Section \ref{sec : dP1 : Definitions And Geometry}), following Algorithm \ref{alg : dP1 : Find Basis Of Picard Group Over Finite Field}, there are three choices for the seventh basis element. Two of them allow us to find an eighth basis element not intersecting the previous seven; the third does not. In our implementation, if the basis built by the first part of Algorithm \ref{alg : dP1 : Find Basis Of Picard Group Over Finite Field} has only seven elements, then we remove the seventh element, and restart the first loop of the algorithm from the index immediately \emph{after} the index of the seventh basis element we just removed. The two ``good'' choices of the seventh generator are guaranteed to be in this right tail of \fontTextVariable{ECListFq}, for if they had appeared earlier, then they would have been added to \fontTextVariable{basis} as the seventh generator, instead of the ``bad'' choice.
\end{remark}



\subsection{Step \ref{itm : dP1 : Step Galois Gens}: Construct elements of the Galois group}

In this step we construct elements $\sigma_{i} \in \galoisGroup(K \fieldExtension K_{0})$. Let $G$ be the subgroup generated by the $\sigma_{i}$. In Steps \ref{itm : dP1 : Step Matrix Reps}--\ref{itm : dP1 : Step Subgroup Matrix Rep} we compute the subgroup $\rho_{K}(G) \subgroupeq \generalLinear(9,\integers)$ using a matrix Galois representation $\rho_{K}$. If the $\sigma_{i}$ form a generating set for $\galoisGroup(K \fieldExtension K_{0})$, then $G = \galoisGroup(K \fieldExtension K_{0})$. If $K$ is a splitting field for $X$, then
\begin{align*}
G
&\isomorphism
\rho_{K}(G)
&
&\text{and}
&
\rho_{K}(\galoisGroup(K \fieldExtension K_{0}))
&\isomorphism
\rho_{\separableClosure{K_{0}}}(\galoisGroup(\separableClosure{K_{0}} \fieldExtension K_{0}))
\end{align*}
If both hypotheses hold, then
\begin{align*}
G
\isomorphism
\rho_{\separableClosure{K_{0}}}(\galoisGroup(\separableClosure{K_{0}} \fieldExtension K_{0}))
\end{align*}
As discussed in Remark \ref{rmk : dP1 : Galois Group And Field Extension}, if all 240 exceptional curves of $X$ are defined over $K$, then the equality $\order G = [K : K_{0}]$ certifies that $K$ is a splitting field of $X$ and that $\rho_{K}(G) \isomorphism \galoisGroup(K \fieldExtension K_{0})$.

The construction of a candidate splitting field $K$ of $X$ in Step \ref{itm : dP1 : Step Construct Splitting Field} produces a finite list of polynomials $p_{0},\ldots,p_{z}$. By definition, each $p_{i}$ was irreducible over the field $K_{i}$. We defined $\theta_{i}$ to be a root of $p_{i}$, and we adjoined $\theta_{i}$ to $K_{i}$ to form $K_{i + 1}$. Thus the field $K$ is obtained by starting from $K_{0}$ and adjoining the roots $\theta_{i}$, iteratively:
\begin{align*}
K
=
K_{0}[\theta_{0}]\cdots[\theta_{z}]
\end{align*}
It follows that an element $\sigma$ of the Galois group $\galoisGroup(K \fieldExtension K_{0})$ is completely determined by its action on the $\theta_{i}$. By Galois theory, $\sigma$ must send each $\theta_{i}$ to a root of the minimal polynomial of $\theta_{i}$ over $K_{0}$. In our construction, we will define Galois elements $\sigma$ that send each $\theta_{i}$ to a root of the minimal polynomial of $\theta_{i}$ over $K_{i}$, that is, of the polynomial $p_{i}$. In the cases we study, this is sufficient to obtain a generating set for $\galoisGroup(K \fieldExtension K_{0})$. Because the Galois group of a finite extension of fields is transitive, we may send each $\theta_{i}$ to any root of its minimal polynomial over $K_{0}$, and hence to any root of $p_{i}$. In our construction, we assume that these choices can be made independently for each $\theta_{i}$.

