%
%
%   Chapter 3 : The inverse Galois problem for algebraic K3 surfaces
%
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\chapter{The inverse {G}alois problem for algebraic {K}3 surfaces}
\label{sec : K3}

In this chapter we use the theory in Chapter \ref{sec : Intro} and the results in Chapter \ref{sec : GLnZ} to put an upper bound on the set of possible algebraic Brauer groups modulo constants associated to a nice variety. In particular, this bound applies to algebraic K3 surfaces defined over $\rationals$. We construct an explicit algebraic K3 surface with geometric Picard rank $2$ that gives an affirmative answer to the inverse Brauer problem (Question \ref{q : Intro : Inverse Brauer Problem}) when the target group is $\integers/2 \integers$, the only possible nontrivial algebraic Brauer group modulo constants under these hypotheses.



\section{Definitions and geometry}
\label{sec : K3 : Definitions And Geometry}

Let $k$ be a field, and let $X$ be a $k$-variety. Let $\fontSheaf{O}_{X}$ denote the structure sheaf of $X$, let $\omega_{X}$ denote the canonical sheaf of $X$, and let $K_{X}$ denote the class of $\omega_{X}$ in the Picard group $\picardGroup X$ of $X$.

\begin{definition}
The variety $X$ is an \fontDefWord{algebraic K3 surface} if it is smooth, projective, of dimension $2$, and
\begin{align*}
\omega_{X}
&\isomorphism
\fontSheaf{O}_{X}
&
\cohomology[1]{X,\fontSheaf{O}_{X}}
&\isomorphism
0
\end{align*}
\end{definition}

Note that the condition $\omega_{X} \isomorphism \fontSheaf{O}_{X}$ implies $K_{X} = 0$.

For the remainder of this section, let $k$ be a field, and let $X$ be an algebraic K3 surface defined over $k$.

Let us show that $\picardGroup X$ has no nontrivial torsion.% Begin footnote.
\footnote{The following argument summarizes \cite{Huybrechts2015}*{Section 1.2}, culminating in Remark 2.5.} % End footnote.
The \fontDefWord{Euler characteristic} of a coherent sheaf $\fontSheaf{F}$ on $X$, denoted $\eulerCharacteristic(X,\fontSheaf{F})$, is% Begin footnote.
\footnote{This definition holds for any projective $k$-scheme $X$. See \cite{Hartshorne1977}*{Exercise III.5.1}.} % End footnote.
\begin{align*}
\eulerCharacteristic(X,\fontSheaf{F})
=
\sum_{i \geq 0} (-1)^{i} h^{i}(X,\fontSheaf{F})
\end{align*}
where $h^{i}(X,\fontSheaf{F}) = \dim_{k} \cohomology[i]{X,\fontSheaf{F}}$. Let $D$ be a divisor on $X$. The Riemann--Roch theorem states that
\begin{align}
\eulerCharacteristic(X,D)
=
\frac{1}{2} (D,D - K_{X}) + \eulerCharacteristic(X,\fontSheaf{O}_{X})%
\label{eq : K3 : Riemann Roch General}
\end{align}
where $(-,-)$ denotes the intersection pairing on $X$. The definition of an algebraic K3 surface immediately gives $h^{0}(X,\fontSheaf{O}_{X}) = 1$ and $h^{1}(X,\fontSheaf{O}_{X}) = 0$. Serre duality implies $\cohomology[2]{X,\fontSheaf{O}_{X}} \isomorphism \cohomology[0]{X,\omega_{X}}^{*}$, so $h^{2}(X,\fontSheaf{O}_{X}) = 1$. All higher cohomology groups are $0$. Therefore $\eulerCharacteristic(X,\fontSheaf{O}_{X}) = 2$. Moreover, as noted above, for an algebraic K3 surface $X$, $K_{X} = 0$. Substituting these results into \eqref{eq : K3 : Riemann Roch General}, we get
\begin{align}
\eulerCharacteristic(X,D)
=
\frac{1}{2} (D,D) + 2%
\label{eq : K3 : Riemann Roch K3}
\end{align}

Suppose $D$ is torsion, that is, $\ell D = 0$ for some $\ell \in \integersPositive$. Then (bi)linearity of the intersection pairing implies $(D,D) = 0$, so \eqref{eq : K3 : Riemann Roch K3} reduces to
\begin{align*}
\eulerCharacteristic(X,D)
=
2
\end{align*}
Thus either $D$ or $-D$ is effective. Without loss of generality, suppose that $D$ is effective.% Begin footnote.
\footnote{If $-D$ is effective, simply run this entire argument on $-D$.} % End footnote.
If there exists a nonzero $s \in \cohomology[0]{X,D}$, then for all $\ell \in \integersPositive$, $s^{\ell} \in \cohomology[0]{X,\ell D}$ is also nonzero, and the zero sets of $s$ and $s^{\ell}$ coincide. By contrapositive, if $\ell D = 0$ for some $\ell \in \integersPositive$ (true because $D$ is torsion), then $D = 0$, as desired.

