The Stacks project

30.3 Vanishing of cohomology

We have seen that on an affine scheme the higher cohomology groups of any quasi-coherent sheaf vanish (Lemma 30.2.2). It turns out that this also characterizes affine schemes. We give two versions.

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Lemma 30.3.1. Let $X$ be a scheme. Assume that

  1. $X$ is quasi-compact,

  2. for every quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ we have $H^1(X, \mathcal{I}) = 0$.

Then $X$ is affine.

Proof. Let $x \in X$ be a closed point. Let $U \subset X$ be an affine open neighbourhood of $x$. Write $U = \mathop{\mathrm{Spec}}(A)$ and let $\mathfrak m \subset A$ be the maximal ideal corresponding to $x$. Set $Z = X \setminus U$ and $Z' = Z \cup \{ x\} $. By Schemes, Lemma 26.12.4 there are quasi-coherent sheaves of ideals $\mathcal{I}$, resp. $\mathcal{I}'$ cutting out the reduced closed subschemes $Z$, resp. $Z'$. Consider the short exact sequence

\[ 0 \to \mathcal{I}' \to \mathcal{I} \to \mathcal{I}/\mathcal{I}' \to 0. \]

Since $x$ is a closed point of $X$ and $x \not\in Z$ we see that $\mathcal{I}/\mathcal{I}'$ is supported at $x$. In fact, the restriction of $\mathcal{I}/\mathcal{I'}$ to $U$ corresponds to the $A$-module $A/\mathfrak m$. Hence we see that $\Gamma (X, \mathcal{I}/\mathcal{I'}) = A/\mathfrak m$. Since by assumption $H^1(X, \mathcal{I}') = 0$ we see there exists a global section $f \in \Gamma (X, \mathcal{I})$ which maps to the element $1 \in A/\mathfrak m$ as a section of $\mathcal{I}/\mathcal{I'}$. Clearly we have $x \in X_ f \subset U$. This implies that $X_ f = D(f_ A)$ where $f_ A$ is the image of $f$ in $A = \Gamma (U, \mathcal{O}_ X)$. In particular $X_ f$ is affine.

Consider the union $W = \bigcup X_ f$ over all $f \in \Gamma (X, \mathcal{O}_ X)$ such that $X_ f$ is affine. Obviously $W$ is open in $X$. By the arguments above every closed point of $X$ is contained in $W$. The closed subset $X \setminus W$ of $X$ is also quasi-compact (see Topology, Lemma 5.12.3). Hence it has a closed point if it is nonempty (see Topology, Lemma 5.12.8). This would contradict the fact that all closed points are in $W$. Hence we conclude $X = W$.

Choose finitely many $f_1, \ldots , f_ n \in \Gamma (X, \mathcal{O}_ X)$ such that $X = X_{f_1} \cup \ldots \cup X_{f_ n}$ and such that each $X_{f_ i}$ is affine. This is possible as we've seen above. By Properties, Lemma 28.27.3 to finish the proof it suffices to show that $f_1, \ldots , f_ n$ generate the unit ideal in $\Gamma (X, \mathcal{O}_ X)$. Consider the short exact sequence

\[ \xymatrix{ 0 \ar[r] & \mathcal{F} \ar[r] & \mathcal{O}_ X^{\oplus n} \ar[rr]^{f_1, \ldots , f_ n} & & \mathcal{O}_ X \ar[r] & 0 } \]

The arrow defined by $f_1, \ldots , f_ n$ is surjective since the opens $X_{f_ i}$ cover $X$. We let $\mathcal{F}$ be the kernel of this surjective map. Observe that $\mathcal{F}$ has a filtration

\[ 0 = \mathcal{F}_0 \subset \mathcal{F}_1 \subset \ldots \subset \mathcal{F}_ n = \mathcal{F} \]

