The Stacks project

Proof. Combining Limits, Proposition 32.5.4 and Lemma 36.36.9 this reduces to the case where $X$ is Noetherian (small detail omitted). Assume $X$ is Noetherian. Recall that $X \times X$ is covered by the affine opens $U \times V$ for affine opens $U$, $V$ of $X$, see Schemes, Section 26.17. Hence to show that the diagonal $\Delta : X \to X \times X$ is affine, it suffices to show that $U \cap V = \Delta ^{-1}(U \times V)$ is affine for all affine opens $U$, $V$ of $X$, see Morphisms, Lemma 29.11.3. In particular, it suffices to show that the inclusion morphism $j : U \to X$ is affine if $U$ is an affine open of $X$. By Cohomology of Schemes, Lemma 30.3.4 it suffices to show that $R^1j_*\mathcal{G} = 0$ for any quasi-coherent $\mathcal{O}_ U$-module $\mathcal{G}$. By Proposition 36.8.3 (this is where we use that we've reduced to the Noetherian case) we can represent $Rj_*\mathcal{G}$ by a complex $\mathcal{H}^\bullet$ of quasi-coherent $\mathcal{O}_ X$-modules. Assume

is nonzero in order to get a contradiction. Then we can find a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ and a map

such that the composition with the projection onto $H^1(\mathcal{H}^\bullet )$ is nonzero. Namely, we can write $\mathop{\mathrm{Ker}}(\mathcal{H}^1 \to \mathcal{H}^2)$ as the filtered union of its coherent submodules by Properties, Lemma 28.22.3 and then one of these will do the job. Next, we choose a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ and a surjection $\mathcal{E} \to \mathcal{F}$ using the resolution property of $X$. This produces a map in the derived category

which is nonzero on cohomology sheaves and hence nonzero in $D(\mathcal{O}_ X)$. By adjunction, this is the same thing as a map

nonzero in $D(\mathcal{O}_ U)$. Since $\mathcal{E}$ is finite locally free this is the same thing as a nonzero element of

where $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{O}_ X)$ is the dual finite locally free module. However, this group is zero by Cohomology of Schemes, Lemma 30.2.2 which is the desired contradiction. (If in doubt about the step using duals, please see the more general Cohomology, Lemma 20.50.5.) $\square$