One can show that the geometric Picard rank of $X$ is an integer in the set $\{1,\ldots,20,22\}$. Combined with what we have just shown, it follows that $\picardGroup X \isomorphism \integers^{n}$ for some $n \in \{1,\ldots,20,22\}$.

By definition, an algebraic K3 surface $X$ is a projective $k$-variety. Thus, the Picard group $\picardGroup X$ has a nontrivial class containing a hyperplane, denote it $H$, defined over $k$. For any Galois extension $K$ of $k$, the divisor $H$ is also defined on $X_{K}$, so the class in $\picardGroup X_{K}$ containing $H$ is nontrivial. Because this class is defined over the base field $k$, it is fixed by the action of the Galois group $\galoisGroup(K \fieldExtension k)$. Consider the representation $\rho_{K} : \galoisGroup(K \fieldExtension k) \rightarrow \automorphismGroup(\picardGroup X_{K}) \isomorphism \generalLinear(n,\integers)$, where the isomorphism is induced by choosing a basis of $\picardGroup X_{K}$.% Begin footnote.
\footnote{See Section \ref{sec : Intro : Galois Representations On The Picard Group}.} % End footnote.
Let $G \subgroupeq \generalLinear(n,\integers)$ denote the image of $\rho_{K}$ under this isomorphism. Let $G$ act on $\integers^{n}$ by left multiplication. Then we may view $\integers^{n}$ as a $\integers G$-module. Tensoring up to $\rationals$, we conclude that the existence of a class containing the hyperplane $H$ defined over $k$ corresponds to a nontrivial eigenspace of $\rationals^{n}$ associated to the eigenvalue $1$. This restricts the allowed images of Galois representations on $\picardGroup X_{\separableClosure{k}}$.

By definition, $X$ is a nice surface. For any field extension $K \fieldExtension k$, the base change $X_{K}$ is also a nice surface. Thus, there exists an intersection pairing on the Picard group $\picardGroup X_{K}$. When the field extension $K \fieldExtension k$ is Galois, the action of $\galoisGroup(K \fieldExtension k)$ on $\picardGroup X_{K}$ preserves the intersection form. This further restricts the allowed images of Galois representations on $\picardGroup X_{\separableClosure{k}}$.

In order to bound from above the rank of geometric Picard groups, we will use of the following result.

\begin{theorem}[See \cite{VarillyAlvarado2015}*{Theorem 2.12}]
\label{thm : Background : Bound From Frobenius}
Let $K$ be a number field, let $R$ be a discrete valuation ring of $K$, and let $k \isomorphism \fontField{F}_{q}$ be the residue field of $R$. Fix an algebraic closure $\algebraicClosure{K}$ of $K$, and let $\overline{R}$ be the integral closure of $R$ in $\algebraicClosure{K}$. Let $\fontIdeal{p} \in \overline{R}$ be a nonzero prime such that $\overline{k} \defeq \overline{R}/\fontIdeal{p}$ is an algebraic closure of $k$. Let $\ell \neq \characteristic k$ be prime. Let $\mathcal{X} \rightarrow R$ be a smooth, proper morphism of relative dimension $2$ such that the surfaces $\mathcal{X}_{\algebraicClosure{K}}$ and $\mathcal{X}_{\overline{k}}$ are geometrically integral. Let $\Phi_{\mathcal{X}_{k}}^{*}$ be the map
\begin{align*}
\Phi_{\mathcal{X}_{k}}^{*}
:
\cohomologyEtale[2]{\mathcal{X}_{\overline{k}},\rationals_{\ell}}
\rightarrow
\cohomologyEtale[2]{\mathcal{X}_{\overline{k}},\rationals_{\ell}}
\end{align*}
constructed from the absolute Frobenius map on $\mathcal{X}_{k}$.

Then there exist natural, injective homomorphisms
\begin{align*}
\neronSeveriGroup \mathcal{X}_{\algebraicClosure{K}} \tensor \rationals_{\ell}
\hookrightarrow
\neronSeveriGroup \mathcal{X}_{\overline{k}} \tensor \rationals_{\ell}
\hookrightarrow
\cohomologyEtale[2]{\mathcal{X}_{\overline{k}},\rationals_{\ell}(1)}
\end{align*}
of $\rationals_{\ell}$-inner product spaces, where the second map respects the action of the Galois group $\galoisGroup(\overline{K} \fieldExtension k)$. It follows that the rank of $\picardGroup \mathcal{X}_{\algebraicClosure{K}}$ is bounded from above by the number of eigenvalues, counted with multiplicity, of $\Phi_{\mathcal{X}_{k}}^{*}$ of the form $q$ times a root of unity.
\end{theorem}



\section{Results: Candidate algebraic {B}rauer groups}

In this section, we present the candidate algebraic Brauer groups modulo constants for algebraic K3 surfaces of geometric Picard ranks $1$ to $6$.