Under these hypotheses, for each $i \in \{1,\ldots,z\}$, we may define $\sigma_{i} \in \galoisGroup(K \fieldExtension K_{0})$ by specifying its action on each $\theta_{j}$ according to the following rules:
\begin{enumerate}
\item for $j = i$, $\sigma_{i}$ sends $\theta_{i}$ to a \emph{different} root of $p_{i}$, that is, $p_{i}(\sigma_{i}(\theta_{i})) = 0$ and $\sigma_{i}(\theta_{i}) \neq \theta_{i}$;
\item for $j < i$, $\sigma_{i}$ fixes $\theta_{j}$; and
\item for $j > i$, $\sigma_{i}$ sends $\theta_{j}$ to a root of $\sigma_{i}(p_{j})$.
\end{enumerate}

\begin{remark}
In our application of this procedure to the del Pezzo surfaces D4 and D8, we chose irreducible factors $p_{i}$ in Step \ref{itm : dP1 : Step Construct Splitting Field} that were relatively simple (so roots other than $\theta_{i}$ are easily specified) and, if possible, symmetric in the roots $\theta_{j}$ with $j < i$ (so $\sigma_{j}(p_{i})$ is relatively easy to determine). This is not necessary, and indeed, while we have not yet done so, it seems possible to automate this procedure in \fontComputer{Magma}. Such automation would facilitate handling the multi-termed polynomials $p_{i}$ arising in the analysis of the del Pezzo surface D13. (See Section \ref{sec : Code : D13 Output}.)
\end{remark}



\subsection{Step \ref{itm : dP1 : Step Matrix Reps}: Compute matrix representations of these Galois elements}
\label{sec : dP1 : Step Matrix Reps}

In this step, we compute explicit matrix representations for each Galois element $\sigma_{i} \in \galoisGroup(K \fieldExtension K_{0})$ defined in Step \ref{itm : dP1 : Step Galois Gens}.

