The usual cohomology theory on schemes uses injective or flasque resolutions of quasi-coherent sheaves. Hence it uses Axiom of Choice. However, if the base scheme is a noetherian separated scheme, the usual cohomology coincides with Cech cohomology which seems to be more constructive than that. So I think it's likely that we can construct the cohomology theory without Axiom of Choice on such schemes. Is my guess right?. Stated more clearly, I'm asking if the usual cohomology theory on schemes can be proved without Axiom of Choice if the base schemes are restricted to noetherian separated schemes. By the usual cohomology theory on schemes, I mean, for example, the one written in the Hartshorne's book.

In particular can we prove the following assertions without Axiom of Choice?

1) We can define the group $H^i(X, \mathcal F)$ for a quasi-coherent sheaf $\mathcal F$ on a noetherian separated scheme $X$ without Axiom of Choice.

2) It has the long cohomology sequence for every exact short sequence of quasi-cohherent sheaves over such a scheme.

3) $H^i(X, \mathcal F) = 0$ for $i \gt 0$ if $\mathcal F$ is flasque.

4) $H^i(X, \mathcal F) = 0$ for $i \gt 0$ if $X$ is a noetherian affine scheme

5) It satisfies Theorem 5.1, Theorem 5.2 and Proposition 5.3 of Hartshorne's book.

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    $\begingroup$ I'm not entirely clear what the question in the OP is. If, however, it is "How much of the cohomology theory of noetherian separated schemes relies on AC?", this seems a real and appropriate-for-MO question. Or am I missing something? $\endgroup$ Apr 17, 2014 at 4:19
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    $\begingroup$ I'll point out the (very well-known) fact that $\Pi^1_2$ sentences don't depend on AC, so any very concrete consequence of cohomological reasoning will not require choice to prove. But presumably you are interested in abstract results specifically about cohomology, which is again why you need to specify what exactly you want to do. $\endgroup$ Apr 17, 2014 at 21:34
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    $\begingroup$ (In particular, specifying what you are asking about - besides just being a good idea - will show that you've put serious thought into the question, and I suspect for a variety of reasons that doubt about this is part of the reason for the votes to close.) $\endgroup$ Apr 17, 2014 at 21:46
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    $\begingroup$ Again, what do you mean by "the usual cohomology theory on schemes?" Do you have specific results in mind? "Cohomology theory" is very big! Certainly some results will require choice, and some results won't; what exactly are you asking about? $\endgroup$ Apr 18, 2014 at 1:56
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    $\begingroup$ "The one written in [book]" is far too broad (unless if a single theorem in the book fails to hold without choice, then you would say the answer is "no"). In order to make this an unambiguously real question, you have to actually tell us what you are asking: what results specifically do you want to remain true? In particular, your comment on David Speyer's answer (". . . affirmative or not . . .") implies that you're looking for a yes/no answer; but for such an answer to exist, you have to actually ask a yes/no question! $\endgroup$ Apr 18, 2014 at 3:43

1 Answer 1


This is basically a comment, but I don't want it to get lost in the conversation about defining the question: The Godemont resolution of a sheaf (of abelian groups, on any topological space) can be defined without making any choices. If you define sheaf cohomology groups as the cohomology of the Godemont resolution, I would guess that you can prove whatever you consider to be the primary results about them without use of choice.

In response to the comment below: Interesting, you are right! The construction literally makes sense without choice, but it doesn't give a long exact sequence. Indeed, if $X$ is a discrete space and $\mathcal{E}$ a sheaf on $X$, then $Gode(\mathcal{E}) \cong \mathcal{E}$, so the Godemont resolution stops in one step and all higher cohomology vanishes. However, as Blass shows, in the absence of choice, if we want long exact sequences, we must define $H^1(X, \mathcal{G})$ to be nonzero for discrete $X$ in some cases.

The fundamental issue is that, without choice, we can have a collection of exact sequences $0 \to A_i \to B_i \to C_i \to 0$, indexed by $i \in I$, so that $0 \to \prod A_i \to \prod B_i \to \prod C_i \to 0$ is not exact. Interesting!

$\def\cA{\mathcal{A}}\def\cB{\mathcal{B}}\def\cC{\mathcal{C}}$ As discussed in comments below, let $X$ be quasi-compact, let $\cA$ be a flasque sheaf on $X$ and let $0 \to \cA \overset{\alpha}{\longrightarrow} \cB \overset{\beta}{\longrightarrow} \cC \to 0$ be exact. Then I think that $0 \to \cA(X) \to \cB(X) \to \cC(X) \to 0$ is exact.

I am assuming that we can prove that sheaves form an abelian category in the first place, and that surjectivity means surjectivity on stalks. I am also only checking surjectivity of $\cB(X) \to \cC(X)$, since that is the hard case in the presence of choice.

Let $c \in \cC(X)$.