We used the algorithm developed in Chapter \ref{sec : GLnZ} to produce a database with exactly one representative from each $\integers$-conjugacy class of finite subgroup of $\generalLinear(n,\integers)$. For each subgroup $G$ in this database, we computed which subgroups have a nontrivial eigenspace associated to the eigenvalue $1$. Of these subgroups, we also recorded which are cyclic. We summarize the results in the following theorem.

\begin{theorem}
Let $n \in \{1,\ldots,6\}$, and let $G$ be a finite subgroup of $\generalLinear(n,\integers)$ with nontrivial eigenspace associated to the eigenvalue $1$. Then there exists a finite subgroup $G' \subgroupeq \generalLinear(n - 1,\integers)$ such that
\begin{align*}
\cohomology[1]{G,\integers^{n}}
\isomorphism
\cohomology[1]{G',\integers^{n - 1}}
\end{align*}
Moreover, for each first cohomology group $B$ produced by such a finite subgroup $G \subgroupeq \generalLinear(n,\integers)$, there exists a cyclic such finite subgroup $C \subgroupeq \generalLinear(n,\integers)$ such that
\begin{align*}
\cohomology[1]{C,\integers^{n}}
\isomorphism
B
\end{align*}
\end{theorem}

For convenience, these candidate groups are presented in Table \ref{tab : K3 : Candidate Algebraic Brauer Groups}. The candidate cohomology groups for rank $n$ are all groups in rows $1,\ldots,n$.
\begin{table}[t]
\centering
\begin{tabular}{c l *{5}{r}}
\hline\hline
Rank	&	\multicolumn{6}{c}{Invariant factors* of new $\brauerGroupAlgebraic X/\brauerGroupConstant X$}	\\
	&	\multicolumn{6}{c}{$[d_{1}]\cdots[d_{z}] \leftrightsquigarrow (\integers/d_{1} \integers) \oplus \ldots \oplus (\integers/d_{z} \integers)$}	\\
\hline
$1$	&	$[\;]$	\\
$2$	&	$[2]$	\\
$3$	&	$[2]^{2}$	&	$[3]$	\\
$4$	&	$[2]^{3}$	&	$[2][3]$		&	$[4]$	\\
$5$	&	$[2]^{4}$	&	$[2]^{2}[3]$	&	$[2][4]$		&	$[3]^{2}$		&	$[5]$	\\
$6$	&	$[2]^{5}$	&	$[2]^{3}[3]$	&	$[2]^{2}[4]$	&	$[2] [3]^{2}$	&	$[2][5]$	&	$[3][4]$	\\
\hline
\end{tabular}
\caption{Candidate algebraic Brauer groups modulo constant classes for nice $k$-surfaces of geometric Picard 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 : K3 : Candidate Algebraic Brauer Groups}
\end{table}

\begin{corollary}
\label{cor : K3 : Algebraic Brauer Group Modulo Constants}
Let $n \in \{1,\ldots,6\}$, and let $X$ be a nice $k$-variety such that $\picardGroup X_{\separableClosure{k}} \isomorphism \integers^{n}$. Then $\brauerGroupAlgebraic X/\brauerGroupConstant X$ is isomorphic to one of the groups in Table \ref{tab : K3 : Candidate Algebraic Brauer Groups}, in the row corresponding to rank $n$ or smaller.
\end{corollary}

In particular, Corollary \ref{cor : K3 : Algebraic Brauer Group Modulo Constants} holds for algebraic K3 surfaces.



\section{Cohomology of finite cyclic groups}

In this section we prove a general result about the cohomology of finite cyclic groups,% Begin footnote.
\footnote{For a more extensive development of group cohomology, see, for example, Dummit and Foote \cite{Dummit2004}*{Section 17.2}.} % End footnote.
which we use in our construction of explicit K3 surfaces below.