First use Remark \ref{rmk : Matching Bases Picard Group} to identify the basis $B$ of exceptional curves $E_{1},\ldots,E_{9}$ on $X_{K}$ corresponding to the basis $\overline{B}$ of exceptional curves $\overline{E}_{1},\ldots,\overline{E}_{9}$ on $X_{\fontField{F}_{q}}$ chosen in Step \ref{itm : dP1 : Step Basis Pic}. Next, fix a Galois element $\sigma_{i}$ defined in Step \ref{itm : dP1 : Step Galois Gens}.% Begin footnote.
\footnote{Perform the subsequent analysis for each $\sigma_{i}$.} % End footnote.
For each $E$ in $B$, compute the $9 \times 1$ matrix $\fontMatrixRep{M}_{B}(\sigma_{i}(E))$ as follows:
\begin{enumerate}
\setcounter{enumi}{-1}
\item\label{itm : dP1 : Matrix Reps : 7 Tuple} Start with the $7$-tuple $(a_{0},\ldots,a_{6}) \in K^{7}$ identified with $E$ in Step \ref{itm : dP1 : Step Construct Splitting Field}.
\item\label{itm : dP1 : Matrix Reps : theta To sigma theta} For each $j \in \{0,\ldots,z\}$, replace each occurrence of $\theta_{j}$ with the explicit expression for $\sigma_{i}(\theta_{j})$ computed in Step \ref{itm : dP1 : Step Galois Gens}. These replacements must be done simultaneously for all $j$---see Remark \ref{rmk : dP1 : Simultaneous Versus Iterative Replacement} below. This yields the point in $K^{7}$ identified with $\sigma_{i}(E)$.
\item\label{itm : dP1 : Matrix Reps : theta To Overline theta} For each $j \in \{0,\ldots,z\}$, replace each occurrence of $\theta_{j}$ with $\overline{\theta}_{j}$. This yields the point in $\fontField{F}_{q}^{7}$ identified with $\overline{\sigma_{i}(E)}$.
\item\label{itm : dP1 : Matrix Reps : Coefficients} Because $\overline{B} = (\overline{E}_{1},\ldots,\overline{E}_{9})$ is a basis of $\picardGroup X_{\fontField{F}_{q}}$, there exist unique $\alpha_{i,j} \in \integers$ such that
\begin{align}
\overline{\sigma_{i}(E)}
=
\sum_{j = 1}^{9} \alpha_{i,j} \overline{E}_{j}%
\label{eq : dP1 : Computing Coefficients For sigma E}
\end{align}
Compute the values of the $\alpha_{i,j}$ using the intersection pairing. More precisely, for each $k \in \{1,\ldots,9\}$, apply the intersection pairing $(-,\overline{E}_{k})_{X_{\fontField{F}_{q}}}$ to both sides of \eqref{eq : dP1 : Computing Coefficients For sigma E}, yielding nine equations. In each equation, on the left side, the value $(\overline{\sigma_{i}(E)},\overline{E}_{k})_{X_{\fontField{F}_{q}}}$ can be computed;% Begin footnote.
\footnote{If $\overline{\sigma_{i}(E)} = \overline{E}_{k}$, then their intersection number is $-1$. Otherwise, their intersection is a finite collection of points, which when counted appropriately equals their intersection number. In the latter case, one may compute this number in Magma using the command \fontComputer{Degree(Intersection($C$,$C_{k}$))}, where $C$ and $C_{k}$ are the curves on $X_{\fontField{F}_{q}}$ (as opposed to $7$-tuples in $\fontField{F}_{q}^{7}$) corresponding to $\overline{\sigma_{i}(E)}$ and $\overline{E}_{k}$, respectively.} % End footnote.
and on the right side, the values $(\overline{E}_{j},\overline{E}_{k})_{X_{\fontField{F}_{q}}}$ are known.% Begin footnote.
\footnote{These intersection numbers are precisely those we computed when constructing the basis $(\overline{E}_{1},\ldots,\overline{E}_{9})$ in Step \ref{itm : dP1 : Step Basis Pic}, and presented in the intersection matrix for the basis $(E_{j})_{j = 1}^{9}$ given in \eqref{eq : dP1 : Intersection Matrix For Basis Of ECs}, which by \eqref{eq : dP1 : Flat Scheme Intersection Numbers} equals the intersection matrix for the basis $(\overline{E}_{j})_{j = 1}^{9}$.} % End footnote.
Thus we have nine equations in the nine unknowns $\alpha_{i,j}$ (recall we have fixed $i$), which we may solve using conventional means. Arrange the results in a $9 \times 1$ column matrix of integers:
\begin{align*}
\fontMatrixRep{M}_{\overline{B}}(\overline{\sigma_{i}(E)})
=
\begin{bmatrix}
\alpha_{i,1}	\\
\vdots		\\
\alpha_{i,9}
\end{bmatrix}
\end{align*}
By \eqref{eq : dP1 : Flat Scheme Intersection Numbers}, we obtain the same $9 \times 1$ matrix if we perform this last step with $\sigma_{i}(E)$ and the basis $B = (E_{1},\ldots,E_{9})$. That is,
\begin{align*}
\fontMatrixRep{M}_{B}(\sigma_{i}(E))
=
\fontMatrixRep{M}_{\overline{B}}(\overline{\sigma_{i}(E)})
\end{align*}
\end{enumerate}
For each $\sigma_{i}$, collect the results of this procedure into a $9 \times 9$ matrix
\begin{align*}
\fontMatrixRep{M}_{B,B}(\sigma_{i})
=
\begin{bmatrix}
|	&		&	|	\\
\fontMatrixRep{M}_{B}(\sigma_{i}(E_{1}))	&	\ldots	&	\fontMatrixRep{M}_{B}(\sigma_{i}(E_{9}))	\\
|	&		&	|
\end{bmatrix}
\end{align*}
This matrix is invertible, so $\fontMatrixRep{M}_{B,B}(\sigma_{i}) \in \generalLinear(9,\integers)$.

