The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

36.29 Theorem of the cube

The following lemma tells us that the diagonal of the Picard functor is representable by locally closed immersions under the assumptions made in the lemma.

Lemma 36.29.1. Let $f : X \to S$ be a flat, proper morphism of finite presentation. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. For a morphism $g : T \to S$ consider the base change diagram

\[ \xymatrix{ X_ T \ar[d]_ p \ar[r]_ q & X \ar[d]^ f \\ T \ar[r]^ g & S } \]

Assume $\mathcal{O}_ T \to p_*\mathcal{O}_{X_ T}$ is an isomorphism for all $g : T \to S$. Then there exists an immersion $j : Z \to S$ of finite presentation such that a morphism $g : T \to S$ factors through $Z$ if and only if there exists a finite locally free $\mathcal{O}_ T$-module $\mathcal{N}$ with $p^*\mathcal{N} \cong q^*\mathcal{E}$.

Proof. Observe that the fibres $X_ s$ of $f$ are connected by our assumption that $H^0(X_ s, \mathcal{O}_{X_ s}) = \kappa (s)$. Thus the rank of $\mathcal{E}$ is constant on the fibres. Since $f$ is open (Morphisms, Lemma 28.24.9) and closed we conclude that there is a decomposition $S = \coprod S_ r$ of $S$ into open and closed subschemes such that $\mathcal{E}$ has constant rank $r$ on the inverse image of $S_ r$. Thus we may assume $\mathcal{E}$ has constant rank $r$. We will denote $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}, \mathcal{O}_ X)$ the dual rank $r$ module.

By cohomology and base change (more precisely by Derived Categories of Schemes, Lemma 35.26.4) we see that $E = Rf_*\mathcal{E}$ is a perfect object of the derived category of $S$ and that its formation commutes with arbitrary change of base. Similarly for $E' = Rf_*\mathcal{E}^\vee $. Since there is never any cohomology in degrees $< 0$, we see that $E$ and $E'$ have (locally) tor-amplitude in $[0, b]$ for some $b$. Observe that for any $g : T \to S$ we have $p_*(q^*\mathcal{E}) = H^0(Lg^*E)$ and $p_*(q^*\mathcal{E}^\vee ) = H^0(Lg^*E')$. Let $j : Z \to S$ and $j' : Z' \to S$ be immersions of finite presentation constructed in Derived Categories of Schemes, Lemma 35.27.4 for $E$ and $E'$ with $a = 0$ and $r = r$; these are roughly speaking characterized by the property that $H^0(Lj^*E)$ and $H^0((j')^*E')$ are finite locally free modules compatible with pullback.

Let $g : T \to S$ be a morphism. If there exists an $\mathcal{N}$ as in the lemma, then, using the projection formula Cohomology, Lemma 20.45.2, we see that the modules

\[ p_*(q^*\mathcal{L}) \cong p_*(p^*\mathcal{N}) \cong \mathcal{N} \otimes _{\mathcal{O}_ T} p_*\mathcal{O}_{X_ T} \cong \mathcal{N}\quad \text{and similarly }\quad p_*(q^*\mathcal{E}^\vee ) \cong \mathcal{N}^\vee \]

are finite locally free modules of rank $r$ and remain so after any further base change $T' \to T$. Hence in this case $T \to S$ factors through $j$ and through $j'$. Thus we may replace $S$ by $Z \times _ S Z'$ and assume that $f_*\mathcal{E}$ and $f_*\mathcal{E}^\vee $ are finite locally free $\mathcal{O}_ S$-modules of rank $r$ whose formation commutes with arbitrary change of base (small detail omitted).

