
## 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 $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-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 is a locally closed subscheme $Z \subset S$ such that a morphism $g : T \to S$ factors through $Z$ if and only if there exists an invertible $\mathcal{O}_ T$-module $\mathcal{N}$ with $p^*\mathcal{N} \cong q^*\mathcal{L}$.

Proof. By cohomology and base change (more precisely by Derived Categories of Schemes, Lemma 35.26.4) we see that $E = Rf_*\mathcal{L}$ is a perfect object of the derived category of $S$ and that its formation commutes with arbitrary change of base. Similarly for $E' = Rf_*\mathcal{L}^{\otimes -1}$. 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 $p_*(q^*\mathcal{L}) = H^0(Lg^*E)$ and $p_*(q^*\mathcal{L}^{\otimes -1}) = H^0(Lg^*E')$. Let $j : Z \to S$ and $j' : Z' \to S$ be the locally closed immersions constructed in Derived Categories of Schemes, Lemma 35.27.4 for $E$ and $E'$ with $a = 0$; these are characterized by the property that $H^0(Lj^*E)$ and $H^0((j')^*E')$ are invertible 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

$p_*(q^*\mathcal{L}) \cong p_*(p^*\mathcal{N}) \cong \mathcal{N} \otimes _{\mathcal{O}_ T} p_*\mathcal{O}_{X_ T} \cong \mathcal{N}\quad \text{and similarly }\quad p_*(q^*\mathcal{L}^{\otimes -1}) \cong \mathcal{N}^{\otimes -1}$

are invertible and remain invertible 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 $f_*\mathcal{L}$ and $f_*\mathcal{L}^{\otimes -1}$ are invertible $\mathcal{O}_ S$-modules whose formation commutes with arbitrary change of base.

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 $0$)

$p_*(q^*\mathcal{L}) \otimes _{\mathcal{O}_ T} p_*(q^*\mathcal{L}^{\otimes -1}) \longrightarrow \mathcal{O}_ T$

is an isomorphism. Conversely, if this cup product map is an isomorphism, then we see that locally on $T$ we have sections $\sigma$ in $p_*(q^*\mathcal{L})$ and $\sigma '$ in $p_*(q^*\mathcal{L}^{\otimes -1})$ whose product is $1$. Thinking of $\sigma$ as a section of $q^*\mathcal{L}$ on $X_ T$ and $\sigma '$ as a section of $q^*\mathcal{L}^{\otimes -1}$ on $X_ T$ with $\sigma \cdot \sigma ' = 1$, we conclude that $\sigma : \mathcal{O}_{X_ T} \to q^*\mathcal{L}$ is an isomorphism. In other words, we see that $p^*p_*q^*\mathcal{L} \cong q^*\mathcal{L}$. But the condition that the cupproduct is nonzero picks out an open subscheme and the proof is complete. $\square$

Lemma 36.29.2. Let $f : X \to S$ and $\mathcal{L}$ be as in Lemma 36.29.1. If moreover the geometric fibres of $f$ are integral, then $Z$ is closed in $S$.

Proof. We first do a standard argument to reduce to the Noetherian case. Namely, the question is local on $S$, hence we may assume that $S = \mathop{\mathrm{Spec}}(R)$ is affine. Then we write $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ as a filtered colimit with $R_ i$ of finite type over $\mathbf{Z}$. Set $S_ i = \mathop{\mathrm{Spec}}(R_ i)$. For some $i$ there exists a flat proper morphism $f_ i : X_ i \to S_ i$ and an invertible $\mathcal{O}_{X_ i}$-module $\mathcal{L}_ i$ whose base change to $S$ gives back $f : X \to S$ and $\mathcal{L}$. See Limits, Lemmas 31.10.1, 31.8.7, 31.13.1, and 31.10.3. Pick $i \in I$. By Lemmas 36.24.5 and 36.25.7 the set $E \subset S_ i$ of points where the fibres of $f_ i$ are geometrically integral is constructible. Since $S \to S_ i$ maps into $E$ by assumption (and Lemmas 36.24.2 and 36.25.2) after increasing $i$ we may assume the fibres of $f_ i$ are geometrically irreducible, see Limits, Lemma 31.4.10. By Derived Categories of Schemes, Lemma 35.26.4 $Rf_{i, *}\mathcal{O}_{X_ i}$ is a perfect object of $D(\mathcal{O}_{S_ i})$ whose formation commutes with arbitrary base change. Let $T \subset S_ i$ be the locally closed subscheme of $S_ i$ constructed in Derived Categories of Schemes, Lemma 35.27.4 for $Rf_{i, *}\mathcal{O}_{X_ i}$ with $a = 0$. By our assumption that $f_*\mathcal{O}_ X = \mathcal{O}_ S$ universally we see that $S \to S_ i$ factors through $T$. Set $Y = X_ i \times _{S_ i} T \to T$ and $\mathcal{M} = \mathcal{L}_ i|_ Y$. By construction the morphism $g : Y \to T$ satisfies $g_*\mathcal{O}_ Y = \mathcal{O}_ T$ universally and we have a cartesian diagram

$\xymatrix{ \mathcal{L} & X \ar[r] \ar[d]_ f & Y \ar[d]^ g & \mathcal{M} \\ & S \ar[r] & T }$

Thus if we can prove the lemma for $g$ and $\mathcal{M}$, then it follows for $f$ and $\mathcal{L}$. Since $T$ is Noetherian, we have reduced to the Noetherian case.

Assume $S$ is Noetherian. Since $Z$ is a locally closed subscheme of a Noetherian scheme it suffices to show that $Z$ is closed under specialization in order to prove that it is closed. By Properties, Lemma 27.5.10 and base change we see that it suffices to prove the lemma in case $S$ is the spectrum of a dvr $A$. In other words, suppose we have a flat proper morphism $X \to \mathop{\mathrm{Spec}}(A)$ with integral scheme theoretic fibres $X_\eta$ (generic), $X_0$ (closed) and an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ whose restriction to $X_\eta$ is trivial. Goal: show that $\mathcal{L}$ is trivial. This follows from Divisors, Lemma 30.28.1. However, we can prove this special case directly as follows: take a trivializing section $s \in \Gamma (X_\eta , \mathcal{L}_\eta )$. After replacing $s$ by $\pi ^ n s$ if necessary ($\pi \in A$ a uniformizer) 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. $\square$

Lemma 36.29.3. 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 36.29.4. 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 36.29.3 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 36.29.5. 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 Künneth (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 36.29.4 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 36.29.6 (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 36.29.2 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 36.29.5 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 $y$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BEZ. Beware of the difference between the letter 'O' and the digit '0'.