Let $G \defeq \langle{}\sigma_{0},\ldots,\sigma_{z}\rangle{}$ be the subgroup of $\galoisGroup(K \fieldExtension K_{0})$ generated by the $\sigma_{i}$. Using the matrices $\fontMatrixRep{M}_{B,B}(\sigma_{i})$, we may define a matrix representation
\begin{align*}
\rho
:
G
\rightarrow
\generalLinear(9,\integers)
\end{align*}
by expressing each $\sigma \in G$ as an expression $R(\sigma_{0},\ldots,\sigma_{z})$ in the $\sigma_{i}$ and mapping $\sigma \mapsto R(\fontMatrixRep{M}_{B,B}(\sigma_{0}),\ldots,\fontMatrixRep{M}_{B,B}(\sigma_{z}))$. In Step \ref{itm : dP1 : Step Subgroup Matrix Rep} we use this representation to determine whether $G = \galoisGroup(K \fieldExtension K_{0})$.

\begin{remark}
\label{rmk : dP1 : Simultaneous Versus Iterative Replacement}
According to our construction of $\sigma_{i}$ in Step \ref{itm : dP1 : Step Galois Gens}, for $j > i$, $\sigma_{i}$ sends $\theta_{j}$ to a root of the polynomial $\sigma_{i}(\theta_{j})$. A priori, this root can involve all the roots $\theta_{0},\ldots,\theta_{z}$ (see Tables \ref{tab : dP1 : D4 : Galois Generators} and \ref{tab : dP1 : D8 : Galois Generators}). By ``simultaneous'' replacement, we mean to contrast with a general ``iterative'' replacement of $\theta_{j}$ by $\sigma_{i}(\theta_{j})$. For example, consider the case of the del Pezzo surface D8 and the Galois element $\sigma_{0}$. (See Table \ref{tab : dP1 : D8 : Galois Generators}.) An iterative replacement from $j = 0$ to $j = z$ would replace all occurrences of $\theta_{1}$ by $\sigma_{0}(\theta_{1}) = \theta_{2}$, then replace all occurrences of $\theta_{2}$ (including those just introduced by the replacement $\theta_{1} \mapsto \sigma_{0}(\theta_{1}) = \theta_{2}$) by $\sigma_{0}(\theta_{2}) = \theta_{1}$. The overall effect is to fix all occurrences of $\theta_{1}$ in the original expression, which is incorrect. In practice, we can achieve correct simultaneous replacement by performing Steps \ref{itm : dP1 : Matrix Reps : theta To sigma theta} and \ref{itm : dP1 : Matrix Reps : theta To Overline theta} in tandem: Specifically, for each $j \in \{1,\ldots,z\}$, replace $\theta_{j}$ with a modified form of the explicit expression for $\sigma_{i}(\theta_{j})$ in which each $\theta_{k}$ in the expression is replaced by $\overline{\theta}_{k}$. In this way, ``new'' $\theta_{k}$ introduced by the replacements $\theta_{j} \mapsto \sigma_{i}(\theta_{j})$ are not modified by subsequent replacements.
\end{remark}



\subsection{Step \ref{itm : dP1 : Step Subgroup Matrix Rep}: Compute the subgroup generated by these matrix representations}

Compute the image of the matrix representation $\rho$ of $G$ defined in Step \ref{itm : dP1 : Step Matrix Reps}. By construction, $\rho(G) \subgroupeq \rho(\galoisGroup(K \fieldExtension K_{0}))$. Also by construction, the field extension $K \fieldExtension K_{0}$ is Galois and minimal such that all exceptional curves on $X$ are defined over $K$. Thus by Remark \ref{rmk : dP1 : Galois Group And Field Extension}, $G = \galoisGroup(K \fieldExtension K_{0})$ if and only if
\begin{align*}
\order \rho(G)
=
[K : K_{0}]
=
\prod_{i = 0}^{z} \deg p_{i}
\end{align*}
If equality does not hold, return to Step \ref{itm : dP1 : Step Galois Gens} and construct another element of the Galois group. Find its matrix representation in Step \ref{itm : dP1 : Step Matrix Reps}, and include that new matrix in the computation of $\rho(G)$.

\begin{remark}
For the surfaces D4 and D8, the Galois elements $\sigma_{0},\ldots,\sigma_{z}$ constructed from the polynomials $p_{0},\ldots,p_{z}$ used to define the field extension $K \fieldExtension K_{0}$ were sufficient to generate the full Galois group $\galoisGroup(K \fieldExtension K_{0})$.
\end{remark}