In this sitation if $g : T \to S$ be a morphism and there exists an $\mathcal{N}$ as in the lemma, then the map (cup product in degree $0$)

\[ p_*(q^*\mathcal{E}) \otimes _{\mathcal{O}_ T} p_*(q^*\mathcal{E}^\vee ) \longrightarrow \mathcal{O}_ T \]

is a perfect pairing. Conversely, if this cup product map is a perfect pairing, then we see that locally on $T$ we may choose a basis of sections $\sigma _1, \ldots , \sigma _ r$ in $p_*(q^*\mathcal{E})$ and $\tau _1, \ldots , \tau _ r$ in $p_*(q^*\mathcal{E}^\vee )$ whose products satisfy $\sigma _ i \tau _ j = \delta _{ij}$. Thinking of $\sigma _ i$ as a section of $q^*\mathcal{E}$ on $X_ T$ and $\tau _ j$ as a section of $q^*\mathcal{E}^\vee $ on $X_ T$, we conclude that

\[ \sigma _1, \ldots , \sigma _ r : \mathcal{O}_{X_ T}^{\oplus r} \longrightarrow q^*\mathcal{E} \]

is an isomorphism with inverse given by

\[ \tau _1, \ldots , \tau _ r : q^*\mathcal{E} \longrightarrow \mathcal{O}_{X_ T}^{\oplus r} \]

In other words, we see that $p^*p_*q^*\mathcal{E} \cong q^*\mathcal{E}$. But the condition that the cupproduct is nondegenerate picks out a retrocompact open subscheme (namely, the locus where a suitable determinant is nonzero) and the proof is complete. $\square$

The lemma above in particular tells us, that if a vector bundle is trivial on fibres for a proper flat family of proper spaces, then it is the pull back of a vector bundle. Let's spell this out a bit.

Lemma 36.29.2. Let $f : X \to S$ be a flat, proper morphism of finite presentation such that $f_*\mathcal{O}_ X = \mathcal{O}_ S$ and this remains true after arbitrary base change. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. Assume

  1. $\mathcal{E}|_{X_ s}$ is isomorphic to $\mathcal{O}_{X_ s}^{\oplus r_ s}$ for all $s \in S$, and

  2. $S$ is reduced.

Then $\mathcal{E} = f^*\mathcal{N}$ for some finite locally free $\mathcal{O}_ S$-module $\mathcal{N}$.

Proof. Namely, in this case the locally closed immersion $j : Z \to S$ of Lemma 36.29.1 is bijective and hence a closed immersion. But since $S$ is reduced, $j$ is an isomorphism. $\square$

Lemma 36.29.3. Let $f : X \to S$ be a proper flat morphism of finite presentation. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume

  1. $S$ is the spectrum of a valuation ring,

  2. $\mathcal{L}$ is trivial on the generic fibre $X_\eta $ of $f$,

  3. the closed fibre $X_0$ of $f$ is integral,

  4. $H^0(X_\eta , \mathcal{O}_{X_\eta })$ is equal to the function field of $S$.

Then $\mathcal{L}$ is trivial.

Proof. Write $S = \mathop{\mathrm{Spec}}(A)$. We will first prove the lemma when $A$ is a discrete valuation ring (as this is the case most often used in practice). Let $\pi \in A$ be a uniformizer. Take a trivializing section $s \in \Gamma (X_\eta , \mathcal{L}_\eta )$. After replacing $s$ by $\pi ^ n s$ if necessary we can assume that $s \in \Gamma (X, \mathcal{L})$. If $s|_{X_0} = 0$, then we see that $s$ is divisible by $\pi $ (because $X_0$ is the scheme theoretic fibre and $X$ is flat over $A$). Thus we may assume that $s|_{X_0}$ is nonzero. Then the zero locus $Z(s)$ of $s$ is contained in $X_0$ but does not contain the generic point of $X_0$ (because $X_0$ is integral). This means that the $Z(s)$ has codimension $\geq 2$ in $X$ which contradicts Divisors, Lemma 30.15.3 unless $Z(s) = \emptyset $ as desired.

