Can we construct cohomolgy theory on noetherian separated schemes without Axiom of Choice? 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.
 A: 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 $\beta(b) = c$. $\square$