\subsection{Step \ref{itm : dP1 : Step Brauer Group}: Compute the Brauer group modulo constants of the del Pezzo surface}

Once we have a faithful matrix representation of $\galoisGroup(K \fieldExtension K_{0})$, we may compute the isomorphism class of the Brauer group of $X$ modulo constants by the isomorphism
\begin{align*}
\brauerGroup X/\brauerGroupConstant X
\isomorphism
\cohomology[1]{\galoisGroup(\separableClosure{K_{0}} \fieldExtension K_{0}),\picardGroup X_{\separableClosure{K_{0}}}}
\isomorphism
\cohomology[1]{\rho(G),\integers^{9}}
\end{align*}
as discussed in Section \ref{sec : Intro : Brauer Groups}.





\section{Results}
\label{sec : dP1 : Results}

Using Dolgachev's classification of del Pezzo surfaces of degree $1$ by automorphism group \cite{Dolgachev2012}*{Table 8.14}, we chose representatives from the types 4, 8, and 13, all defined over $K_{0} = \rationals$. We denote these representative surfaces by D4, D8, and D13, respectively. For each surface, we applied the approach in Section \ref{sec : dP1 : Computing Galois Representations}. (For the surface D13, we achieved only a partial analysis.)

In this section we present, for each surface, the following:
\begin{enumerate}
\item explicit polynomials $p_{i}$ used to build a tower $K = K_{z} \fieldExtension \ldots \fieldExtension K_{1} \fieldExtension K_{0}$ of field extensions, where $K$ is a splitting field of $X$;
\item explicit generators of the Galois group $\galoisGroup(K \fieldExtension K_{0})$; and
\item the isomorphism class of the Brauer group of $X$ modulo constants, viewing $X$ as defined over intermediate fields $L$ of $K \fieldExtension K_{0}$, including $L = K_{0}$.
\end{enumerate}

Throughout this section, we take $K_{0} = \rationals$. The polynomial $f$ defining each surface belongs to the graded ring $\rationals[x_{0},x_{1},x_{2},x_{3}]$, where the indeterminates $x_{0},x_{1},x_{2},x_{3}$ have weights $1,1,2,3$, respectively, and $f$ is smooth and homogeneous of (weighted) degree $6$. Hence the zero set $\zeroSet(f)$ defines a del Pezzo surface of degree $1$ in the weighted projective space $\projective_{\rationals,[1,1,2,3]}^{3}$. Following the notation in Section \ref{sec : dP1 : Computing Galois Representations}, each irreducible polynomial $p_{i}$ is defined over the field $K_{i}$, and $\theta_{i}$ denotes a root of $p_{i}$.



\subsection{Surface D4}

\begin{theorem}
Let $X = \zeroSet(f)$ be the type-4 degree-$1$ del Pezzo surface defined by
\begin{align*}
f
=
x_{3}^{2} - x_{2}^{2} - x_{0} (x_{0}^{5} + x_{1}^{5})
\end{align*}
Then $X$ has splitting field $K = \rationals(\theta_{0})(\theta_{1})(\theta_{2},\theta_{3},\theta_{4})(\theta_{5})$ of degree $240$ over $\rationals$, where each $\theta_{i}$ is a root of the polynomial $p_{i}$, given below:% Begin footnote.
\footnote{The polynomials $p_{2},p_{3},p_{4}$ depend only on $\theta_{0}$ and $\theta_{1}$, so we may think of adjoining the roots $\theta_{2},\theta_{3},\theta_{4}$ to $K_{2}$ simultaneously.}% End footnote.
\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*}