\begin{lemma}
\label{lem : K3 : Cohomology Finite Cyclic Group}
Let $G$ be a finite cyclic group of order $m$, let $\sigma$ be a generator of $G$, let $N = \sum_{i = 1}^{m} \sigma^{i} \in \integers G$, and let $A$ be a $\integers G$-module. Then
\begin{align}
\cohomology[i]{G,A}
\isomorphism
\begin{dcases*}
A^{G}				&	if $i = 0$					\\
{}_{N}A/(\sigma - 1) A	&	if $i \in 2 \integersPositive - 1$	\\
A^{G}/N A				&	if $i \in 2 \integersPositive$
\end{dcases*}%
\label{eq : K3 : Cohomology Finite Cyclic Group}
\end{align}
\end{lemma}

\begin{proof}
Let $1 = 1 e$ denote the multiplicative identity element of the group ring $\integers G$. Direct computation shows that
\begin{align*}
N (\sigma - 1)
=
(\sigma - 1) N
=
\sigma^{m} - 1
=
0
\end{align*}
as elements of $\integers G$. It follows that
\begin{align*}
\cdots
\xrightarrow{\sigma - 1}
\integers G
\xrightarrow{N}
\integers G
\xrightarrow{\sigma - 1}
\integers G
\xrightarrow{\augmentationMap}
\integers
\rightarrow
0
\end{align*}
is a projective (in fact, a free) resolution of $\integers$ by $\integers G$-modules, where $\sigma - 1$ and $N$ denote the maps from $\integers G$ to itself induced by left-multiplication by these elements, and $\augmentationMap$ denotes the augmentation map
\begin{align*}
\augmentationMap
:
\integers G
&\rightarrow
\integers
\\
\sum_{g \in G} \alpha_{g} g
&\mapsto
\sum_{g \in G} \alpha_{g}
\end{align*}
Applying the contravariant functor $\Hom_{\integers G}(-,A)$ to this resolution and dropping the $\Hom_{\integers G}(\integers,A)$ term gives the cochain complex
\begin{align*}
0
\rightarrow
\Hom_{\integers G}(\integers G,A)
\xrightarrow{\sigma - 1}
\Hom_{\integers G}(\integers G,A)
\xrightarrow{N}
\Hom_{\integers G}(\integers G,A)
\xrightarrow{\sigma - 1}
\cdots
\end{align*}
where $\sigma - 1$ and $N$ denote precomposition with the corresponding map on $\integers G$, introduced above.

Note that a $\integers G$-module homomorphism $\integers G \rightarrow A$ is completely determined by the image of $1 = 1 e \in \integers G$ (in fact, by the image of $1 g$ for any $g \in G$), and that this element may be mapped to any $a \in A$. Hence
\begin{align*}
\Hom_{\integers G}(\integers G,A)
\isomorphism
A
\end{align*}
Using this identification, we may view the preceding cochain complex as
\begin{align}
0
\rightarrow
A
\xrightarrow{\sigma - 1}
A
\xrightarrow{N}
A
\xrightarrow{\sigma - 1}
\cdots%
\label{eq : K3 : Finite Cyclic Group Cochain Complex Intermediate}
\end{align}
We compute
\begin{align*}
\ker(\sigma - 1)
&\isomorphism
A^{G}
&
\ker N
&\isomorphism
{}_{N}A
\\
\image N
&=
N A
&
\image(\sigma - 1)
&=
(\sigma - 1) A
\end{align*}
where
\begin{align*}
A^{G}
=
\left\{a \in A \st \forall g \in G,g a = a\right\}
\end{align*}
is the subgroup of $A$ fixed pointwise by all elements $g \in G$, and
\begin{align*}
{}_{N}A
=
\left\{a \in A \st N a = 0\right\}
\end{align*}
is the subgroup of $A$ annihilated by $N$. Computing the cohomology at each module (after the initial $0$) in \eqref{eq : K3 : Finite Cyclic Group Cochain Complex Intermediate}, we get the results in \eqref{eq : K3 : Cohomology Finite Cyclic Group}.
\end{proof}

We apply Lemma \ref{lem : K3 : Cohomology Finite Cyclic Group} to two particular finite subgroups of $\generalLinear(2,\integers)$. The subgroup in Corollary \ref{cor : K3 : Cohomology 100-1} will be relevant to our construction of an algebraic K3 surface over $\rationals$ with geometric Picard rank $2$ whose algebraic Brauer group modulo constants is isomorphic to $\integers/2 \integers$. The subgroup in Corollary \ref{cor : K3 : Cohomology 0110} illuminates our understanding of $\integers$- and $\rationals$-conjugacy.% Begin footnote.
\footnote{See Example \ref{exm : Cyclic Groups Order 2}.}

\begin{corollary}
\label{cor : K3 : Cohomology 100-1}
Let $G \subgroupeq \generalLinear(2,\integers)$ be the subgroup generated by the matrix
\begin{align*}
\begin{pmatrix}
1	&	0	\\
0	&	-1
\end{pmatrix}
\end{align*}
Let $G$ act on $\integers^{2}$ by left-multiplication. Then $\cohomology[1]{G,\integers^{2}} \isomorphism \integers/2 \integers$.
\end{corollary}

