[III Proposition 2.1.12, EGA]

Lemma 30.8.1. Let $R$ be a ring. Let $n \geq 0$ be an integer. We have

$H^ q(\mathbf{P}^ n, \mathcal{O}_{\mathbf{P}^ n_ R}(d)) = \left\{ \begin{matrix} (R[T_0, \ldots , T_ n])_ d & \text{if} & q = 0 \\ 0 & \text{if} & q \not= 0, n \\ \left(\frac{1}{T_0 \ldots T_ n} R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}]\right)_ d & \text{if} & q = n \end{matrix} \right.$

as $R$-modules.

Proof. We will use the standard affine open covering

$\mathcal{U} : \mathbf{P}^ n_ R = \bigcup \nolimits _{i = 0}^ n D_{+}(T_ i)$

to compute the cohomology using the Čech complex. This is permissible by Lemma 30.2.6 since any intersection of finitely many affine $D_{+}(T_ i)$ is also a standard affine open (see Constructions, Section 27.8). In fact, we can use the alternating or ordered Čech complex according to Cohomology, Lemmas 20.23.3 and 20.23.6.

The ordering we will use on $\{ 0, \ldots , n\}$ is the usual one. Hence the complex we are looking at has terms

$\check{\mathcal{C}}_{ord}^ p(\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) = \bigoplus \nolimits _{i_0 < \ldots < i_ p} (R[T_0, \ldots , T_ n, \frac{1}{T_{i_0} \ldots T_{i_ p}}])_ d$

Moreover, the maps are given by the usual formula

$d(s)_{i_0 \ldots i_{p + 1}} = \sum \nolimits _{j = 0}^{p + 1} (-1)^ j s_{i_0 \ldots \hat i_ j \ldots i_{p + 1}}$

see Cohomology, Section 20.23. Note that each term of this complex has a natural $\mathbf{Z}^{n + 1}$-grading. Namely, we get this by declaring a monomial $T_0^{e_0} \ldots T_ n^{e_ n}$ to be homogeneous with weight $(e_0, \ldots , e_ n) \in \mathbf{Z}^{n + 1}$. It is clear that the differential given above respects the grading. In a formula we have

$\check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) = \bigoplus \nolimits _{\vec{e} \in \mathbf{Z}^{n + 1}} \check{\mathcal{C}}^\bullet (\vec{e})$

where not all summands on the right hand side occur (see below). Hence in order to compute the cohomology modules of the complex it suffices to compute the cohomology of the graded pieces and take the direct sum at the end.

Fix $\vec{e} = (e_0, \ldots , e_ n) \in \mathbf{Z}^{n + 1}$. In order for this weight to occur in the complex above we need to assume $e_0 + \ldots + e_ n = d$ (if not then it occurs for a different twist of the structure sheaf of course). Assuming this, set

$NEG(\vec{e}) = \{ i \in \{ 0, \ldots , n\} \mid e_ i < 0\} .$

With this notation the weight $\vec{e}$ summand $\check{\mathcal{C}}^\bullet (\vec{e})$ of the Čech complex above has the following terms

$\check{\mathcal{C}}^ p(\vec{e}) = \bigoplus \nolimits _{i_0 < \ldots < i_ p, \ NEG(\vec{e}) \subset \{ i_0, \ldots , i_ p\} } R \cdot T_0^{e_0} \ldots T_ n^{e_ n}$

In other words, the terms corresponding to $i_0 < \ldots < i_ p$ such that $NEG(\vec{e})$ is not contained in $\{ i_0 \ldots i_ p\}$ are zero. The differential of the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ is still given by the exact same formula as above.

Suppose that $NEG(\vec{e}) = \{ 0, \ldots , n\}$, i.e., that all exponents $e_ i$ are negative. In this case the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ has only one term, namely $\check{\mathcal{C}}^ n(\vec{e}) = R \cdot \frac{1}{T_0^{-e_0} \ldots T_ n^{-e_ n}}$. Hence in this case

$H^ q(\check{\mathcal{C}}^\bullet (\vec{e})) = \left\{ \begin{matrix} R \cdot \frac{1}{T_0^{-e_0} \ldots T_ n^{-e_ n}} & \text{if} & q = n \\ 0 & \text{if} & \text{else} \end{matrix} \right.$

The direct sum of all of these terms clearly gives the value

$\left(\frac{1}{T_0 \ldots T_ n} R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}]\right)_ d$

in degree $n$ as stated in the lemma. Moreover these terms do not contribute to cohomology in other degrees (also in accordance with the statement of the lemma).

Assume $NEG(\vec{e}) = \emptyset$. In this case the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ has a summand $R$ corresponding to all $i_0 < \ldots < i_ p$. Let us compare the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ to another complex. Namely, consider the affine open covering

$\mathcal{V} : \mathop{\mathrm{Spec}}(R) = \bigcup \nolimits _{i \in \{ 0, \ldots , n\} } V_ i$

where $V_ i = \mathop{\mathrm{Spec}}(R)$ for all $i$. Consider the alternating Čech complex

$\check{\mathcal{C}}_{ord}^\bullet (\mathcal{V}, \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$

By the same reasoning as above this computes the cohomology of the structure sheaf on $\mathop{\mathrm{Spec}}(R)$. Hence we see that $H^ p( \check{\mathcal{C}}_{ord}^\bullet (\mathcal{V}, \mathcal{O}_{\mathop{\mathrm{Spec}}(R)}) ) = R$ if $p = 0$ and is $0$ whenever $p > 0$. For these facts, see Lemma 30.2.1 and its proof. Note that also $\check{\mathcal{C}}_{ord}^\bullet (\mathcal{V}, \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ has a summand $R$ for every $i_0 < \ldots < i_ p$ and has exactly the same differential as $\check{\mathcal{C}}^\bullet (\vec{e})$. In other words these complexes are isomorphic complexes and hence have the same cohomology. We conclude that

