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\title{The theorem of the cube}
\author{Pedro A. Castillejo}
\date{27.05.2014}

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\begin{document}

\maketitle

\begin{abstract}
In this talk of the seminar Abelian Varieties (FU Berlin, SS 2014) we present the theorem of the cube, a purely algebraic result that results very important in the study of Abelian Varieties. Following the notes \textit{Abelian Varieties}, by Gerard van der Geer and Ben Moonen, we complete and correct some of the proofs and we try to prove the theorem. There is a missing step in the proof that I was not able to understand.

We offer a section of prerequisites to make the script as self-contained as possible, and we discuss some important corollaries of the theorem. 

\end{abstract}

\section{Prerequisites}

Here we write some important definitions and facts (all of them with proof or an exact reference) that will be later used.

\begin{definition} Let $f: X \rightarrow Y$ be a morphism of schemes.
\begin{enumerate}
\item $f$ is \textbf{projective} if it factors into a closed immersion $i: X \rightarrow \mathbb{P}^n_Y$ for some $n$, followed by the projection $\mathbb{P}^n_Y \rightarrow Y$, where we define $\mathbb{P}^n_Y$ as $\mathbb{P}^n_{\mathbb{Z}} \times_{\mathbb{Z}} Y$.
\item $f$ is \textbf{locally of finite type} if there exists a covering of $Y$ by open affine subsets $V_i = \spec(B_i)$ such that for each $i$, $f^{-1}(V_i)$ can be covered by open affine subsets $U_{ij} = \spec(A_{ij})$, where each $A_{ij}$ is a finitely generated $B_i$-algebra. $f$ is of \textbf{finite type} if in addition each $f^{-1}(V_i)$ can be covered by a finite number of the $U_{ij}$.
\item $f$ is \textbf{separated} if the diagonal morphism $\Delta: X \rightarrow X \times_Y X$ is a closed immersion. We also say that $X$ is separated over $Y$.
\item $f$ is \textbf{closed} if the image of any closed subset is closed. $f$ is \textbf{universally closed} if it is closed, and for any morphism $Y' \rightarrow Y$, the extension $f': X' \rightarrow Y'$ is also closed.
\item $f$ is \textbf{proper} if it separated, of finite type, and universally closed.

\end{enumerate}
\end{definition}

\begin{proposition} We have that proper morphisms are stable under base extension (see for example Hartshorne, Cor. II.4.8 c)).
\end{proposition}

\begin{definition} Let $y \in Y$, and let $k(y)$ be the residue field of $y$, $\spec(k(y)) \rightarrow Y$ the natural morphism. Then the \textbf{fibre} of the morphism $f$ over the point $y$ is the scheme $X \times_Y \spec(k(y))$.
\end{definition}

\begin{remark} As topological spaces, $X_y \cong f^{-1}(y)$.

We can see the morphism $f: X \rightarrow Y$ as a family of schemes of the shape $X_y$ parametrized by the points of $Y$.
\end{remark}

\begin{remark} Given $X$, $Y$ $S$-schemes, then we can consider $X \times_S Y$. Given $y \in Y$, we can define the subscheme $X_y$ as $X \times_S \spec(k(y))$, where $k(y)$ is the residue field. Caution: Note that this \textit{is not} the fibre of any morphism, since we don't have in general a morphism $X \rightarrow Y$. This has to be seen as the topological subset $X \times_S \{y\}$, since they are isomorphic as topological spaces.  \textit{But} (mixing a little bit the notations), we have that the fibre of the projection $p: X \times_S Y \rightarrow Y$ over the point $y$ is just $(X \times_S Y) \times_Y \spec(k(y)) \cong X \times_S \spec(k(y))$. Therefore this abuse of notation should be read as $(X \times_S Y)_y$, and we will also call this the \textbf{fibre} of $X \times_S Y$ over the point $y \in Y$, and will denote it by $X_y$ when there is no risk of confusion.

Note that here we have the natural morphism $id \times \iota: X_y \rightarrow X \times_S Y$ coming from $\iota: \spec(k(y)) \rightarrow Y$. Then, if we have a line bundle $L$ over $X \times_S Y$, we will denote $L_y$ for the restriction $L_{|X \times \{y\}}$, i.e. for the pullback $(id \times \iota)^*L$.
\end{remark}

We will be also interested in the fibre of line bundles, specially in the case of line bundles defined over a product variety $X \times Y$.

\begin{definition}
Let $X \times Y$ be a variety, and let $x \in X$. Let $L$ be a line bundle over $X \times Y$. Then the \textbf{fibre} of $L$ over the point $x$ is the restriction $L_{|\{x\} \times Y}$, i.e. the pullback $(\iota \times id)^*L$, where $\iota \times id: \{ x \} \times Y \rightarrow X \times Y$ is the obvious morphism. When there is no risk of confusion, we will denote $L_x = L_{| \{x\}\times Y}$, and $L_y = L_{|X \times \{y\} }$.
\end{definition}

\begin{definition} Let $k$ be a field. A $k$-scheme is \textbf{geometrically integral} if for some algebraically closed field $K$ containing $k$ the scheme $X_K := X \times_k K$ is integral, i.e. irreducible and reduced. By EGA IV, if it holds for some alg. closed field $K$ over $k$, then $X_K$ is integral for every field $K$ containing $k$.

