The Stacks project


Lemma 36.20.1. Let $A$ be a ring. Let $X = \mathbf{P}^ n_ A = \text{Proj}(S)$ where $S = A[X_0, \ldots , X_ n]$. With $P$ as in ( and $R$ as in ( the functor

\[ - \otimes _ R^\mathbf {L} P : D(R) \longrightarrow D_\mathit{QCoh}(\mathcal{O}_ X) \]

is an $A$-linear equivalence of triangulated categories sending $R$ to $P$.

Proof. To prove that our functor is fully faithful it suffices to prove that $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(P, P)$ is zero for $i \not= 0$ and equal to $R$ for $i = 0$, see Differential Graded Algebra, Lemma 22.35.6. As in the proof of Lemma 36.18.2 we see that

\[ \mathop{\mathrm{Ext}}\nolimits ^ i_ X(P, P) = H^ i(X, P^\wedge \otimes P) = \bigoplus \nolimits _{0 \leq a, b \leq n} H^ i(X, \mathcal{O}_ X(a - b)) \]

By the computation of cohomology of projective space (Cohomology of Schemes, Lemma 30.8.1) we find that these $\mathop{\mathrm{Ext}}\nolimits $-groups are zero unless $i = 0$. For $i = 0$ we recover $R$ because this is how we defined $R$ in ( By Differential Graded Algebra, Lemma 22.35.5 our functor has a right adjoint, namely $R\mathop{\mathrm{Hom}}\nolimits (P, -) : D_\mathit{QCoh}(\mathcal{O}_ X) \to D(R)$. Since $P$ is a generator for $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 36.16.3 we see that the kernel of $R\mathop{\mathrm{Hom}}\nolimits (P, -)$ is zero. Hence our functor is an equivalence of triangulated categories by Derived Categories, Lemma 13.7.2. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 36.20: An example equivalence

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Your email address will not be published.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 0BQU. Beware of the difference between the letter 'O' and the digit '0'.