$H^ q(\check{\mathcal{C}}^\bullet (\vec{e})) = \left\{ \begin{matrix} R \cdot T_0^{e_0} \ldots T_ n^{e_ n} & \text{if} & q = 0 \\ 0 & \text{if} & \text{else} \end{matrix} \right.$

in the case that $NEG(\vec{e}) = \emptyset$. The direct sum of all of these terms clearly gives the value

$(R[T_0, \ldots , T_ n])_ d$

in degree $0$ as stated in the lemma. Moreover these terms do not contribute to cohomology in other degrees (also in accordance with the statement of the lemma).

To finish the proof of the lemma we have to show that the complexes $\check{\mathcal{C}}^\bullet (\vec{e})$ are acyclic when $NEG(\vec{e})$ is neither empty nor equal to $\{ 0, \ldots , n\}$. Pick an index $i_{\text{fix}} \not\in NEG(\vec{e})$ (such an index exists). Consider the map

$h : \check{\mathcal{C}}^{p + 1}(\vec{e}) \to \check{\mathcal{C}}^ p(\vec{e})$

given by the rule that for $i_0 < \ldots < i_ p$ we have

$h(s)_{i_0 \ldots i_ p} = \left\{ \begin{matrix} 0 & \text{if} & p \not\in \{ 0, \ldots , n - 1\} \\ 0 & \text{if} & i_{\text{fix}} \in \{ i_0, \ldots , i_ p\} \\ s_{i_{\text{fix}} i_0 \ldots i_ p} & \text{if} & i_{\text{fix}} < i_0 \\ (-1)^ a s_{i_0 \ldots i_{a - 1} i_{\text{fix}} i_ a \ldots i_ p} & \text{if} & i_{a - 1} < i_{\text{fix}} < i_ a \\ (-1)^ p s_{i_0 \ldots i_ p} & \text{if} & i_ p < i_{\text{fix}} \end{matrix} \right.$

Please compare with the proof of Lemma 30.2.1. This makes sense because we have

$NEG(\vec{e}) \subset \{ i_0, \ldots , i_ p\} \Leftrightarrow NEG(\vec{e}) \subset \{ i_{\text{fix}}, i_0, \ldots , i_ p\}$

The exact same (combinatorial) computation1 as in the proof of Lemma 30.2.1 shows that

$(hd + dh)(s)_{i_0 \ldots i_ p} = s_{i_0 \ldots i_ p}$

Hence we see that the identity map of the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ is homotopic to zero which implies that it is acyclic. $\square$

[1] For example, suppose that $i_0 < \ldots < i_ p$ is such that $i_{\text{fix}} \not\in \{ i_0, \ldots , i_ p\}$ and that $i_{a - 1} < i_{\text{fix}} < i_ a$ for some $1 \leq a \leq p$. Then we have
\begin{align*} & (dh + hd)(s)_{i_0 \ldots i_ p} \\ & = \sum \nolimits _{j = 0}^ p (-1)^ j h(s)_{i_0 \ldots \hat i_ j \ldots i_ p} + (-1)^ a d(s)_{i_0 \ldots i_{a - 1} i_{\text{fix}} i_ a \ldots i_ p}\\ & = \sum \nolimits _{j = 0}^{a - 1} (-1)^{j + a - 1} s_{i_0 \ldots \hat i_ j \ldots i_{a - 1} i_{\text{fix}} i_ a \ldots i_ p} + \sum \nolimits _{j = a}^ p (-1)^{j + a} s_{i_0 \ldots i_{a - 1} i_{\text{fix}} i_ a \ldots \hat i_ j \ldots i_ p} + \\ & \sum \nolimits _{j = 0}^{a - 1} (-1)^{a + j} s_{i_0 \ldots \hat i_ j \ldots i_{a - 1} i_{\text{fix}} i_ a \ldots i_ p} + (-1)^{2a} s_{i_0 \ldots i_ p} + \sum \nolimits _{j = a}^ p (-1)^{a + j + 1} s_{i_0 \ldots i_{a - 1} i_{\text{fix}} i_ a \ldots \hat i_ j \ldots i_ p} \\ & = s_{i_0 \ldots i_ p} \end{align*}
as desired. The other cases are similar.

Comments (6)

Comment #2696 by Zhang on

Reference: EGA, Chapitre III, "Étude cohomologique des faisceaux cohérents", Proposition (2.1.12)

Comment #3244 by Aravind Asok on

In the case $NEG(\vec(e)) = \{0,\ldots,n\}$, I think the expression $\frac{1}{T^{-e_0}\cdots T^{-e_n}}$ should be $\frac{1}{T_0^{-e_0}\cdots T_n^{-e_n}}$ (twice). In the case $NEG(\vec(e)) = \emptyset$ (near the end), $R \cdot T^{e_0}\cdots T^{e_n}$ should be $R \cdot T_0^{e_0} \cdots T_n^{e_n}$.

Comment #3343 by on

Actually, I think we should perhaps replace this whole proof by thinking about the Koszul complex discussed in Section 15.28. But for now I have corrected the typos you pointed out, see here. Thanks!

Comment #6667 by WhatJiaranEatsTonight on

I think there are some little problem for the discussion of the last case. If $NEG(e)$ has only one integral then its zero cech complex is not zero.(But undoubtly the differential is injective in this case).

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