A \textbf{$k$-variety} $X$ is a $k$-scheme of finite type and geometrically integral. If it is proper over $k$, we say that $X$ is \textbf{complete}.
\end{definition}

\begin{remark}
Let $X$ and $Y$ be $k$-varieties, with $X$ complete. Then the fibre $X_y$ (coming from $X \times_k Y$) is also complete (over $k(y)$). Indeed, here $X_y = X \times_k \spec(k(y))$, and we have that $X$ is of finite type over $k$, so it will also be of finite type over $k(y)$ (note that this is just a finite extension of $k$, and if $A_{ij}$ is a finitely generated $k$-algebra, then of course $A_{ij} \otimes_k k(y)$ is a finitely generated $k(y)$-algebra), and it is also geometrically closed, because if $K$ is an alg. closed overfield of $k(y)$, then it is also an overfield of $k$. (c.f. Qing Liu, Ex. III.2.14) Finally, since properness is stable under base change (c.f. Prop. 1), then $X_y$ is proper over $k(y)$.
\end{remark}

\begin{proposition} We have that if $X$ is proper over a ring $A$, then the ring of global sections $\mathcal{O}(X)$ is integral over $A$ (c.f. Qing Liu, Prop. 3.18 of chapter 3). As a corollary we obtain that if $X$ is complete over a (not necessarily alg. closed) field $k$, then $\mathcal{O}(X)=k$ (Qing Liu, Cor. 3.21).
\end{proposition}

\begin{proposition}\label{trivial invertible sheaf}
Let $X$ be a complete variety. An invertible sheaf $L$ over $X$ is trivial if and only if both it and its dual have non trivial global sections, i.e. $H^0(M) \neq 0 \neq H^0(M^{-1})$. (Milne, AG, Prop. 13.3)
\end{proposition}



\begin{definition}
Let $f: X \rightarrow Y$ be a morphism of schemes, and let $\mathcal{F}$ be an $\mathcal{O}_X$-module. We say that $\mathcal{F}$ is \textbf{flat} over $Y$ at a point $x \in X$, if the stalk $\mathcal{F}_x$ is a flat $\mathcal{O}_{Y, f(x)}$-module. We say that $\mathcal{F}$ is flat over $Y$ if it is flat on every point of $X$, and we say that $X$ is flat over $Y$ if $\mathcal{O}_X$ is.
\end{definition}

\begin{remark}
Note that with this definition, it is clear that locally free sheaves are flat.
\end{remark}

\begin{proposition}
Flatness is stable under base extension. (c.f. Hartshorne, III.9.2)
\end{proposition}

\begin{proposition}\label{cohomology commutes with base extension}
Let $f: X \rightarrow Y$ be a separated morphism of finite type of noetherian schemes, and let $\mathcal{F}$ be a quasicoherent sheaf on $X$. Let $u: Y' \rightarrow Y$ be a flat morphism of noetherian schemes.
$$
\xymatrix{
X' \ar[d]^{g} \ar[r]^{v}  & X  \ar[d]^{f} \\
Y' \ar[r]^u & Y}
$$
Then (c.f. Hartshorne, III.9.3) for all $i \geq 0$ there are natural isomorphisms
$$ u^*R^if_*(\mathcal{F}) \cong R^ig_*(v^* \mathcal{F}).$$

\end{proposition}

\begin{proposition}[Künneth formula]
If $X, Y$ are locally noetherian schemes of finite type over $k$, then
$$ H^n( X \times Y, \O_{X \times Y}) \cong \bigoplus_{i+j = n} H^i(X, \O_X) \otimes H^j(Y, \O_Y)$$
(c.f. EGA III$_{2}$, Theorem 6.7.8)
\end{proposition}

\begin{definition}
Let $Y$ be a top. space. A function $\phi: Y \rightarrow \mathbb{Z}$ is \textbf{upper semicontinuous} if for each $y \in Y$, there is an open neighbourhood $U$ of $y$ st for all $y' \in U$, $\phi(y') \leq \phi(y)$. Intuitively: this means that $\phi$ gets bigger in special points.
\end{definition}

For example, given a curve $X$, define $\phi(x) = 0$ if $x$ is regular or generic, and $\phi(x) = 1$ if it is singular. Then $\phi$ is upper semicontinuous, because we know that the set of singular points is closed.