\begin{proof}
Note that $G$ is cyclic of order $2$. In the setting of Lemma \ref{lem : K3 : Cohomology Finite Cyclic Group}, take
\begin{align*}
\sigma
&=
\begin{pmatrix}
1	&	0	\\
0	&	-1
\end{pmatrix}
&
G
&=
\langle \sigma \rangle
&
A
&=
\integers^{2}
\end{align*}
Then
\begin{align*}
\ker(\sigma - 1)
&\isomorphism
A^{G}
=
\Span
\begin{pmatrix}
1	\\
0
\end{pmatrix}
&
\ker N
&\isomorphism
{}_{N} A
=
\Span
\begin{pmatrix}
0	\\
1
\end{pmatrix}
\\
\image N
&=
N A
=
2 \Span
\begin{pmatrix}
1	\\
0
\end{pmatrix}
&
\image(\sigma - 1)
&=
(\sigma - 1) A
=
2 \Span
\begin{pmatrix}
0	\\
1
\end{pmatrix}
\end{align*}
so
\begin{align*}
\cohomology[i]{G,\integers^{2}}
\isomorphism
\begin{dcases*}
\integers			&	if $i = 0$	\\
\integers/2 \integers	&	if $i \in \integersPositive$
\end{dcases*}
\end{align*}
In particular, $\cohomology[1]{G,\integers^{2}} \isomorphism \integers/2 \integers$.
\end{proof}



\begin{corollary}
\label{cor : K3 : Cohomology 0110}
Let $G \subgroupeq \generalLinear(2,\integers)$ be the subgroup generated by the matrix
\begin{align*}
\begin{pmatrix}
0	&	1	\\
1	&	0
\end{pmatrix}
\end{align*}
Let $G$ act on $\integers^{2}$ by left-multiplication. Then $\cohomology[1]{G,\integers^{2}} \isomorphism 0$.
\end{corollary}

\begin{proof}
Note that $G$ is cyclic of order $2$. In the setting of Lemma \ref{lem : K3 : Cohomology Finite Cyclic Group}, take
\begin{align*}
\sigma
&=
\begin{pmatrix}
0	&	1	\\
1	&	0
\end{pmatrix}
&
G
&=
\langle \sigma \rangle
&
A
&=
\integers^{2}
\end{align*}
Then
\begin{align*}
\ker(\sigma - 1)
&\isomorphism
A^{G}
=
\Span
\begin{pmatrix}
1	\\
1
\end{pmatrix}
&
\ker N
&\isomorphism
{}_{N} A
=
\Span
\begin{pmatrix}
1	\\
-1
\end{pmatrix}
\\
\image N
&=
N A
=
\Span
\begin{pmatrix}
1	\\
1
\end{pmatrix}
&
\image(\sigma - 1)
&=
(\sigma - 1) A
=
\Span
\begin{pmatrix}
1	\\
-1
\end{pmatrix}
\end{align*}
so
\begin{align*}
\cohomology[i]{G,\integers^{2}}
\isomorphism
\begin{dcases*}
\integers	&	if $i = 0$	\\
0		&	if $i \in \integersPositive$
\end{dcases*}
\end{align*}
In particular, $\cohomology[1]{G,\integers^{2}} \isomorphism 0$.
\end{proof}



\section{Results: Explicit {K}3 surfaces}

In this section we construct an algebraic K3 surface $X$ over $\rationals$ with geometric Picard rank $2$, whose algebraic Brauer group modulo constants is isomorphic to $\integers/2 \integers$, the only nontrivial candidate target group in that rank.

Our approach constructs $X$ as a quartic in $\projective^{3}_{\rationals}$ around a curve $E$ defined over a quadratic extension of $\rationals$. Under the intuition that the geometric Picard rank of $X$ ``wants'' to be as small as possible, this curve $E$ represents the only element in $\picardGroup X_{\separableClosure{\rationals}}$ other than the hyperplane class. Hence $X$ has geometric Picard rank $2$. The natural action of the absolute Galois group fixes $H$ and sends $E$ to $-E$, so the image of the Galois representation on $\picardGroup X_{\separableClosure{\rationals}}$ is the subgroup $G$ in Corollary \ref{cor : K3 : Cohomology 100-1}. Hence the algebraic Brauer group of $X$ modulo constants is $\integers/2 \integers$, as desired.



\subsection{General construction}

Let $i$ be a root of an irreducible degree-$2$ polynomial over $\rationals$---for example, the polynomial $t^{2} + 1$. Let $g_{1} \in \rationals[x_{0},\ldots,x_{3}]$ and $g_{2} \in \rationals(i)[x_{0},\ldots,x_{3}]$ be homogeneous quadratic polynomials, where at least one coefficient of $g_{2}$ involves $i$ nontrivially. In general, the curve
\begin{align*}
E
\defeq
\zeroSet(g_{1}) \cap \zeroSet(g_{2})
\end{align*}
in $\projective^{3}_{\rationals(i)}$ is smooth with arithmetic genus $1$.

