Lemma 42.32.5. Let $Y$ be a scheme. Let $\mathcal{L}_ i$, $i = 1, 2$ be invertible $\mathcal{O}_ Y$-modules. Let $s$ be a global section of $\mathcal{L}_1 \otimes _{\mathcal{O}_ X} \mathcal{L}_2$. Denote $i : D \to X$ the zero scheme of $s$. Then there exists a commutative diagram
\[ \xymatrix{ D_1 \ar[r]_{i_1} \ar[d]_{p_1} & L \ar[d]^ p & D_2 \ar[l]^{i_2} \ar[d]^{p_2} \\ D \ar[r]^ i & Y & D \ar[l]_ i } \]
and sections $s_ i$ of $p^*\mathcal{L}_ i$ such that the following hold:
$p^*s = s_1 \otimes s_2$,
$p$ is of finite type and flat of relative dimension $1$,
$D_ i$ is the zero scheme of $s_ i$,
$D_ i \cong \underline{\mathop{\mathrm{Spec}}}(\text{Sym}^*(\mathcal{L}_{1 - i}^{\otimes -1})|_ D))$ over $D$ for $i = 1, 2$,
$p^{-1}D = D_1 \cup D_2$ (scheme theoretic union),
$D_1 \cap D_2$ (scheme theoretic intersection) maps isomorphically to $D$, and
$D_1 \cap D_2 \to D_ i$ is the zero section of the line bundle $D_ i \to D$ for $i = 1, 2$.
Moreover, the formation of this diagram and the sections $s_ i$ commutes with arbitrary base change.
Proof.
Let $p : X \to Y$ be the relative spectrum of the quasi-coherent sheaf of $\mathcal{O}_ Y$-algebras
\[ \mathcal{A} = \left(\bigoplus \nolimits _{a_1, a_2 \geq 0} \mathcal{L}_1^{\otimes -a_1} \otimes _{\mathcal{O}_ Y} \mathcal{L}_2^{\otimes -a_2}\right)/\mathcal{J} \]
where $\mathcal{J}$ is the ideal generated by local sections of the form $st - t$ for $t$ a local section of any summand $\mathcal{L}_1^{\otimes -a_1} \otimes \mathcal{L}_2^{\otimes -a_2}$ with $a_1, a_2 > 0$. The sections $s_ i$ viewed as maps $p^*\mathcal{L}_ i^{\otimes -1} \to \mathcal{O}_ X$ are defined as the adjoints of the maps $\mathcal{L}_ i^{\otimes -1} \to \mathcal{A} = p_*\mathcal{O}_ X$. For any $y \in Y$ we can choose an affine open $V \subset Y$, say $V = \mathop{\mathrm{Spec}}(B)$, containing $y$ and trivializations $z_ i : \mathcal{O}_ V \to \mathcal{L}_ i^{\otimes -1}|_ V$. Observe that $f = s(z_1z_2) \in A$ cuts out the closed subscheme $D$. Then clearly
\[ p^{-1}(V) = \mathop{\mathrm{Spec}}(B[z_1, z_2]/(z_1 z_2 - f)) \]
Since $D_ i$ is cut out by $z_ i$ everything is clear.
$\square$
Comments (2)
Comment #4404 by awllower on
Comment #4506 by Johan on