Noetherian condition of Serre's theorem on affineness

Question

I am currently reading the proof of Hartshorne Theorem 3.7, which is a famous result of Serre. It stated the following: Let $X$ be a noetherian scheme, then the following are equivalent:

1. $X$ is affine.
2. $H^i(X,\mathfrak{F})=0$ for all quasi-coherent sheaf $\mathfrak{F}$ and all $i>0$.
3. $H^1(X,\mathfrak{I})=0$ for all coherent sheaves of ideals $\mathfrak{I}$.

It is clear that to prove 3 implies 1, we need quasi-compactness, which is implied by the noetherian condition. However, why can't we replace the noetherian condition by quasi-compactness since I don't think you need the full noetherian condition for the implication 3 implies 1.

I understand that Hartshorne used noetherian to prove 1 implies 2. However, as he stated in Remark 3.5.1, noetherian for this direction is not necessary, with proof in EGA 3.

Answer 1

I guess you are right. This is 01XF in the stacks project, or EGA II, Théorème (5.2.1).