Let $h_{1} \in \rationals[x_{0},\ldots,x_{3}]$ be another homogeneous quadratic polynomial, let
\begin{align*}
f
\defeq
g_{1} h_{1} + \normMap_{\rationals(i) \fieldExtension \rationals}(g_{2})
\end{align*}
and let
\begin{align*}
X
\defeq
\zeroSet(f)
\end{align*}
For a sufficiently general choice of $h_{1}$, (i) the polynomial $f$ is nonsingular, so $X$ is smooth and hence an algebraic K3 surface over $\rationals$; and (ii) the geometric Picard rank of $X$ is as small as possible, namely, $2$.

To see this last assertion, recall that $\picardGroup X$ always contains a hyperplane class, denote it $H$. By construction, $X$ contains the curve $E$, which is not linearly equivalent to any multiple of $H$, as may be verified by showing that the determinant of their intersection matrix is nonzero. More precisely, we may compute intersection numbers with $H$ by noting that $H$ embeds $X$ into $\projective^{3}_{\rationals}$, so
\begin{align*}
(H,H)
&=
\deg H
=
4
&
(H,E)
&=
\deg E
=
4
\end{align*}
Using the adjunction formula, we compute the self-intersection number of $E$ to be% Begin footnote.
\footnote{Recall that $X$ is an algebraic K3 surface, so by definition, $K_{X} = 0$; and $E$ has arithmetic genus $1$.}% End footnote.
\begin{align*}
(E,E)
=
(E,E + K_{X})
=
2 \arithmeticGenus(E) - 2
=
0
\end{align*}
Hence the intersection matrix for $H,E$ is
\begin{align*}
\begin{pmatrix}
4	&	4	\\
4	&	0
\end{pmatrix}
\end{align*}
which has determinant $-16 \neq 0$. Hence $\picardGroup X_{\separableClosure{\rationals}}$ has at least rank $2$.

Consider the natural action of the Galois group $\galoisGroup(\rationals(i) \fieldExtension \rationals)$, which is isomorphic to the cyclic group of order $2$, on $\picardGroup X_{\rationals(i)}$. The action of the nontrivial element $\sigma \in \galoisGroup(\rationals(i) \fieldExtension \rationals)$ on the generators $E,H$ of $\picardGroup X_{\rationals(i)}$ is
\begin{align}
\sigma(H)
=
H
&&
\sigma(E)
=
\zeroSet(g_{1}) \cap \zeroSet(\sigma(g_{2}))%
\label{eq : K3 : GC sigma H}
\end{align}
Because
\begin{align*}
X \cap \zeroSet\left(\normMap_{\rationals[i] \fieldExtension \rationals}(g_{2})\right)
=
E \cup \sigma(E)
\end{align*}
it follows that $2 H$ is linearly equivalent to $E + \sigma(E)$, so that
\begin{align}
\sigma(E)
=
2 H - E%
\label{eq : K3 : GC sigma E}
\end{align}
In particular, in $\picardGroup X_{\separableClosure{\rationals}}$,
\begin{align*}
\sigma(E - H)
=
\sigma(E) + \sigma(H)
=
(2 H - E) + H
=
H - E
=
-(E - H)
\end{align*}

Assuming that the geometric Picard rank of $X$ is $2$,% Begin footnote.
\footnote{We will prove this using Theorem \ref{thm : Background : Bound From Frobenius} for the specific algebraic K3 surface in Theorem \ref{thm : K3 : Specific K3 ABG Z/2Z}.} % End footnote.
the list $H,E - H$ is a basis of $\picardGroup X_{\separableClosure{\rationals}}$. The preceding computations show that this choice of basis identifies the nontrivial Galois element $\sigma \in \galoisGroup(\rationals(i) \fieldExtension \rationals)$ with the matrix
\begin{align*}
\begin{pmatrix}
1	&	0	\\
0	&	-1
\end{pmatrix}
\end{align*}
Therefore
\begin{align*}
\brauerGroupAlgebraic X/\brauerGroupConstant X
\isomorphism
\cohomology[1]{\galoisGroup(\separableClosure{\rationals} \fieldExtension \rationals),\picardGroup X_{\separableClosure{\rationals}}}
\isomorphism
\cohomology[1]{\galoisGroup(\rationals(i) \fieldExtension \rationals),\picardGroup X_{\rationals(i)}}
\isomorphism
\integers/2 \integers
\end{align*}
as desired.



\subsection{Specific example}

The following theorem gives explicit equations for an algebraic K3 surface over $\rationals$, following the general construction above.

\begin{theorem}
\label{thm : K3 : Specific K3 ABG Z/2Z}
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}

\begin{proof}
Define the curves $H$ and $E$ on $X$ as explained in the general construction above. We must show that (i) the surface $X$ and the curves $H$ and $E$ are smooth, and (ii) that $H,E$ is a basis of $\picardGroup X_{\separableClosure{\rationals}}$.