Proof in the general case. Since the valuation ring $A$ is coherent (Algebra, Example 10.89.2) we see that $H^0(X, \mathcal{L})$ is a coherent $A$-module, see Derived Categories of Schemes, Lemma 35.29.1. Equivalently, $H^0(X, \mathcal{L})$ is a finitely presented $A$-module (Algebra, Lemma 10.89.4). Since $H^0(X, \mathcal{L})$ is torsion free (by flatness of $X$ over $A$), we see from More on Algebra, Lemma 15.104.3 that $H^0(X, \mathcal{L}) = A^{\oplus n}$ for some $n$. By flat base change (Cohomology of Schemes, Lemma 29.5.2) we have

\[ K = H^0(X_\eta , \mathcal{O}_{X_\eta }) \cong H^0(X_\eta , \mathcal{L}_\eta ) = H^0(X, \mathcal{L}) \otimes _ A K \]

where $K$ is the fraction field of $A$. Thus $n = 1$. Pick a generator $s \in H^0(X, \mathcal{L})$. Let $\mathfrak m \subset A$ be the maximal ideal. Then $\kappa = A/\mathfrak m = \mathop{\mathrm{colim}}\nolimits A/\pi $ where this is a filtered colimit over nonzero $\pi \in \mathfrak m$ (here we use that $A$ is a valuation ring). Thus $X_0 = \mathop{\mathrm{lim}}\nolimits X \times _ S \mathop{\mathrm{Spec}}(A/\pi )$. If $s|_{X_0}$ is zero, then for some $\pi $ we see that $s$ restricts to zero on $X \times _ S \mathop{\mathrm{Spec}}(A/\pi )$, see Limits, Lemma 31.4.7. But if this happens, then $\pi ^{-1} s$ is a global section of $\mathcal{L}$ which contradicts the fact that $s$ is a generator of $H^0(X, \mathcal{L})$. Thus $s|_{X_0}$ is not zero. Let $Z(s) \subset X$ be the zero scheme of $s$. Since $s|_{X_0}$ is not zero and since $X_0$ is integral, we see that $Z(s)_0 \subset X_0$ is an effective Cartier divisor. Since $f$ is proper and $S$ is local, every point of $Z(s)$ specializes to a point of $Z(s)_0$. Thus by Divisors, Lemma 30.18.9 part (3) we see that $Z(s)$ is a relative effective Cartier divisor, in particular $Z(s) \to S$ is flat. Hence if $Z(s)$ were nonemtpy, then $Z(s)_\eta $ would be nonempty which contradicts the fact that $s|_{X_\eta }$ is a trivialization of $\mathcal{L}_\eta $. Thus $Z(s) = \emptyset $ as desired. $\square$

Lemma 36.29.4. Let $f : X \to S$ and $\mathcal{E}$ be as in Lemma 36.29.1 and in addition assume $\mathcal{E}$ is an invertible $\mathcal{O}_ X$-module. If moreover the geometric fibres of $f$ are integral, then $Z$ is closed in $S$.

Proof. Since $j : Z \to S$ is of finite presentation, it suffices to show: for any morphism $g : \mathop{\mathrm{Spec}}(A) \to S$ where $A$ is a valuation ring with fraction field $K$ such that $g(\mathop{\mathrm{Spec}}(K)) \in j(Z)$ we have $g(\mathop{\mathrm{Spec}}(A)) \subset j(Z)$. See Morphisms, Lemma 28.6.5. This follows from Lemma 36.29.3 and the characterization of $j : Z \to S$ in Lemma 36.29.1. $\square$

Lemma 36.29.5. Consider a commutative diagram of schemes

\[ \xymatrix{ X' \ar[rr] \ar[dr]_{f'} & & X \ar[dl]^ f \\ & S } \]

with $f' : X' \to S$ and $f : X \to S$ satisfying the hypotheses of Lemma 36.29.1. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module and let $\mathcal{L}'$ be the pullback to $X'$. Let $Z \subset S$, resp. $Z' \subset S$ be the locally closed subscheme constructed in Lemma 36.29.1 for $(f, \mathcal{L})$, resp. $(f', \mathcal{L}')$ so that $Z \subset Z'$. If $s \in Z$ and

\[ H^1(X_ s, \mathcal{O}) \longrightarrow H^1(X'_ s, \mathcal{O}) \]

is injective, then $Z \cap U = Z' \cap U$ for some open neighbourhood $U$ of $s$.