\begin{theorem}[Semicontinuity] Let $f: X \rightarrow Y$ be a projective morphism of noetherian schemes, and let $\mathcal{F}$ be a coherent sheaf on $X$, flat over $Y$. Then for each $i \geq 0$, the function
$$ h^i(y, \mathcal{F}) = \dim_{k(y)} H^i(X_y, \mathcal{F}_y)$$
is an upper semicontinuous functions on $Y$. In particular, the set $\{ y \in Y| h^i(y, \mathcal{F}) \geq n\}$ is closed for each $n$ (see Hartshorne, III.12.8).
\end{theorem}

\begin{corollary}[Grauert]
With the same hypotheses as the theorem, suppose furthermore that $Y$ is integral, and that for some $i$, the function $h^i(y, \mathcal{F})$ is constant on $Y$. Then $R^if_*(\mathcal{F})$ is locally free on $Y$, and for every $y$ the natural map
$$ R^if_*(\mathcal{F}) \otimes k(y) \rightarrow H^i(X_y, \mathcal{F}_y)$$
is an isomorphism, where $R^if_*(\mathcal{F})$ is the sheaf associated to the presheaf $V \mapsto H^i(f^{-1}(V), \mathcal{F}_{|f^{-1}(V)})$.
\end{corollary}





\section{The theorem of the cube}

The aim of this section is to clarify the contents of the notes as much as possible.

\begin{theorem}
Let $X$ and $Y$ be varieties. Suppose $X$ is complete. Let $L$ and $M$ be two line bundles on $X \times Y$. If for all closed points $y \in Y$ we have $L_y \cong M_y$, there exists a line bundle $N$ on $Y$ such that $L \cong M \otimes p^*N$, where $p= pr_Y: X \times Y \rightarrow Y$ is the projection.

\end{theorem}

\begin{proof}
We have that $L_y \otimes M_y^{-1}$ is the trivial bundle of $X_y$. By remark 3, $X_y$ is complete, so we can apply prop. 2 to conclude that $H^0(X_y, L_y \otimes M_y^{-1}) \cong k(y)$.

By Grauert's corollary, this implies that $p_*(L \times M^{-1}) \otimes k(y)$ is isomorphic to $k(y)$, and therefore $p_*(L \otimes M^{-1})$ is a line bundle over $Y$ which we denote $N$. Note that we just now that it has rank 1, so it may not be trivial. Indeed, in general, it will not be trivial.

So if we prove that $L \otimes M^{-1} \cong p^*N$, we will be done. For this, we will show that the natural map
$$ \alpha : p^*p_* (L \otimes M^{-1}) \rightarrow L \otimes M^{-1}$$
is an isomorphism.

For this we proceed in two steps: first we restrict $\alpha$ to the fibres $X_y$ (where $y\in Y$ is a closed point) and see that there is trivial, and secondly we see that this implies that $L \otimes M^{-1} \cong p^*(p_*(L\otimes M^{-1}))$. Caution: note that this is different from saying that if two line bundles are isomorphic in the fibres, then they are isomorphic: first we need to fix a morphism, and afterwards we have to check that this morphism is an isomorphism in the fibres (as we will do now).

So consider the commutative diagram:
$$
\xymatrix{
X_y \ar[d]^{\pi_{y}} \ar[r]^-{\phi}  & X \times Y \ar[d]^{p} \\
\spec(k(y)) \ar@{^{(}->}[r]^-\iota & Y}
$$
where $\phi = id_{X \times Y} \times \iota$. How does $\alpha$ looks like in the fibre? Note that since the diagram commutes, $\phi^*p^*p_* (L \otimes M^{-1}) = \pi_y^* \iota^* p_* (L \otimes M^{-1})$, and by Prop. \ref{cohomology commutes with base extension}, $\iota^*p_* (L \otimes M^{-1}) \cong \pi_{y*} \phi^* (L \otimes M^{-1})$. Hence the restriction is $\phi^*(\alpha): \pi_y^* \pi_{y*} (L \otimes M^{-1})_{|y} \rightarrow (L \otimes M^{-1})_{|y}$.

Note that the pushforward is by definition $\pi_{y*}(L \otimes M^{-1}) (U) = H^0(X_y, L_y \otimes M^{-1}_y)$ for every non empty $U$, since we are going to a single point. But we have already seen that this is isomorphic to $k(y)$, so $\pi_y^* \pi_{y*} (L \otimes M^{-1}) \cong \pi_y^* \mathcal{O}_{k(y)} \cong \mathcal{O}_{X_y} \otimes_{\mathcal{O}_{X_y}} \mathcal{O}_{X_y} \cong \mathcal{O}_{X_y}$.