so that each subquotient $\mathcal{F}_ i/\mathcal{F}_{i - 1}$ is isomorphic to a quasi-coherent sheaf of ideals. Namely we can take $\mathcal{F}_ i$ to be the intersection of $\mathcal{F}$ with the first $i$ direct summands of $\mathcal{O}_ X^{\oplus n}$. The assumption of the lemma implies that $H^1(X, \mathcal{F}_ i/\mathcal{F}_{i - 1}) = 0$ for all $i$. This implies that $H^1(X, \mathcal{F}_2) = 0$ because it is sandwiched between $H^1(X, \mathcal{F}_1)$ and $H^1(X, \mathcal{F}_2/\mathcal{F}_1)$. Continuing like this we deduce that $H^1(X, \mathcal{F}) = 0$. Therefore we conclude that the map

\[ \xymatrix{ \bigoplus \nolimits _{i = 1, \ldots , n} \Gamma (X, \mathcal{O}_ X) \ar[rr]^{f_1, \ldots , f_ n} & & \Gamma (X, \mathcal{O}_ X) } \]

is surjective as desired. $\square$

Note that if $X$ is a Noetherian scheme then every quasi-coherent sheaf of ideals is automatically a coherent sheaf of ideals and a finite type quasi-coherent sheaf of ideals. Hence the preceding lemma and the next lemma both apply in this case.

sloganreference

Lemma 30.3.2. Let $X$ be a scheme. Assume that

  1. $X$ is quasi-compact,

  2. $X$ is quasi-separated, and

  3. $H^1(X, \mathcal{I}) = 0$ for every quasi-coherent sheaf of ideals $\mathcal{I}$ of finite type.

Then $X$ is affine.

Proof. By Properties, Lemma 28.22.3 every quasi-coherent sheaf of ideals is a directed colimit of quasi-coherent sheaves of ideals of finite type. By Cohomology, Lemma 20.19.1 taking cohomology on $X$ commutes with directed colimits. Hence we see that $H^1(X, \mathcal{I}) = 0$ for every quasi-coherent sheaf of ideals on $X$. In other words we see that Lemma 30.3.1 applies. $\square$

We can use the arguments given above to find a sufficient condition to see when an invertible sheaf is ample. However, we warn the reader that this condition is not necessary.

Lemma 30.3.3. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume that

  1. $X$ is quasi-compact,

  2. for every quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ there exists an $n \geq 1$ such that $H^1(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$.

Then $\mathcal{L}$ is ample.

Proof. This is proved in exactly the same way as Lemma 30.3.1. Let $x \in X$ be a closed point. Let $U \subset X$ be an affine open neighbourhood of $x$ such that $\mathcal{L}|_ U \cong \mathcal{O}_ U$. Write $U = \mathop{\mathrm{Spec}}(A)$ and let $\mathfrak m \subset A$ be the maximal ideal corresponding to $x$. Set $Z = X \setminus U$ and $Z' = Z \cup \{ x\} $. By Schemes, Lemma 26.12.4 there are quasi-coherent sheaves of ideals $\mathcal{I}$, resp. $\mathcal{I}'$ cutting out the reduced closed subschemes $Z$, resp. $Z'$. Consider the short exact sequence

\[ 0 \to \mathcal{I}' \to \mathcal{I} \to \mathcal{I}/\mathcal{I}' \to 0. \]

For every $n \geq 1$ we obtain a short exact sequence

\[ 0 \to \mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \to \mathcal{I}/\mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \to 0. \]

By our assumption we may pick $n$ such that $H^1(X, \mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$. Since $x$ is a closed point of $X$ and $x \not\in Z$ we see that $\mathcal{I}/\mathcal{I}'$ is supported at $x$. In fact, the restriction of $\mathcal{I}/\mathcal{I'}$ to $U$ corresponds to the $A$-module $A/\mathfrak m$. Since $\mathcal{L}$ is trivial on $U$ we see that the restriction of $\mathcal{I}/\mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}$ to $U$ also corresponds to the $A$-module $A/\mathfrak m$. Hence we see that $\Gamma (X, \mathcal{I}/\mathcal{I'} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = A/\mathfrak m$. By our choice of $n$ we see there exists a global section $s \in \Gamma (X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n})$ which maps to the element $1 \in A/\mathfrak m$. Clearly we have $x \in X_ s \subset U$ because $s$ vanishes at points of $Z$. This implies that $X_ s = D(f)$ where $f \in A$ is the image of $s$ in $A \cong \Gamma (U, \mathcal{L}^{\otimes n})$. In particular $X_ s$ is affine.

