Lemma 48.15.1. Let $Y$ be a quasi-compact and quasi-separated scheme. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ Y$-module of rank $n + 1$ with determinant $\mathcal{L} = \wedge ^{n + 1}(\mathcal{E})$. Let $f : X = \mathbf{P}(\mathcal{E}) \to Y$ be the projection. Let $a$ be the right adjoint for $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ of Lemma 48.3.1. Then there is an isomorphism

$c : f^*\mathcal{L}(-n - 1)[n] \longrightarrow a(\mathcal{O}_ Y)$

In particular, if $\mathcal{E} = \mathcal{O}_ Y^{\oplus n + 1}$, then $X = \mathbf{P}^ n_ Y$ and we obtain $a(\mathcal{O}_ Y) = \mathcal{O}_ X(-n - 1)[n]$.

Proof. In (the proof of) Cohomology of Schemes, Lemma 30.8.4 we constructed a canonical isomorphism

$R^ nf_*(f^*\mathcal{L}(-n - 1)) \longrightarrow \mathcal{O}_ Y$

Moreover, $Rf_*(f^*\mathcal{L}(-n - 1))[n] = R^ nf_*(f^*\mathcal{L}(-n - 1))$, i.e., the other higher direct images are zero. Thus we find an isomorphism

$Rf_*(f^*\mathcal{L}(-n - 1)[n]) \longrightarrow \mathcal{O}_ Y$

This isomorphism determines $c$ as in the statement of the lemma because $a$ is the right adjoint of $Rf_*$. By Lemma 48.4.4 construction of the $a$ is local on the base. In particular, to check that $c$ is an isomorphism, we may work locally on $Y$. In other words, we may assume $Y$ is affine and $\mathcal{E} = \mathcal{O}_ Y^{\oplus n + 1}$. In this case the sheaves $\mathcal{O}_ X, \mathcal{O}_ X(-1), \ldots , \mathcal{O}_ X(-n)$ generate $D_\mathit{QCoh}(X)$, see Derived Categories of Schemes, Lemma 36.16.3. Hence it suffices to show that $c : \mathcal{O}_ X(-n - 1)[n] \to a(\mathcal{O}_ Y)$ is transformed into an isomorphism under the functors

$F_{i, p}(-) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{O}_ X(i), (-)[p])$

for $i \in \{ -n, \ldots , 0\}$ and $p \in \mathbf{Z}$. For $F_{0, p}$ this holds by construction of the arrow $c$! For $i \in \{ -n, \ldots , -1\}$ we have

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{O}_ X(i), \mathcal{O}_ X(-n - 1)[n + p]) = H^ p(X, \mathcal{O}_ X(-n - 1 - i)) = 0$

by the computation of cohomology of projective space (Cohomology of Schemes, Lemma 30.8.1) and we have

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{O}_ X(i), a(\mathcal{O}_ Y)[p]) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*\mathcal{O}_ X(i), \mathcal{O}_ Y[p]) = 0$

because $Rf_*\mathcal{O}_ X(i) = 0$ by the same lemma. Hence the source and the target of $F_{i, p}(c)$ vanish and $F_{i, p}(c)$ is necessarily an isomorphism. This finishes the proof. $\square$

Comment #8772 by Bogdan on

It seems that the proof never uses that $Y$ is noetherian.

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).