Lemma 29.6.3. Let $f : X \to Y$ be a morphism of schemes. Let $Z \subset Y$ be the scheme theoretic image of $f$. If $f$ is quasi-compact then

1. the sheaf of ideals $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$ is quasi-coherent,

2. the scheme theoretic image $Z$ is the closed subscheme determined by $\mathcal{I}$,

3. for any open $U \subset Y$ the scheme theoretic image of $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is equal to $Z \cap U$, and

4. the image $f(X) \subset Z$ is a dense subset of $Z$, in other words the morphism $X \to Z$ is dominant (see Definition 29.8.1).

Proof. Part (4) follows from part (3). To show (3) it suffices to prove (1) since the formation of $\mathcal{I}$ commutes with restriction to open subschemes of $Y$. And if (1) holds then in the proof of Lemma 29.6.1 we showed (2). Thus it suffices to prove that $\mathcal{I}$ is quasi-coherent. Since the property of being quasi-coherent is local we may assume $Y$ is affine. As $f$ is quasi-compact, we can find a finite affine open covering $X = \bigcup _{i = 1, \ldots , n} U_ i$. Denote $f'$ the composition

$X' = \coprod U_ i \longrightarrow X \longrightarrow Y.$

Then $f_*\mathcal{O}_ X$ is a subsheaf of $f'_*\mathcal{O}_{X'}$, and hence $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f'_*\mathcal{O}_{X'})$. By Schemes, Lemma 26.24.1 the sheaf $f'_*\mathcal{O}_{X'}$ is quasi-coherent on $Y$. Hence we win. $\square$

## Comments (2)

Comment #4285 by Dario Weißmann on

typo: $\mathcal{I}=\text{Ker}(\mathcal{O}_Y\to \mathcal{O}_{X'})$ is missing an $f'_{\ast}$

There are also:

• 9 comment(s) on Section 29.6: Scheme theoretic image

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