Let $\mathcal{U}$ be the set of open sets $U$ in $X$ so that there exists $b \in \cB(U)$ with $\beta(b) = c|_U$. By one possible definition of surjectivity, $\bigcup_{U \in \mathcal{U}} U = X$. By quasi-compactness, there is some finite list of sets $U_1$, $U_2$, ..., $U_n$ in $\mathcal{U}$ with $\bigcup U_i = X$. I'll use the stadnard shorthand $U_{ij} = U_i \cap U_j$, etc.

Choose (finitely many choices) elements $b_i$ in $\cB(U_i)$ with $\beta(b_i) = c|_{U_i}$. Define $a_{ij} = b_i|_{U_{ij}} - b_j|_{U_{ij}}$; note that $\beta(a_{ij})=0$ so $a_{ij} \in \cA(U_{ij})$. Observe also that we have the Cech cocycle condition $$a_{ij}|_{U_{ijk}} + a_{jk}|_{U_{ijk}} + a_{ki}|_{U_{ijk}} = 0 \quad (\dagger)$$ and $a_{ij} = - a_{ji}$.

Lemma Given $a_{ij}$ obeying $(\dagger)$ and $a_{ij} = - a_{ji}$, we can find $a_i \in \cA(U_i)$ with $$a_i|_{U_{ij}} - a_j|_{U_{ij}} = a_{ij} \quad (\ast)$$

This is the proof I came up with when I did this assignment, but I couldn't find a source that does it this way so I'm writing it up.

Proof We show by induction on $m$ that we can construct $a_1$, $a_2$, ..., $a_m$ so that $(\ast)$ holds whenever $1 \leq i < j \leq m$. The base case, $m=1$, is vacuously true and the case $m=n$ is the desired claim.

Suppose that $a_1$, ..., $a_{m-1}$ have been constructed. For $i < m$, set $a'_i = a_i|_{U_{im}}- a_{im}$. Then $$a'_i|_{U_{ijm}} - a'_j|_{U_{ijm}} = a_i|_{U_{ijm}} - a_{im}|_{U_{ijm}} - a_j|_{U_{ijm}} + a_{jm}|_{U_{ijm}} =$$ $$ (a_i|_{U_i} - a_j|_{U_j})|_{U_{ijm}} - a_{im}|_{U_{ijm}} + a_{jm}|_{U_{ijm}} =a_{ij}|_{U_{ijm}} - a_{im}|_{U_{ijm}} + a_{jm}|_{U_{ijm}} =0$$ where the last two equalitites are the inductive hypothesis and $(\dagger)$.

So, by the sheaf condition, there is an element $a'$ in $\cA \left( \bigcup_{i < m} U_{im} \right)$ defined by $a'|_{U_{im}} = a'_i$. By flasqueness, we can choose (just one choice!) $a_m \in \cA(U_m)$ which restricts to $a'$ on $\bigcup_{i < m} U_{im}$. We then compute $$a_i - a_m|_{U_{im}} = a_i|_{U_{im}} - a'|_{U_{im}} = a_i|_{U_{im}} - a'_i$$ $$=a_i|_{U_{im}} - \left( a_i|_{U_{im}} - a_{im} \right) = a_{im}. \quad \square$$

Now, note that $$\left( b_i - \alpha(a_i) \right)|_{U_{ij}} - \left( b_j - \alpha(a_j) \right)|_{U_{ij}} = b_i|_{U_{ij}} - b_j|_{U_{ij}} - \alpha(a_{ij}) = 0.$$ So (by the sheaf condition) there is $b \in \cB$ so that $b|_{U_i} = b_i$. Then $\beta(b)|_{U_i} = c|_{U_i}$, and we conclude (sheaf condition one more time!) that $\beta(b) = c$. $\square$

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    $\begingroup$ [If you define sheaf cohomology groups as the cohomology of the Godemont resolution, I would guess that you can prove whatever you consider to be the primary results about them without use of choice.] Godement uses Axiom of Choice in his method. Please read this question. $\endgroup$ Apr 17, 2014 at 23:29
  • $\begingroup$ Could you tell me whether my question is affirmative or not? With an explanation, of course. $\endgroup$ Apr 18, 2014 at 2:35
  • $\begingroup$ Still thinking about it. I think you might be able to construct cohomology of sheaves of abelian groups on any quasi-compact topological space. My thinking is that the Godemont construction is defined without choice, is flasque without choice, and thus shows that every sheaf has a flasque resolution. The answers to the question you linked suggest (but don't state) that Hartshorne 2.1.16b is true without choice on quasi-compact spaces. I think that this should be enough to show that you can define cohomology using any flasque resolution. $\endgroup$ Apr 18, 2014 at 13:42
  • $\begingroup$ I also hope Andreas Blass may show up here to give his thoughts; this is much more up his alley than mine. $\endgroup$ Apr 18, 2014 at 13:43
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    $\begingroup$ PS @MakotoKato You obviously have a passion for thinking about foundational issues in algebraic geometry in the failure of choice. This is perfectly reasonable niche for an academic researcher and, as far as I can tell, one that isn't taken. Have you considered starting writing your own papers on the subject, and perhaps eventually a monograph? $\endgroup$ Apr 18, 2014 at 13:46

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