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When $X$ is assumed to be a Noetherian, the answer is affirmative (Hartshorne Exercise III 3.6). It comes down to showing that injectives in $\mathfrak{Qco} (X)$ are flasque. The proof can be given by showing that if $\mathcal{I}\in\mathfrak{Qco}(X)$ is injective and $U\subset X$ is open, then $\mathcal{I}|_{U}\in \mathfrak{Qco}(U)$ is still injective. Then by covering $X$ with open affines $U_i=\operatorname{Spec} (A_i)$, we have $\mathcal{I}|_{U_i}\cong \widetilde{M_i}$ for some $A_i$-module $M_i$. In addition $M_i$ must actually be injective since taking global sections of a quasi-coherent sheaf over an affine scheme is exact. Finally Proposition III 3.4 then tells us that $\mathcal{I}|_{U_i}$ is flasque and from here we can just glue local sections together to satisfy the flasque condition for $\mathcal{I}$

This particular proof uses the Noetherian hypothesis a few times, and most crucially in the application of Proposition III 3.4.

So the question remains: Is this still true if we remove the Noetherian hypothesis?

Edit: The application of Proposition III 3.4 does not require the Noetherian hypothesis, as pointed out below. On the other hand when $X$ is not Noetherian, it will not be true in general that the restriction of an injective quasicoherent sheaf to an open subscheme will still be an injective quasicoherenet sheaf.

Edit 2: Here's a construction from Stacks Project that might be useful in finding a counterexample: https://stacks.math.columbia.edu/tag/0273

If $I$ is an injective $A$-module, then $\tilde{I}$ is an injective sheaf in Qcoh(Spec A) (c.f. Hartshorne III Ex. 3.6a). Thus the higher cohomology of $\tilde{I}$ in Qcoh(Spec A) is 0. The Stacks Project construction almost seems to do the trick, but not quite. We'd need to show that the restriction of $\tilde{I}$ to U is still injective, or at the very least still has vanishing H^1. This is known to be true for a locally Noetherian scheme, but in the Stacks example we only have a topologically Noetherian scheme. Perhaps the result could be extended...

Nathan Lowry
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    If you're only concerned about III.3.4, then this extends to the non-noetherian case. See for instance Stacks or this MSE question. Are there other specific places in which you're worried about the noetherian hypothesis? – KReiser Feb 22 '20 at 20:53
  • I was originally concerned "$\mathcal{I}|_{U_i}$ is flasque for an open cover $X=\cup U_i$ implies $\mathcal{I}$ is flasque" would require the Noetherian hypothesis, but I managed to write down a proof without it using Zorn's lemma. The other place we use the hypothesis is in the very first step: restricting an injective quasicoherent to an open subscheme is still an injective quasicoherent. The only proofs I can imagine certainly use the Noetherian hypothesis since they will require the push forward of a quasicoherent sheaf to be quasicoherent, which we cannot guarantee in general. – Nathan Lowry Feb 22 '20 at 22:10
  • @KReiser We used the noetherian hypothesis in the application of Proposition III 3.4, but your links are for Propisition III 2.4. – Nathan Lowry Feb 22 '20 at 22:18
  • No, they handle the case of III.3.4 as well via the equivalence of categories between $Qcoh(\operatorname{Spec} R)$ and $R$-mod (this is completely general and does not require a noetherian hypothesis). As for restriction to an open subscheme, could you elaborate on where you need the pushforwards? – KReiser Feb 22 '20 at 22:38
  • @KReiser Ahh, I see now, thanks! The proof I had in mind would use the adjunction between $i_{\ast}$ and $i^{-1}$, but it appears the statement is actually false when $X$ is not Noetherian. See the comments here – Nathan Lowry Feb 22 '20 at 23:01
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    See also this blog for a list of facts on non noetherian schemes : https://www.math.columbia.edu/~dejong/wordpress/?p=1982 – Roland Feb 23 '20 at 15:46
  • @Roland So this blog post claims that the statement is false. That's very surprising to me, as I had been expecting the statement to be true. Do you happen to have a source for a counterexample? Thanks! – Nathan Lowry Feb 23 '20 at 19:17
  • Unfortunately no, I did not manage to find one. – Roland Feb 23 '20 at 23:03

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