By hypothesis, we have that $(L \otimes M^{-1})_{|y} \cong \mathcal{O}_{X_y}$, so writing everything together we have that $\phi^*(\alpha)$ is an isomorphism.

We now go for the second step. It is enough to show that given a morphism $f: \mathcal{E} \rightarrow \mathcal{O}_{X \times Y}$ s.t. $f_y$ is an isomorphism for every closed point $y$, then $\mathcal{E}$ is trivial.

Let $\tilde{x} \in X \times Y$ be a closed point. Since we can see $(X \times Y)$ as an algebraic family of the fibres $X_y = (X \times Y)_y$ parametrized by $Y$, there is a closed point $y \in Y$ s.t. $x$ "lies" in $X_y$, so to say, that there exists a point that maps to $\tilde{x}$ via $\phi: X_y \rightarrow X \times Y$. We call this point $x$.

We want to show that $\mathcal{E}_{\tilde{x}} = 0$. We have that 
$$ 0 = (\mathcal{E}_y)_x = (\phi^*\mathcal{E})_x = (\phi^{-1} \mathcal{E} \otimes_{\phi^{-1}\mathcal{O}_{X \times Y}} \mathcal{O}_{X_y})_x = \mathcal{E}_{\tilde{x}} \otimes_{\mathcal{O}_{X \times Y, \tilde{x}}} \mathcal{O}_{X_y, x}
$$
and if $I$ is the kernel of $\O_{X \times Y, \tilde{x}} \rightarrow (\phi_* \O_{X_y})_{\tilde{x}} \cong \O_{X_y, x}$, then we have
$$0 = \mathcal{E}_{\tilde{x}} \otimes_{\mathcal{O}_{X \times Y, \tilde{x}}} \mathcal{O}_{X_y, x} \cong \mathcal{E}_{\tilde{x}} \otimes_{\mathcal{O}_{X \times Y, \tilde{x}}} \mathcal{O}_{X \times Y, \tilde{x}}/I \cong \mathcal{E}_{\tilde{x}}/ I \mathcal{E}_{\tilde{x}}$$
and Nakayama's lemma implies that $\mathcal{E}_{\tilde{x}} =0$.

\end{proof}

This theorem induces an equivalence relation in the set $\Pic(X\times Y)$, and therefore induces the following

\begin{definition}
We say that a line bundle $L$ over $X \times Y$ is \textbf{trivial (in the fibres) over $Y$} if it is the pullback of a line bundle on $Y$. This is equivalent to being trivial (in the usual sense) over $Y$, i.e. that $L_y \cong \O_{X_y}$. We say that two line bundles are \textbf{isomorphic (in the fibres) over $Y$} if $L \otimes M^{-1}$ is trivial in the fibres.
\end{definition}

\begin{theorem}[See-saw Principle] Let $X$ and $Y$ be as above, and $L, M$ two line bundles on $X \times Y$ isomorphic over $Y$. If in addition we have $L_x = M_x$ for some $x \in X(k)$, then $L \cong M$, i.e. $L$ and $M$ are isomorphic in the usual sense.

\end{theorem}

\begin{proof}
We have that $L \cong M \otimes pr_Y^* N$ for some line bundle $N$ over $Y$. If we restrict to $\{ x \} \times Y$, we have that $M_x = L_x \cong M_x \otimes (pr_Y^* N)_x$, so $(pr_Y^*N)_x$ is trivial. Hence, $N$ is trivial, because we have the following commutative diagram:
$$
\xymatrix{
Y \ar@{.>}[dr]|f \ar[drr]^{i} \ar[ddr]_{\psi} \\
& Y_x \ar[d]^{\pi_{x}} \ar[r]_{\phi}  & X \times Y \ar[d]^{pr_X} \\
&\spec(k(x)) \ar@{^{(}->}[r]^-\iota & X}
$$
where $\psi$ is just the structure morphism of $Y$ (note that $x$ is $k$-rational!), and $i$ arises from the fibre product of $\iota \circ \psi$ and the identity on $Y$.
Since the diagram is commutative, we have that $id_Y = pr_Y \circ \phi \circ f$, and $f^* \O_{Y_x} = \O_Y$, so
$$ \O_Y = f^* \O_{Y_x} \cong f^* (\phi^* pr_Y^*N) = id_Y^* N = N $$
and therefore $N$ is trivial. And we are done, because $pr_Y^* \O_Y \cong \O_{X \times Y} \Rightarrow L \cong M$.

\end{proof}

Now we want to prove the main theorem of the talk, the theorem of the cube. It says (under certain conditions) that given a line bundle $L$ over $X \times Y \times Z$, if it is trivial over $\{ x \} \times Y \times Z$, $X \times \{y \} \times Z$ and $X \times Y \times \{z \}$ for some $k$-rational points $x, y$ and $z$, then the line bundle $L$ is trivial.