Table \ref{tab : dP1 : D4 : Galois Generators} presents elements $\sigma_{j}$ of the Galois group $\galoisGroup(K \fieldExtension \rationals)$ as defined by their action on the generators $\theta_{i}$ of $K \fieldExtension \rationals$. We extend these actions by $\rationals$-algebra linearity to define $\sigma_{j}$ on all of $K$. In defining the Galois elements $\sigma_{3}$ and $\sigma_{5}$ we use a primitive $3$rd root of unity and a primitive $5$th root of unity, which we denote $\zeta_{3}$ and $\zeta_{5}$, respectively. Specifically, in terms of the adjoined roots $\theta_{i}$, we define
\begin{align}
\zeta_{3}
&=
\frac{1}{12} \left((2 \theta_{0} - 5) \theta_{1} - (\theta_{0} + 5)\right) \theta_{2} \theta_{4} - \frac{1}{2}%
\label{eq : dP1 : D4 : Primitive 3 Root Of 1}
\\
\zeta_{5}
&=
\frac{1}{36} \left((5 \theta_{0} - 10) \theta_{1} + 6 \theta_{0}\right) \theta_{2} \theta_{4} - \frac{1}{4} (\theta_{0} + 1)%
\label{eq : dP1 : D4 : Primitive 5 Root Of 1}
\end{align}
\afterpage{% Begin afterpage.
\clearpage
\begin{landscape}
\begin{table}[t]
\centering
\scriptsize
\begin{tabular}{r|*{6}{c}}
			&	$\theta_{0}$			&	$\theta_{1}$			&	$\theta_{2}$			&	$\theta_{3}$			&	$\theta_{4}$			&	$\theta_{5}$	\\
\hline
$\sigma_{0}$	&	$-\theta_{0}$			&	$\displaystyle\frac{1}{2} (13 \theta_{0} - 29) \theta_{1}$	&	$\displaystyle\left(\frac{1}{4} (11 \theta_{0} - 25) \theta_{1} + \theta_{0} - 3\right) \theta_{4}$	&	$\displaystyle\left(\frac{1}{8} (11 \theta_{0} - 25) \theta_{1} - \frac{1}{2}\right) \theta_{3}$	&	$\theta_{2}$	&	$\displaystyle\frac{1}{72} \left((47 \theta_{0} - 75) \theta_{1} + 6 (17 \theta_{0} + 25)\right) \theta_{2} \theta_{4} \theta_{5}$	\\
$\sigma_{1}$	&	$\theta_{0}$			&	$-\theta_{1}$			&	$\displaystyle\left(\frac{1}{4} (3 \theta_{0} - 5) \theta_{1} + \theta_{0} + 3\right) \theta_{2}$	&	$\displaystyle-\frac{1}{4} \left((2 \theta_{0} - 5) \theta_{1} - (3 \theta_{0} + 1)\right) \theta_{3}$	&	$\displaystyle\left(\frac{1}{4} (11 \theta_{0} - 25) \theta_{1} + \theta_{0} - 3\right) \theta_{4}$	&	$\displaystyle\frac{1}{8} \left(-(3 \theta_{0} - 15) \theta_{1} + 2 (11 \theta_{0} + 27)\right) \theta_{5}$	\\
$\sigma_{2}$	&	$\theta_{0}$			&	$\theta_{1}$			&	$-\theta_{2}$			&	$\theta_{3}$			&	$\theta_{4}$			&	$\theta_{5}$	\\
$\sigma_{3}$	&	$\theta_{0}$			&	$\theta_{1}$			&	$\theta_{2}$			&	$\zeta_{3} \theta_{3}$	&	$\theta_{4}$			&	$\theta_{5}$	\\
$\sigma_{4}$	&	$\theta_{0}$			&	$\theta_{1}$			&	$\theta_{2}$			&	$\theta_{3}$			&	$-\theta_{4}$			&	$\theta_{5}$	\\
$\sigma_{5}$	&	$\theta_{0}$			&	$\theta_{1}$			&	$\theta_{2}$			&	$\theta_{3}$			&	$\theta_{4}$			&	$\zeta_{5} \theta_{5}$
\end{tabular}
\caption{Defining actions for elements $\sigma_{i}$ of the Galois group $\galoisGroup(K \fieldExtension \rationals)$ for the del Pezzo surface of degree $1$ called D4. The action of each $\sigma_{i}$ is given on each root $\theta_{j}$ used in the construction of the field extension $K \fieldExtension \rationals$. Extend these actions by $\rationals$-algebra linearity to define $\sigma_{i}$ on all of $K$. The symbols $\zeta_{3}$ and $\zeta_{5}$ denote a primitive $3$rd and $5$th root of unity, respectively. We define these in terms of the adjoined roots $\theta_{j}$ in Equations \eqref{eq : dP1 : D4 : Primitive 3 Root Of 1} and \eqref{eq : dP1 : D4 : Primitive 5 Root Of 1}.}
\label{tab : dP1 : D4 : Galois Generators}
\end{table}
\end{landscape}
}% End afterpage.
The corresponding matrix representations $\rho_{K}(\sigma_{i}) \in \generalLinear(9,\integers)$ appear in Section \ref{sec : Code : D4 Output}.
\begin{itemize}
\item View $X$ as a surface over $\rationals$. Then $\brauerGroup X/\brauerGroupConstant X \isomorphism 0$.

