The Stacks project

Proof. Let $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$. If $\mathcal{I}$ is quasi-coherent then we just take $Z$ to be the closed subscheme determined by $\mathcal{I}$, see Lemma 29.2.3. This works by Schemes, Lemma 26.4.6. In general the same lemma requires us to show that there exists a largest quasi-coherent sheaf of ideals $\mathcal{I}'$ contained in $\mathcal{I}$. This follows from Lemma 29.4.2. $\square$

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

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. $\square$

Proof. Let $z \in \mathop{\mathrm{Spec}}(R) = V \subset Y$ be an affine open neighbourhood of $z$. By Lemma 29.6.3 the intersection $Z \cap V$ is the scheme theoretic image of $f^{-1}(V) \to V$. Hence we may replace $Y$ by $V$ and assume $Y = \mathop{\mathrm{Spec}}(R)$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Say $X = U_1 \cup \ldots \cup U_ n$ is a finite affine open covering. Write $U_ i = \mathop{\mathrm{Spec}}(A_ i)$. Let $I = \mathop{\mathrm{Ker}}(R \to A_1 \times \ldots \times A_ n)$. By Lemma 29.6.3 again we see that $Z$ corresponds to the closed subscheme $\mathop{\mathrm{Spec}}(R/I)$ of $Y$. If $\mathfrak p \subset R$ is the prime corresponding to $z$, then we see that $I_{\mathfrak p} \subset R_{\mathfrak p}$ is not an equality. Hence (as localization is exact, see Algebra, Proposition 10.9.12) we see that $R_{\mathfrak p} \to (A_1)_{\mathfrak p} \times \ldots \times (A_ n)_{\mathfrak p}$ is not zero. Hence one of the rings $(A_ i)_{\mathfrak p}$ is not zero. Hence there exists an $i$ and a prime $\mathfrak q_ i \subset A_ i$ lying over a prime $\mathfrak p_ i \subset \mathfrak p$. By Algebra, Lemma 10.50.2 we can choose a valuation ring $A \subset K = \kappa (\mathfrak q_ i)$ dominating the local ring $R_{\mathfrak p}/\mathfrak p_ iR_{\mathfrak p} \subset \kappa (\mathfrak q_ i)$. This gives the desired diagram. Some details omitted. $\square$

Proof. The scheme theoretic inverse image of $Z_2$ in $Y_1$ is a closed subscheme of $Y_1$ through which $f_1$ factors. Hence $Z_1$ is contained in this. This proves the lemma. $\square$

Proof. This is true because the reduced induced scheme structure on $\overline{f(X)}$ is clearly the smallest closed subscheme of $Y$ through which $f$ factors, see Schemes, Lemma 26.12.7. $\square$

Proof. The assumption that $V$ is retrocompact in $Y$ (Topology, Definition 5.12.1) means that $V \to Y$ is a quasi-compact morphism. By Schemes, Lemma 26.21.14 the morphism $s : V \to X$ is quasi-compact. Hence the construction of the scheme theoretic image $Y'$ of $s$ commutes with restriction to opens by Lemma 29.6.3. In particular, we see that $Y' \cap f^{-1}(V)$ is the scheme theoretic image of a section of the separated morphism $f^{-1}(V) \to V$. Since a section of a separated morphism is a closed immersion (Schemes, Lemma 26.21.11), we conclude that $Y' \cap f^{-1}(V) \to V$ is an isomorphism as desired. $\square$