Before we prove (and state correctly) this, we need a lemma.

\begin{lemma}
Let $X$ and $Y$ be varieties, with $X$ complete. For a line bundle $L$ on $X \times Y$, the set $\{ y \in Y| L_y \mbox{ is trivial over } Y \}$ is closed in $Y$.
\end{lemma}

\begin{proof}
By Prop. \ref{trivial invertible sheaf}, we have that
$$ \{ y \in Y| L_y \mbox{ is trivial} \} = \{ y \in Y| h^0(L_y) > 0 \} \cap \{ y \in Y| h^0(L_y^{-1}) > 0 \}. $$
Since the functions $y \mapsto h^0(L_y)$ and $y \mapsto h^0(L^{-1})$ are upper semi-continuous (c.f. Semicontinuity theorem), the sets on the right hand side are closed and we are done.

\end{proof}

There is a refinement of this lemma:

\begin{lemma}
Let $X$ be a complete variety over $k$, $Y$ a $k$-scheme, and let $L$ be a line bundle on $X \times Y$. Then there exists a closed subscheme $ Y_0 \hookrightarrow Y$ which is the  maximal subscheme of $Y$ over which $L$ is trivial. More concretely:
\begin{enumerate}
\item Triviality: the restriction of $L$ to $X \times Y_0$ is the pullback of a line bundle on $Y_0$.
\item Maximality: for every morphism $\phi : Z \rightarrow Y$ such that $(id_X \times \phi)^*(L)$ is trivial over $Z$, then $\phi$ factors through $Y_0$. Note than in particular, if we take $Z$ as closed subschemes of $Y$, we are asking for $L$ to be trivial over this closed subscheme, so this $Y_0$ is the maximal one.

\end{enumerate}

\end{lemma}

Now we go for the big theorem:

\begin{theorem}
Let $X$ and $Y$ be complete varieties and let $Z$ be a connected, locally noetherian scheme. Let $x \in X(k)$, $y \in Y(k)$ and let $z$ be a point of $Z$. If $L$ is a line bundle on $X \times Y \times Z$ whose restriction to $\{ x \} \times Y \times Z$, $X \times \{y \} \times Z$ and to $X \times Y \times \{z \}$ is trivial, then $L$ is trivial.

\end{theorem}

\begin{proof}
Since the projection $X \times Y \times Z \rightarrow Z$ is flat, we can see $L$ as a family of line bundles parametrized by $Z$. Now let $Z'$ be the maximal closed subscheme of $Z$ over which $L$ is trivial, as in the lemma. By the maximality condition, taking $z \hookrightarrow Z$, then this factorizes through $j: Z'\hookrightarrow Z$: in other words, $z \in Z'$, so $Z'$ is non empty. We want to show that $Z' = Z$, and for this it will be enough to show that it is open (because $Z$ is connected and $Z'$ is closed and non-empty).

Let $\zeta$ be a point of $Z'$. Let $\mathfrak{m}$ be the maximal ideal of $\O_{Z, \zeta}$ and $I \subset \O_{Z, \zeta}$ be the ideal defining the germ of $Z'$, i.e. the kernel of $\O_{Z, \zeta} \rightarrow (j_*\O_{Z'})_\zeta$. Note that $I=(0)$ if and only if there is an open subset $\zeta \in V \subset Z$ s.t. $V \subset Z'$, because of the exact sequence $0 \rightarrow i_{!} (\O_U) \rightarrow \O_Z \rightarrow j_* \O_{Z'} \rightarrow 0$, where $i : U \subset Z$ is the complement of $Z'$. (c.f. Hartshorne, Ex. II.1.19) Therefore if we prove that $(I)$ is zero, then $\O_Z \cong j_* \O_{Z'}$, and since $Z'$ is a closed subscheme of $Z$, they must be equal and we will be done.

So assume that $I \neq (0)$. $\O_{Z, \zeta}$ is noetherian because $Z$ is locally noetherian, so $\cap_n \mathfrak{m}^n = (0)$. Let $n$ be the natural number s.t. $I \subset \mathfrak{m}^n$ and $I \nsubseteq \mathfrak{m}^{n+1}$. Write $a_1 = (I, \mathfrak{m}^{n+1})$. Claim: there exists an ideal $a_2$ s.t. $\mathfrak{m}^{n+1} \subset a_2 \subset a_1 \subset \mathfrak{m}^n$ and $\dim_{k(\zeta )}(a_1/a_2) = 1$. Indeed, if we quotient with $\mathfrak{m}^{n+1}$ we have the chain of $k(\zeta)$-vector spaces $0 \subset a_1 \subset \mathfrak{m}^n/\mathfrak{m}^{n+1}$, and since $I \nsubseteq \mathfrak{m}^{n+1}$, the dimension of $a_1/\mathfrak{m}^{n+1}$ is greater than zero. Hence, taking a subspace $\bar{a_2} \subset a_1/\mathfrak{m}^{n+1}$ of codimension one and going back to the chain of $\O_{Z, \zeta}$-modules, we obtain such an $a_2$. By construction, $I \nsubseteq a_2$.