\item Let $L$ be the fixed field of one of the order-$10$ subgroups
\begin{align*}
&\langle\sigma_{2} \sigma_{3},\sigma_{5}\rangle
&
&\langle\sigma_{3} \sigma_{4},\sigma_{5}\rangle
\end{align*}
View $X$ as a surface over $L$. Then $\brauerGroup X_{L}/\brauerGroupConstant X_{L} \isomorphism \integers/5 \integers$.

\item Let $L$ be the fixed field of the order-$5$ subgroup
\begin{align*}
H
&=
\langle\sigma_{5}\rangle
\end{align*}
View $X$ as a surface over $L$. Then $\brauerGroup X_{L}/\brauerGroupConstant X_{L} \isomorphism (\integers/5 \integers)^{2}$.
\end{itemize}
\end{theorem}



\subsection{Surface D8}

\begin{theorem}
Let $X = \zeroSet(f)$ be the type-8 degree-$1$ del Pezzo surface defined by
\begin{align*}
f
=
x_{3}^{2} - x_{2}^{3} - x_{0}^{4} x_{2} - x_{0} x_{1}^{5}
\end{align*}
Then $X$ has splitting field $K = \rationals(\theta_{0})(\theta_{1},\theta_{2})(\theta_{3})$ of degree $160$ over $\rationals$, where each $\theta_{i}$ is a root of the polynomial $p_{i}$, given below:% Begin footnote.
\footnote{The polynomials $p_{1}$ and $p_{2}$ depend only on $\theta_{0}$, so we may think of adjoining the roots $\theta_{1}$ and $\theta_{2}$ to $K_{1}$ simultaneously. We note that, in the polynomial $p_{3}$, $\theta_{1}^{2} \theta_{2}^{2} = \sqrt{5}$.}% End footnote.
\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*}

Table \ref{tab : dP1 : D8 : Galois Generators} presents elements $\sigma_{j}$ of the Galois group $\galoisGroup(K \fieldExtension \rationals)$ as defined by their action on the generators $\theta_{i}$ of $K \fieldExtension \rationals$. We extend these actions by $\rationals$-algebra linearity to define $\sigma_{j}$ on all of $K$. The corresponding matrix representations $\rho_{K}(\sigma_{i}) \in \generalLinear(9,\integers)$ appear in Section \ref{sec : Code : D8 Output}.
\begin{table}[t]
\centering
\begin{tabular}{r|*{4}{c}}
			&	$\theta_{0}$	&	$\theta_{1}$			&	$\theta_{2}$			&	$\theta_{3}$	\\
\hline
$\sigma_{0}$	&	$-\theta_{0}$	&	$\theta_{2}$			&	$\theta_{1}$			&	$\displaystyle\frac{1}{5} (1 + 2 \theta_{0}) \theta_{1}^{2} \theta_{2}^{2} \theta_{3}$	\\
$\sigma_{1}$	&	$\theta_{0}$	&	$\theta_{0} \theta_{1}$	&	$\theta_{2}$			&	$\displaystyle\frac{1}{2} \theta_{0} (\theta_{1}^{2} \theta_{2}^{2} - 1) \theta_{3}$	\\
$\sigma_{2}$	&	$\theta_{0}$	&	$\theta_{1}$			&	$\theta_{0} \theta_{2}$	&	$\displaystyle\frac{1}{2} \theta_{0} (\theta_{1}^{2} \theta_{2}^{2} - 1) \theta_{3}$	\\
$\sigma_{3}$	&	$\theta_{0}$	&	$\theta_{1}$			&	$\theta_{2}$			&	$\displaystyle\frac{1}{4} (\theta_{1}^{3} \theta_{2} + \theta_{1}^{2} \theta_{2}^{2} - \theta_{1} \theta_{3}^{3} - 1) \theta_{3}$
\end{tabular}
\caption{Defining actions for elements $\sigma_{i}$ of the Galois group $\galoisGroup(K \fieldExtension \rationals)$ for the del Pezzo surface of degree $1$ called D8. The action of each $\sigma_{i}$ is given on each root $\theta_{j}$ used in the construction of the field extension $K | \rationals$. Extend these actions by $\rationals$-algebra linearity to define $\sigma_{i}$ on all of $K$.}
\label{tab : dP1 : D8 : Galois Generators}
\end{table}

