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.
$\square$

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