Let $Z_i \subset \spec(\O_{Z, \zeta})$ be the closed subschemes defined by $a_i$, which topologically they are both just $\mathfrak{m}$, since $\bar{\mathfrak{p}} \in \spec( \O_{Z, \zeta}/a_i)$ iff  $\mathfrak{p} \supset a_i \supset \mathfrak{m}^{n+1}$ iff $\mathfrak{p} = \mathfrak{m}$. Let $\spec(A) \subset Z$ be an open subset of $Z$ containing $\zeta$. Note that $A_\zeta = \O_{Z, \zeta}$, so the localization $A \rightarrow A_\zeta$ induces a morphism $Z_i \subset \spec(\O_{Z, \zeta}) \rightarrow \spec(A) \hookrightarrow Z$ for each $i$. Let $\phi_i$ denote the composition $\phi_i: Z_i \hookrightarrow Z$. Since $I \subset a_1$, we have that $\phi_1$ factors through $Z'$, so the restriction of $L$ to $X \times Y \times Z_1$ is trivial. We will show now that the restriction of $L$ to $X \times Y \times Z_2$ is trivial, which implies that $\phi_2$ factors also through $Z'$, but this implies that $I \subset a_2$, which is a contradiction, and therefore we will be done.

First we fix the notation. Let $L_i$ be the restriction $j_i^*(L) := (id_X \times id_Y \times \phi_i)^*L$. Take a trivializing global section $s \in \Gamma(L_1)$, i.e. a global section $s$ s.t. the morphism $\O_{X \times Y \times Z_1} \rightarrow L_1: 1 \mapsto s$ is actually an isomorphism (this can be done because $L_1$ is trivial!). The inclusion $Z_1 \subset Z_2$ induces $i_{12} : X \times Y \times Z_1 \rightarrow X \times Y \times Z_2$. We also have the commutative diagram
$$
\xymatrix{
X \times Y \times Z_1 \ar[d]_{i_{12}} \ar[dr]^{j_1} \\
X \times Y \times Z_2 \ar[r]_{j_2} & X \times Y \times Z}
$$
so we have the natural map $L_2 \rightarrow i_{12*}(i_{12}^*L_2) = i_{12*}L_1$ coming from the identity $i^*_{12} L_2 \rightarrow i^*_{12} L_2$ via the adjunction formula. (c.f. Hartshorne, p. 110) Now, taking global sections, we have a restriction map $\psi: \Gamma(L_2) \rightarrow \Gamma(L_1)$. Claim: $L_2$ is trivial if and only if $s$ can be lifted to a global section $s'$ of $L_2$.

Indeed, suppose first that we have a lift $s'$ of the trivializing global section $s \in \Gamma(L_1)$. Since the underlying point set of both $Z_1$ and $Z_2$ is the same, $X \times Y \times Z_i$ are homeomorphic. We know that if the image $s'(P)$ of $s'$ via $\O_{X \times Y \times Z_2} \rightarrow L_2 \rightarrow (L_2)_P \rightarrow (L_2)_P/\mathfrak{m}_P$, where $\mathfrak{m}_P$ is the maximal ideal of $(\O_{X \times Y \times Z_2})_P$, is zero, then also $s(P)$ is zero. Indeed, just look at the commutative diagram
$$
\xymatrix{
L_2 \ar[r] \ar[d] & i_{12*} L_1 \ar[d] \\
(L_2)_P \ar[d] \ar[r] & (i_{12*} L_1)_P \ar[d] \\
L_{2,P} \otimes_{\O_{X \times Y \times Z_2}} k(P)\ar[r] & (i_{12*} L_1)_P \otimes_{\O_{X \times Y \times Z_2}} k(P) }
$$
and note that in the last row, the tensor product with $k(P)$ is isomorphic to the quotient $L_{2,P}/ \mathfrak{m}_P L_{2, P}$, and something similar in the right hand side.