Proof. We may replace $S$ by $Z'$. After shrinking $S$ to an affine open neighbourhood of $s$ we may assume that $\mathcal{L}' = \mathcal{O}_{X'}$. Let $E = Rf_*\mathcal{L}$ and $E' = Rf'_*\mathcal{L}' = Rf'_*\mathcal{O}_{X'}$. These are perfect complexes whose formation commutes with arbitrary change of base (Derived Categories of Schemes, Lemma 35.26.4). In particular we see that

\[ E \otimes _{\mathcal{O}_ S}^\mathbf {L} \kappa (s) = R\Gamma (X_ s, \mathcal{L}_ s) = R\Gamma (X_ s, \mathcal{O}_{X_ s}) \]

The second equality because $s \in Z$. Set $h_ i = \dim _{\kappa (s)} H^ i(X_ s, \mathcal{O}_{X_ s})$. After shrinking $S$ we can represent $E$ by a complex

\[ \mathcal{O}_ S \to \mathcal{O}_ S^{\oplus h_1} \to \mathcal{O}_ S^{\oplus h_2} \to \ldots \]

see More on Algebra, Lemma 15.70.6 (strictly speaking this also uses Derived Categories of Schemes, Lemmas 35.3.5 and 35.9.7). Similarly, we may assume $E'$ is represented by a complex

\[ \mathcal{O}_ S \to \mathcal{O}_ S^{\oplus h'_1} \to \mathcal{O}_ S^{\oplus h'_2} \to \ldots \]

where $h'_ i = \dim _{\kappa (s)} H^ i(X'_ s, \mathcal{O}_{X'_ s})$. By functoriality of cohomology we have a map

\[ E \longrightarrow E' \]

in $D(\mathcal{O}_ S)$ whose formation commutes with change of base. Since the complex representing $E$ is a finite complex of finite free modules and since $S$ is affine, we can choose a map of complexes

\[ \xymatrix{ \mathcal{O}_ S \ar[r]_ d \ar[d]_ a & \mathcal{O}_ S^{\oplus h_1} \ar[r] \ar[d]_ b & \mathcal{O}_ S^{\oplus h_2} \ar[r] \ar[d]_ c & \ldots \\ \mathcal{O}_ S \ar[r]^{d'} & \mathcal{O}_ S^{\oplus h'_1} \ar[r] & \mathcal{O}_ S^{\oplus h'_2} \ar[r] & \ldots } \]

representing the given map $E \to E'$. Since $s \in Z$ we see that the trivializing section of $\mathcal{L}_ s$ pulls back to a trivializing section of $\mathcal{L}'_ s = \mathcal{O}_{X'_ s}$. Thus $a \otimes \kappa (s)$ is an isomorphism, hence after shrinking $S$ we see that $a$ is an isomorphism. Finally, we use the hypothesis that $H^1(X_ s, \mathcal{O}) \to H^1(X'_ s, \mathcal{O})$ is injective, to see that there exists a $h_1 \times h_1$ minor of the matrix defining $b$ which maps to a nonzero element in $\kappa (s)$. Hence after shrinking $S$ we may assume that $b$ is injective. However, since $\mathcal{L}' = \mathcal{O}_{X'}$ we see that $d' = 0$. It follows that $d = 0$. In this way we see that the trivializing section of $\mathcal{L}_ s$ lifts to a section of $\mathcal{L}$ over $X$. A straightforward topological argument (omitted) shows that this means that $\mathcal{L}$ is trivial after possibly shrinking $S$ a bit further. $\square$

Lemma 36.29.6. Consider $n$ commutative diagrams of schemes

\[ \xymatrix{ X_ i \ar[rr] \ar[dr]_{f_ i} & & X \ar[dl]^ f \\ & S } \]

with $f_ i : X_ i \to S$ and $f : X \to S$ satisfying the hypotheses of Lemma 36.29.1. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module and let $\mathcal{L}_ i$ be the pullback to $X_ i$. Let $Z \subset S$, resp. $Z_ i \subset S$ be the locally closed subscheme constructed in Lemma 36.29.1 for $(f, \mathcal{L})$, resp. $(f_ i, \mathcal{L}_ i)$ so that $Z \subset \bigcap _{i = 1, \ldots , n} Z_ i$. If $s \in Z$ and

\[ H^1(X_ s, \mathcal{O}) \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} H^1(X_{i, s}, \mathcal{O}) \]

is injective, then $Z \cap U = (\bigcap _{i = 1, \ldots , n} Z_ i) \cap U$ (scheme theoretic intersection) for some open neighbourhood $U$ of $s$.