\begin{itemize}
\item View $X$ as a surface over $\rationals$. Then $\brauerGroup X/\brauerGroupConstant X \isomorphism 0$.

\item Let $L$ be the fixed field of one of the order-$10$ subgroups
\begin{align*}
&\langle\sigma_{1}^{2},\sigma_{3}\rangle
&
&\langle\sigma_{2}^{2},\sigma_{3}\rangle
&
&\langle\sigma_{0} \sigma_{1} \sigma_{2},\sigma_{3}\rangle
&
&\langle\sigma_{0},\sigma_{3}\rangle
\end{align*}
or one of the order-$20$ subgroups
\begin{align*}
&\langle\sigma_{1},\sigma_{3}\rangle
&
&\langle\sigma_{2},\sigma_{3}\rangle
\end{align*}
View $X$ as a surface over $L$. Then $\brauerGroup X_{L}/\brauerGroupConstant X_{L} \isomorphism \integers/5 \integers$.

\item Let $L$ be the fixed field of the order-$5$ subgroup
\begin{align*}
\langle\sigma_{3}\rangle
\end{align*}
View $X$ as a surface over $L$. Then $\brauerGroup X_{L}/\brauerGroupConstant X_{L} \isomorphism (\integers/5 \integers)^{2}$.
\end{itemize}
\end{theorem}



\subsection{Surface D13}

\begin{theorem}
Let $X = \zeroSet(f)$ be the type-13 degree-$1$ del Pezzo surface defined by
\begin{align*}
f
=
x_{3}^{2} - x_{2}^{3} - x_{0}^{4} x_{2} - x_{0} (x_{0}^{5} + x_{1}^{5})
\end{align*}
Then $X$ has splitting field $K = \rationals(\theta_{0})(\theta_{1},\theta_{2})(\theta_{3},\theta_{4})$ of degree $2400$ over $\rationals$, where each $\theta_{i}$ is a root of the polynomial $p_{i}$, whose degrees are given below:% Begin footnote.
\footnote{The polynomials $p_{1}$ and $p_{3}$ depend only on $\theta_{0}$, so we may think of adjoining the roots $\theta_{0}$ and $\theta_{3}$ to $K_{1}$ simultaneously. Similarly, the polynomials $p_{2}$ and $p_{4}$ depend only on $\theta_{0}$ and $\theta_{1}$, so we may think of adjoining the roots $\theta_{2}$ and $\theta_{4}$ simultaneously to a suitable intermediate field.}% End footnote.
\begin{align*}
\deg p_{0}
&=
12
&
\deg p_{1}
&=
10
&
\deg p_{2}
&=
2
&
\deg p_{3}
&=
2
&
\deg p_{4}
&=
5
\end{align*}
\end{theorem}

\begin{remark}
In Section \ref{sec : Code : D13 Output} we give a full specification of the polynomials $p_{i}$ used in the construction of the splitting field $K$ for the surface D13. Recalling our analysis of the subgroups of $\weylGroup E_{8}$ in Remark \ref{rmk : dP1 : Orders Of Subgroups Weyl E8 Yielding 5 Torsion}, we note that $[K : \rationals] = 2400$ is consistent with $5$-torsion in $\brauerGroup X/\brauerGroupConstant X$.
\end{remark}