Lemma 54.16.5. Let b : X \to X' be the contraction of an exceptional curve of the first kind E \subset X. Then there is a short exact sequence
0 \to \mathop{\mathrm{Pic}}\nolimits (X') \to \mathop{\mathrm{Pic}}\nolimits (X) \to \mathbf{Z} \to 0
where the first map is pullback by b and the second map sends \mathcal{L} to the degree of \mathcal{L} on the exceptional curve E. The sequence is split by the map n \mapsto \mathcal{O}_ X(-nE).
Proof.
Since E = \mathbf{P}^1_ k we see that the Picard group of E is \mathbf{Z}, see Divisors, Lemma 31.28.5. Hence we can think of the last map as \mathcal{L} \mapsto \mathcal{L}|_ E. The degree of the restriction of \mathcal{O}_ X(E) to E is -1 by definition of exceptional curves of the first kind. Combining these remarks we see that it suffices to show that \mathop{\mathrm{Pic}}\nolimits (X') \to \mathop{\mathrm{Pic}}\nolimits (X) is injective with image the invertible sheaves restricting to \mathcal{O}_ E on E.
Given an invertible \mathcal{O}_{X'}-module \mathcal{L}' we claim the map \mathcal{L}' \to b_*b^*\mathcal{L}' is an isomorphism. This is clear everywhere except possibly at the image point x \in X' of E. To check it is an isomorphism on stalks at x we may replace X' by an open neighbourhood at x and assume \mathcal{L}' is \mathcal{O}_{X'}. Then we have to show that the map \mathcal{O}_{X'} \to b_*\mathcal{O}_ X is an isomorphism. This follows from Lemma 54.3.4 part (4).
Let \mathcal{L} be an invertible \mathcal{O}_ X-module with \mathcal{L}|_ E = \mathcal{O}_ E. Then we claim (1) b_*\mathcal{L} is invertible and (2) b^*b_*\mathcal{L} \to \mathcal{L} is an isomorphism. Statements (1) and (2) are clear over X' \setminus \{ x\} . Thus it suffices to prove (1) and (2) after base change to \mathop{\mathrm{Spec}}(\mathcal{O}_{X', x}). Computing b_* commutes with flat base change (Cohomology of Schemes, Lemma 30.5.2) and similarly for b^* and formation of the adjunction map. But if X' is the spectrum of a regular local ring then \mathcal{L} is trivial by the description of the Picard group in Lemma 54.3.3. Thus the claim is proved.
Combining the claims proved in the previous two paragraphs we see that the map \mathcal{L} \mapsto b_*\mathcal{L} is an inverse to the map
\mathop{\mathrm{Pic}}\nolimits (X') \longrightarrow \mathop{\mathrm{Ker}}(\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (E))
and the lemma is proved.
\square
Comments (0)