Applying the Jacobian criterion to the equations defining $X$, $H$, and $E$, we verify that $X$, $H$, and $E$ are smooth.

To show that $H,E$ is a basis of $\picardGroup X_{\separableClosure{\rationals}}$, recall that in the general construction we have shown that $H,E$ is linearly independent, so $\picardGroup X_{\separableClosure{\rationals}} \geq 2$. To show that $\Rank(\picardGroup X_{\separableClosure{\rationals}}) \leq 2$, we use Theorem \ref{thm : Background : Bound From Frobenius}: Taking $q = 103$ and $\ell = 3$, we find the characteristic polynomial of $\Phi_{\mathcal{X}_{k}}^{*}$ has exactly two eigenvalues of the form $q$ times a root of unity.

It remains to show that the sublattice of $\picardGroup X_{\separableClosure{\rationals}}$ generated by $H,E$ is in fact the full geometric Picard group. We first show that $H$ is primitive in $\picardGroup X_{\separableClosure{\rationals}}$. Suppose there exist $H_{0} \in \picardGroup X_{\separableClosure{\rationals}}$ and $a \in \integers$ such that $H = a H_{0}$. Then
\begin{align*}
4
=
(H,H)
=
(a H_{0},a H_{0})
=
a^{2} (H_{0},H_{0})
\end{align*}
The intersection pairing on an algebraic K3 surface is integral and even, so $(H_{0},H_{0}) \in 2 \integers$. This forces $a = \pm{}1$. Hence $H$ is primitive in $\picardGroup X_{\separableClosure{\rationals}}$.

Because $H$ is primitive, we may extend $H$ to a basis $H,D$ of $\picardGroup X_{\separableClosure{\rationals}}$.% Begin footnote.
\footnote{This follows from Lemma \ref{lem : GLnZ : Pure Submodule}. In rank $2$, we may prove the claim using the (extended) Euclidean algorithm: Let $u = (u_{1},u_{2}) \in \integers^{2}$ be primitive. Then $\gcd(u_{1},u_{2}) = 1$. Hence there exist $v_{1},v_{2} \in \integers$ such that $u_{1} v_{2} - u_{2} v_{1} = 1$. Let $v = (v_{1},v_{2})$. Then
\begin{align*}
v_{2} (u_{1},u_{2}) - u_{2} (v_{1},v_{2})
&=
(1,0)
&
&\text{and}
&
-v_{1} (u_{1},u_{2}) + u_{1} (v_{1},v_{2})
&=
(0,1)
\end{align*}
Hence the sublattice of $\integers^{2}$ generated by $u,v$ is the full lattice, as desired.

For a statement of the analogous result in rank $n$, see Remark \ref{rmk : GLnZ : Pure Submodule Matrix Minors}.} % End footnote.
By definition, there exist $a,b \in \integers$ such that
\begin{align*}
E
=
a H + b D
\end{align*}
Using bilinearity of the intersection pairing, we compute% Begin footnote.
\footnote{In both computations that follow, we use that $(H,H) = 4$.}% End footnote.
\begin{align}
0
=
(E,E)
=
(a H + b D,a H + b D)
=
4 a^{2} + 2 a b (H,D) + b^{2} (D,D)%
\label{eq : K3 : (E,E) For E = a H + b D}
\end{align}
and
\begin{align*}
4
=
(H,E)
=
(H,a H + b D)
=
4 a + b (H,D)
\end{align*}
or equivalently,
\begin{align}
(H,D)
=
\frac{4 (1 - a)}{b}%
\label{eq : K3 : (H,D)}
\end{align}
Substituting \eqref{eq : K3 : (H,D)} into \eqref{eq : K3 : (E,E) For E = a H + b D} and solving for $(D,D)$, we get
\begin{align}
(D,D)
=
-\frac{4 a (2 - a)}{b^{2}}%
\label{eq : K3 : (D,D)}
\end{align}
Using these results, we may construct the intersection matrix for $H,D$,
\begin{align*}
\begin{pmatrix}
4				&	\frac{4 (1 - a)}{b}		\\
\frac{4 (1 - a)}{b}	&	-\frac{4 a (2 - a)}{b^{2}}
\end{pmatrix}
\end{align*}
which has determinant $-\frac{16}{b^{2}}$. Intersection numbers are integers, so this determinant is an integer, and hence $b \in \{\pm{}1,\pm{}2,\pm{}4\}$. Without loss of generality, we may assume that $b > 0$. (If $b < 0$, replace $D$ with $-D$.)