Proof. This lemma is a variant of Lemma 36.29.5 and we strongly urge the reader to read that proof first; this proof is basically a copy of that proof with minor modifications. It follows from the description of (scheme valued) points of $Z$ and the $Z_ i$ that $Z \subset \bigcap _{i = 1, \ldots , n} Z_ i$ where we take the scheme theoretic intersection. Thus we may replace $S$ by the scheme theoretic intersection $\bigcap _{i = 1, \ldots , n} Z_ i$. After shrinking $S$ to an affine open neighbourhood of $s$ we may assume that $\mathcal{L}_ i = \mathcal{O}_{X_ i}$ for $i = 1, \ldots , n$. Let $E = Rf_*\mathcal{L}$ and $E_ i = Rf_{i, *}\mathcal{L}_ i = Rf_{i, *}\mathcal{O}_{X_ i}$. These are perfect complexes whose formation commutes with arbitrary change of base (Derived Categories of Schemes, Lemma 35.26.4). In particular we see that

\[ E \otimes _{\mathcal{O}_ S}^\mathbf {L} \kappa (s) = R\Gamma (X_ s, \mathcal{L}_ s) = R\Gamma (X_ s, \mathcal{O}_{X_ s}) \]

The second equality because $s \in Z$. Set $h_ j = \dim _{\kappa (s)} H^ j(X_ s, \mathcal{O}_{X_ s})$. After shrinking $S$ we can represent $E$ by a complex

\[ \mathcal{O}_ S \to \mathcal{O}_ S^{\oplus h_1} \to \mathcal{O}_ S^{\oplus h_2} \to \ldots \]

see More on Algebra, Lemma 15.70.6 (strictly speaking this also uses Derived Categories of Schemes, Lemmas 35.3.5 and 35.9.7). Similarly, we may assume $E_ i$ is represented by a complex

\[ \mathcal{O}_ S \to \mathcal{O}_ S^{\oplus h_{i, 1}} \to \mathcal{O}_ S^{\oplus h_{i, 2}} \to \ldots \]

where $h_{i, j} = \dim _{\kappa (s)} H^ j(X_{i, s}, \mathcal{O}_{X_{i, s}})$. By functoriality of cohomology we have a map

\[ E \longrightarrow E_ i \]

in $D(\mathcal{O}_ S)$ whose formation commutes with change of base. Since the complex representing $E$ is a finite complex of finite free modules and since $S$ is affine, we can choose a map of complexes

\[ \xymatrix{ \mathcal{O}_ S \ar[r]_ d \ar[d]_{a_ i} & \mathcal{O}_ S^{\oplus h_1} \ar[r] \ar[d]_{b_ i} & \mathcal{O}_ S^{\oplus h_2} \ar[r] \ar[d]_{c_ i} & \ldots \\ \mathcal{O}_ S \ar[r]^{d_ i} & \mathcal{O}_ S^{\oplus h_{i, 1}} \ar[r] & \mathcal{O}_ S^{\oplus h_{i, 2}} \ar[r] & \ldots } \]