Consider the union $W = \bigcup X_ s$ over all $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ for $n \geq 1$ such that $X_ s$ is affine. Obviously $W$ is open in $X$. By the arguments above every closed point of $X$ is contained in $W$. The closed subset $X \setminus W$ of $X$ is also quasi-compact (see Topology, Lemma 5.12.3). Hence it has a closed point if it is nonempty (see Topology, Lemma 5.12.8). This would contradict the fact that all closed points are in $W$. Hence we conclude $X = W$. This means that $\mathcal{L}$ is ample by Properties, Definition 28.26.1. $\square$

There is a variant of Lemma 30.3.3 with finite type ideal sheaves which we will formulate and prove here if we ever need it.

Lemma 30.3.4. Let $f : X \to Y$ be a quasi-compact morphism with $X$ and $Y$ quasi-separated. If $R^1f_*\mathcal{I} = 0$ for every quasi-coherent sheaf of ideals $\mathcal{I}$ on $X$, then $f$ is affine.

Proof. Let $V \subset Y$ be an affine open subscheme. We have to show that $U = f^{-1}(V)$ is affine. The inclusion morphism $V \to Y$ is quasi-compact by Schemes, Lemma 26.21.14. Hence the base change $U \to X$ is quasi-compact, see Schemes, Lemma 26.19.3. Thus any quasi-coherent sheaf of ideals $\mathcal{I}$ on $U$ extends to a quasi-coherent sheaf of ideals on $X$, see Properties, Lemma 28.22.1. Since the formation of $R^1f_*$ is local on $Y$ (Cohomology, Section 20.7) we conclude that $R^1(U \to V)_*\mathcal{I} = 0$ by the assumption in the lemma. Hence by the Leray Spectral sequence (Cohomology, Lemma 20.13.4) we conclude that $H^1(U, \mathcal{I}) = H^1(V, (U \to V)_*\mathcal{I})$. Since $(U \to V)_*\mathcal{I}$ is quasi-coherent by Schemes, Lemma 26.24.1, we have $H^1(V, (U \to V)_*\mathcal{I}) = 0$ by Lemma 30.2.2. Thus we find that $U$ is affine by Lemma 30.3.1. $\square$


Comments (6)

Comment #932 by correction_bot on

"This is a would contradict the fact that all…", delete "is a".

Comment #4237 by Si Yu How on

In the proof of Lemma 01XF, instead of working with a closed point , we can work with an open set where . Then following the exact same proof we can find such that is a principal open subscheme of and .

Since is quasi-compact, it has a finite affine covering . Since , there exists such that is affine open and . Similarly, since , there exists such that is affine open and . By iterating such steps we can get and that each is affine open.

This approach has the advantage that it avoids Zorn's lemma.

Comment #4238 by Si Yu How on

In the above comment, I mean whenever I wrote .

Comment #4416 by on

Dear Si Yu How, I do not understand how the open is chosen? Namely, we need the quotient to have "recognizable" global sections. So the argument doesn't work if is affine open, is arbitrary open, cuts out the complement of , and cuts out the complement of (I do mean intersection here).

Maybe what you mean is to not work with closed points but with finite type points, which are those points of which correspond to closed points of affine opens? (See Lemma 29.15.4). This would work because finite type points are always dense (Lemma 29.15.7). But I actually thought it was fun that you can sometimes get away with only looking at closed points...

The axiom of choice is really key in many of the arguments in the Stacks project. I personally absolutely love the axiom of choice.

Comment #4867 by bangphamkhoa on

In the proof of Lemma 01XF, I think should be quasi-separated because otherwise even if is affine, it is not sure that the restriction of the canonical morphism to is an isomorphism. The hypothesis of being quasi-separated (for example, Noetherian scheme) allows one to use the qcqs lemma.

Comment #4871 by on

@#4867. Sorry, I do not understand your comment. I carefully checked and none of the arguments use quasi-separatedness. Also, I don't understand what you mean by the "qcqs lemma". In addition, if you have a comment about a particular lemma, please leave a comment on the page of the lemma in question.


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