If $b = 1$, then $E = a H + D$, or equivalently, $D = -a H + E$. Thus $H,E$ is a basis for $\picardGroup X_{\separableClosure{\rationals}}$, as desired. We show that the other two possibilities for $b$ lead to a contradiction.

Suppose for the sake of contradiction that $b = 4$. First, by \eqref{eq : K3 : (D,D)}, if $b = 4$, then the self-intersection number for $D$ is
\begin{align*}
(D,D)
=
-\frac{a (2 - a)}{4}
\end{align*}
$(D,D)$ is an integer, so $a$ must be even. Second, if $b = 4$, then $E = a H + 4 D$. Using the Galois action on $H$ and $E$ computed in \eqref{eq : K3 : GC sigma H} and \eqref{eq : K3 : GC sigma E}, we compute
\begin{align*}
\sigma(E)
=
\sigma(a H + 4 D)
=
a \sigma(H) + 4 \sigma(D)
=
a H + 4 \sigma(D)
\end{align*}
and
\begin{align*}
\sigma(E)
=
2 H - E
=
2 H - (a H + 4 D)
=
(2 - a) H - 4 D
\end{align*}
Thus
\begin{align}
a H + 4 \sigma(D)
=
(2 - a) H - 4 D%
\label{eq : K3 : Example GPR 2 ABG Z/2Z : sigma E Equate}
\end{align}
Because $H,D$ is a basis of $\picardGroup X_{\separableClosure{\rationals}}$, there exist $a',b' \in \integers$ such that
\begin{align*}
\sigma(D)
=
a' H + b' D
\end{align*}
Substituting this into \eqref{eq : K3 : Example GPR 2 ABG Z/2Z : sigma E Equate} and simplifying, we get
\begin{align*}
2 (1 - a - 2 a') H - 4 (1 + b') D
=
0
\end{align*}
As a basis, $H,D$ is $\integers$-free, so this last equation implies that
\begin{align*}
2 (1 - a - 2 a')
&=
0
&
&\text{or equivalently}
&
a
=
1 - 2 a'
\end{align*}
Thus $a$ is odd, a contradiction.

Now suppose for the sake of contradiction that $b = 2$. First, note that the self-intersection computation in \eqref{eq : K3 : (E,E) For E = a H + b D} holds for any element of the geometric Picard group. When $b = 2$, this self-intersection number has the form
\begin{align*}
4 a^{2} + 4 a (H,D) + 4 (D,D)
\end{align*}
In particular, for all $C \in \picardGroup X_{\separableClosure{\rationals}}$, $4 \divides (C,C)$. If $C$ is a rational curve on $X$, then $\arithmeticGenus(C) = 0$, so by the adjunction formula,
\begin{align*}
(C,C)
=
(C,C + K_{X})
=
2 \arithmeticGenus(C) - 2
=
-2
\end{align*}
Because $4 \notDivides -2$, it follows that $X$ contains no rational curves.

Second, if $b = 2$, then
\begin{align*}
E
=
a H + 2 D
=
2 a' H + 2 D
\end{align*}
Thus we may define
\begin{align*}
E_{0}
=
a' H + D
\end{align*}
an element of $\picardGroup X_{\separableClosure{\rationals}}$ with the property that $2 E_{0} = E$. We compute
\begin{align*}
4
&=
(H,E)
=
(H,2 E_{0})
=
2 (H,E_{0})
&
&\Leftrightarrow
&
(H,E_{0})
&=
2
\end{align*}
and
\begin{align*}
0
&=
(E,E)
=
(2 E_{0},2 E_{0})
=
4 (E_{0},E_{0})
&
&\Leftrightarrow
&
(E_{0},E_{0})
&=
0
\end{align*}

Because $X$ contains no rational curves and $(E_{0},E_{0}) \geq 0$, either $E_{0}$ or $-E_{0}$ is nef.% Begin footnote.
\footnote{See Huybrechts \cite{Huybrechts2015}*{Chapter 2, Corollary 1.5}.} % End footnote.
Because $E$ is effective, $E_{0} = \frac{1}{2} E$ is nef. Hence there exist a smooth, irreducible elliptic curve $E' \in \picardGroup X_{\separableClosure{\rationals}}$ and $m \in \integersPositive$ such that $E_{0} = m E'$.% Begin footnote.
\footnote{See Huybrechts \cite{Huybrechts2015}*{Chapter 2, Proposition 3.10}.} % End footnote.
Hence
\begin{align*}
m \deg E'
=
m (H,E')
=
(H,m E')
=
(H,E_{0})
=
2
\end{align*}
so
\begin{align*}
\deg E'
=
\frac{2}{m}
\leq
2
\end{align*}
That is, $H$ embeds the elliptic curve $E'$, a curve of genus $1$, into $\projective^{3}_{\separableClosure{\rationals}}$ as a curve of degree less than or equal to $2$, a contradiction.
\end{proof}