Lemma 37.33.1. 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 29.25.10) 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 36.30.4) 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 36.31.4 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 20.52.2, 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. $\square$

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