Now assume that $L_2$ is trivial. Then we have that the sequence with $L_2$ is just
$$ \O_{X \times Y \times Z_2} \rightarrow i_{12*} \O_{X \times Y \times Z_1} \rightarrow 0$$
and we can take global sections and use K\"unneth formula, i.e. $H^0(X \times Y \times Z_i, \O_{X \times Y \times Z_i}) \cong H^0(X \times Y, \O_{X \times Y}) \otimes H^0(Z_i, \O_{Z_i})$ and we have that the induced morphism is just
$$ id_{X \times Y} \otimes \pi: H^0(X \times Y, \O_{X \times Y}) \otimes_k H^0(Z_2, \O_{Z_2}) \rightarrow  H^0(X \times Y, \O_{X \times Y}) \otimes_k H^0(Z_1, \O_{Z_1})$$
where $\pi$ is just the projection given by $\O_{Z, \zeta}/a_1 \rightarrow \O_{Z, \zeta} / a_2$. Hence this is surjective, so we will be able to lift $s$.

So we reduced the problem to finding a lift $s'$ of $s$. For this, it is enough to show that $\Gamma(L_2) \rightarrow \Gamma(L_1)$ is surjective. From the exact sequence
$$ 0 \rightarrow \ker(\psi) \rightarrow L_2 \rightarrow i_{12*}L_1 \rightarrow 0 $$
it is enough to show that $H^1(X \times Y \times Z_2, \ker(\psi)) = 0$.

\textbf{The missing step:} For this, it is enough to show that $H^1(X \times Y, \O_{X \times Y}) = 0$.

Assumed that this is enough, let's prove that it is indeed equal to zero. Let $\xi \in H^1(X \times Y, \O_{X \times Y})$. By hypothesis, the restrictions of $L_2$ to $\{ x \} \times Y \times Z_2$ and $X \times \{y\} \times Z_2$ are trivial. As in the proof of the See-saw principle, since the points $x,y \in X(k)$, we find the maps $i_1 = (id_X, y) : X \hookrightarrow X \times Y$ and $i_2 = (x, id_Y): Y \hookrightarrow X \times Y$. But being trivial implies that $i_1^*( \xi) = 0 = i_2^*(\xi)$, because of the direction of the claim that we haven't proved and the missing step. (Sorry for that) Finally, since $X$ and $Y$ are complete, we have that $H^1(X, \O_X) \otimes H^0(X, \O_X) \cong H^1(X, \otimes \O_X)$, and analog with $\O_Y$. Hence the Künneth formula gives us the following isomorphism
$$ H^1(X \times Y, \O_{X \times Y}) \cong H^1(X, \O_X) \oplus H^1(Y, \O_Y)$$
so $\xi = 0$ and we are done (modulo the missing steps :P).

\end{proof}


\begin{remark}
In the theorem as stated we require $x$ and $y$ to be $k$-rational points of $X$ and $Y$, but we can generalize this. We used this fact in the definition of $i_1$ and $i_2$, but the theorem still holds without these assumptions. The point is that if $k \subset K$ is a field extension then a line bundle $M$ on a $k$-variety $V$ is trivial if and only if the line bundle $M_K$ on $V_K$ is trivial. (Exercise (2.1) of van der Geer and Moonen notes of AV)

\end{remark}

\begin{remark}
The analogous statement for line bundles on a product of two complete varieties is false in general, so to say, if $L$ is a line bundle on $X \times Y$, then we may have $L_x \cong \O_Y$ and $L_y \cong \O_X$, and still we will not have $L \cong \O_{X \times Y}$. For example, consider $X=Y$  an elliptic curve, and take the divisor
$$ D = \Delta_X - (\{ 0 \} \times X) - (X \times \{ 0\})$$
where $\Delta_X \subset X \times X$ is the diagonal. Then $L = \O_{X \times X}(D)$ restricts to the trivial bundle on $\{0\} \times X$ and $X \times \{0\}$, because the divisor $1 \cdot e_X$ is linearly equivalent (from the group law) to a divisor whose support doesn't contain $e_X$. But $L$ is not the trivial bundle: if it were, $L_{| \{P\} \times X} = \O_X(P - e_x) \cong \O_X$, and then we would have that $X$ is a rational curve, and we get a contradiction. (c.f. Hartshorne, Example II.6.10.1)

\end{remark}


\section{Consequences}

This theorem has a lot of consequences which will give us information about line bundles. First, recall that if $f(x) = ax^2 + bx + c \in \mathbb{R}[x]$ is a quadratic form, then the polynomial
$$ f(x+y+z) - f(x+y) - f(x+z) - f(y+z) + f(x) + f(y) + f(z)$$
is constant. We have an analogue of this for line bundles on abelian varieties. Before we state it, let's prove the following corollary which will be helpful.