representing the given map $E \to E_ i$. Since $s \in Z$ we see that the trivializing section of $\mathcal{L}_ s$ pulls back to a trivializing section of $\mathcal{L}_{i, s} = \mathcal{O}_{X_{i, s}}$. Thus $a_ i \otimes \kappa (s)$ is an isomorphism, hence after shrinking $S$ we see that $a_ i$ is an isomorphism. Finally, we use the hypothesis that $H^1(X_ s, \mathcal{O}) \to \bigoplus _{i = 1, \ldots , n} H^1(X_{i, s}, \mathcal{O})$ is injective, to see that there exists a $h_1 \times h_1$ minor of the matrix defining $\oplus b_ i$ which maps to a nonzero element in $\kappa (s)$. Hence after shrinking $S$ we may assume that $(b_1, \ldots , b_ n) : \mathcal{O}_ S^{h_1} \to \bigoplus _{i = 1, \ldots , n} \mathcal{O}_ S^{h_{i, 1}}$ is injective. However, since $\mathcal{L}_ i = \mathcal{O}_{X_ i}$ we see that $d_ i = 0$ for $i = 1, \ldots n$. It follows that $d = 0$ because $(b_1, \ldots , b_ n) \circ d = (\oplus d_ i) \circ (a_1, \ldots , a_ n)$. In this way we see that the trivializing section of $\mathcal{L}_ s$ lifts to a section of $\mathcal{L}$ over $X$. A straightforward topological argument (omitted) shows that this means that $\mathcal{L}$ is trivial after possibly shrinking $S$ a bit further. $\square$

Lemma 36.29.7. Let $f : X \to S$ and $g : Y \to S$ be morphisms of schemes satisfying the hypotheses of Lemma 36.29.1. Let $\sigma : S \to X$ and $\tau : S \to Y$ be sections of $f$ and $g$. Let $s \in S$. Let $\mathcal{L}$ be an invertible sheaf on $X \times _ S Y$. If $(1 \times \tau )^*\mathcal{L}$ on $X$, $(\sigma \times 1)^*\mathcal{L}$ on $Y$, and $\mathcal{L}|_{(X \times _ S Y)_ s}$ are trivial, then there is an open neighbourhood $U$ of $s$ such that $\mathcal{L}$ is trivial over $(X \times _ S Y)_ U$.

Proof. By K√ľnneth (Varieties, Lemma 32.29.1) the map

\[ H^1(X_ s \times _{\mathop{\mathrm{Spec}}(\kappa (s)} Y_ s, \mathcal{O}) \to H^1(X_ s, \mathcal{O}) \oplus H^1(Y_ s, \mathcal{O}) \]

is injective. Thus we may apply Lemma 36.29.6 to the two morphisms

\[ 1 \times \tau : X \to X \times _ S Y \quad \text{and}\quad \sigma \times 1 : Y \to X \times _ S Y \]

to conclude. $\square$

Theorem 36.29.8 (Theorem of the cube). Let $k$ be a field. Let $X, Y, Z$ be varieties with $k$-rational points $x, y, z$. Let $\mathcal{L}$ be an invertible module on $X \times Y \times Z$. If

  1. $\mathcal{L}$ is trivial over $x \times Y \times Z$, $X \times y \times Z$, and $X \times Y \times z$, and

  2. $X$ and $Y$ are geometrically integral and proper over $k$,

then $\mathcal{L}$ is trivial.

Proof. Since $X$ and $Y$ are geometrically integral and proper over $k$ the product $X \times _ k Y$ is geometrically integral and proper over $k$. This implies that $H^0(X \times Y, \mathcal{O}_{X \times Y}) = k$ and that the same remains true after any base change. Thus we may apply Lemma 36.29.1 to the morphism

\[ p : X \times Y \times Z \longrightarrow Z \]

and the invertible module $\mathcal{L}$ to get a locally closed subscheme $Z' \subset Z$ such that $\mathcal{L}|_{X \times Y \times Z'}$ is the pullback of an invertible module $\mathcal{N}$ on $Z'$. By Lemma 36.29.4 we see that $Z' \subset Z$ is a closed subscheme. Hence if $Z'$ contains an open neighbourhood of $z$, then $Z' = Z$ and we see that $\mathcal{L} = p^*\mathcal{N}$. Restricting to $x \times y \times Z$ we find that $\mathcal{N} \cong \mathcal{O}_ Z$ and $\mathcal{L}$ is trivial. To get the desired open neighbourhood of $z$ apply Lemma 36.29.7 to the morphism $p$, the point $z$, and the sections $\sigma : Z \to X \times Z$ and $\tau : Z \to Y \times Z$ given by $x$ and $y$. $\square$


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