Lemma 37.81.3. Let $X$ be a scheme. Let $f : X \to S = \mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))$ be the canonical morphism of Schemes, Lemma 26.6.4. The largest quasi-coherent $\mathcal{O}_ S$-module contained in the kernel of $f^\sharp : \mathcal{O}_ S \to f_*\mathcal{O}_ X$ is zero. If $X$ is quasi-compact, then $f^\sharp$ is injective. In particular, if $X$ is quasi-compact, then $f$ is a dominant morphism.

Proof. Let $M \subset \Gamma (S, \mathcal{O}_ S)$ be the submodule corresponding to the largest quasi-coherent $\mathcal{O}_ S$-module contained in the kernel of $f^\sharp$. Then any element $a \in M$ is mapped to zero by $f^\sharp$. However, $f^\sharp (a)$ is the element of

$\Gamma (S, f_*\mathcal{O}_ X) = \Gamma (X, \mathcal{O}_ X) = \Gamma (S, \mathcal{O}_ S)$

corresponding to $a$ itself! Thus $a = 0$. Hence $M = 0$ which proves the first assertion. Note that this is equivalent to the morphism $f : X \to S$ being scheme-theoretically surjective.

If $X$ is quasi-compact, then $\mathop{\mathrm{Ker}}(f^\sharp )$ is quasi-coherent by Morphisms, Lemma 29.6.3. Hence $\mathop{\mathrm{Ker}}(f^\sharp ) = 0$ and $f^\sharp$ is injective. In this case, $f$ is a dominant morphism by part (4) of Morphisms, Lemma 29.6.3. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).