\begin{corollary}
Let $X$ be an abelian variety, and let $I = \{i_1, \ldots, i_r \} \subset \{ 1,2,3\}$. We denote $p_I: X^3 \rightarrow X$ for the morphism sending $(x_1, x_2, x_3)$ to $x_{i_1} + \ldots + x_{i_r}$, i.e. $p_{12} + p_1 + p_2$, and so on. Let $L$ be a line bundle on $X$. Then the line bundle
$$\begin{array}{rl}
\Theta(L) := & \bigotimes\limits_{I \subset \{ 1,2,3 \}} p_I^* L^{ \otimes (-1)^{1 + \# I}} \\
= & p_{123}^* L \otimes p_{12}^* L^{-1} \otimes p_{13}^* L^{-1} \otimes p_{23}^* L ^{-1} \otimes p_1^*L \otimes p_2^*L \otimes p_3^*L

\end{array}
$$
on $X^3$ is trivial.
\end{corollary}

\begin{proof}
From the big theorem, it is enough to show that the restrictions to $\{0\} \times X \times X$, $X \times \{ 0 \} \times X$ and $X \times X \times \{0\}$ are trivial. We do it for $\{0\} \times X \times X$. Let $j: \{0\} \times X \times X \hookrightarrow X^3$ be the obvious map. Then $j^* p^*_{123} L \cong p^*_{23} L$, and  $j^* p^*_1 L \cong \O_{ \{0 \} \times X \times X}$, and similarly $j^*_{12} L \cong p^*_2 L$. Then, when we substitute, everything cancels and we get that the line bundle is trivial.

\end{proof}

And now we get the analogue of our quadratic form:

\begin{corollary}
Let $Y$ be a scheme and let $X$ be an abelian variety. For every triple $f, g, h$ of morphisms $Y \rightarrow X$ and for every line bundle $L$ on $X$, we have that the bundle
$$(f + g + h)^*L \otimes (f+g)^* L^{-1} \otimes (f+h)^* L^{-1} \otimes (g+h)^* L^{-1} \otimes f^* L \otimes g^*L \otimes h^*L$$
on $Y$ is trivial.

\end{corollary}

\begin{proof}
Consider $(f,g,h): Y \rightarrow X \times X \times X$ and apply the previous corollary. Since from the definition of the fibre product we have that the diagrams 
$$
\xymatrix{
Y \ar@{.>}[dr] \ar[drr]^f \ar[ddr]_g \\
& X \times X \ar[d] \ar[r]  & X  \ar[d] \\
& X \ar[r] & \spec(k) }
$$
commute, so the above line bundle is the pullback of the constant line bundle $\Theta(L)$, and hence it is again constant.

\end{proof}

And we finish with an important corollary of the theorem of the cube, the so called theorem of the square:

\begin{corollary}[Theorem of the Square]
Let $X$ be an abelian variety and let $L$ be a line bundle over it. Then, for all $x, y \in X(k)$,
$$ t^*_{x+y} L \otimes L \cong t^*_x L \otimes t^*_y L.$$

\end{corollary}

\begin{proof}
The statement follows from the previous corollary by taking $f = id_X$, and $g$ and $h$ the constant maps to $x$ and $y$ respectively.

\end{proof}

\begin{remark}
\begin{enumerate}
\item The previous corollary holds more generally:  let $T$ be a $k$-scheme and let $L_T$ be the pullback of $L$ to $X_T$. Then
$$ t^*_{x+y} L_T \otimes L_T \cong t^*_x L_T \otimes t^*_y L_T \otimes pr^*_T ((x+y)^*L \otimes x^*L^{-1} \otimes y^* L^{-1}). $$
\item If we tensor the isomorphism in the corollary with $L^{-2}$ and look at the points $x$ and $e_X$, we obtain the following important fact:

\end{enumerate}

\end{remark}

\begin{corollary}
Given a line bundle $L$ over an abelian variety $X$, then the map 
$$\varphi_L: X(k) \rightarrow Pic(X): x \mapsto [t^*_x L \otimes L^{-1}]$$
is a group homomorphism.

\end{corollary}

This generalizes the well known fact for elliptic curves. (c.f. Hartshorne, IV.1.3.7)


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	\bibitem{notes} van der Geer, Gerard; Moonen, Ben: \textit{Abelian Varieties (preliminary version of the first chapters)}. Available at 
	http://www.math.ru.nl/personal/bmoonen/index.html
	\bibitem{EGA III} Grothendieck, Alexander; Dieudonné, Jean (1963): \textit{Éléments de géométrie algébrique (rédigés avec la collaboration de Jean Dieudonné) : III. Étude cohomologique des faisceaux cohérents, Seconde partie}. Publications Mathématiques de l'IHÉS.
	\bibitem{Hartshorne} Hartshorne, R. (1977): \textit{Algebraic Geometry}. Springer. 
	\bibitem{Liu} Liu, Q. (2002): \textit{Algebraic Geometry and Arithmetic Curves}. Oxford Graduate Texts in Mathematics.
	\bibitem{Milne} Milne, James S. (2012): \textit{Algebraic Geometry, (v 5.22)}. Available at www.jmilne.org/math/
	


\end{thebibliography}





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