The Stacks project

Proof. Clearly (1) implies (2), (2) implies (3), and (3) implies (4). It remains to show (4) implies (1). Assume (4). We begin by reducing to $S$ affine. Being separated is a local on the base (see Schemes, Lemma 26.21.7). Hence, if we can show that whenever $X \to S$ has (4) that the restriction $X_\alpha \to S_\alpha$ has (4) where $S_\alpha \subset S$ is an (affine) open subset and $X_\alpha := f^{-1}(S_\alpha )$, then we will be done. The generic points of the irreducible components of $X_\alpha$ will be the generic points of irreducible components of $X$, since $X_\alpha$ is open in $X$. Therefore, any two distinct dotted arrows in the diagram

would then give two distinct arrows in diagram (32.15.1.1) via the maps $X_\alpha \to X$ and $S_\alpha \to S$, which is a contradiction. Thus we have reduced to the case $S$ is affine. We remark that in the course of this reduction, we prove that if $X \to S$ has (4) then the restriction $U \to V$ has (4) for opens $U \subset X$ and $V \subset S$ with $f(U) \subset V$.

We next wish to reduce to the case $X \to S$ is finite type. Assume that we know (4) implies (1) when $X$ is finite type. Since $S$ is Noetherian and $X$ is locally of finite type over $S$ we see $X$ is locally Noetherian as well (see Morphisms, Lemma 29.15.6). Thus, $X \to S$ is quasi-separated (see Properties, Lemma 28.5.4), and therefore we may apply the valuative criterion to check whether $X$ is separated (see Schemes, Lemma 26.22.2). Let $X = \bigcup _\alpha X_\alpha$ be an affine open cover of $X$. Given any two dotted arrows, in a diagram (32.15.1.1), the image of the closed points of $\mathop{\mathrm{Spec}}A$ will fall in two sets $X_\alpha$ and $X_\beta$. Since $X_\alpha \cup X_\beta$ is open, for topological reasons it must contain the image of $\mathop{\mathrm{Spec}}(A)$ under both maps. Therefore, the two dotted arrows factor through $X_\alpha \cup X_\beta \to X$, which is a scheme of finite type over $S$. Since $X_\alpha \cup X_\beta$ is an open subset of $X$, by our previous remark, $X_\alpha \cup X_\beta$ satisfies (4), so by assumption, is separated. This implies the two given dotted arrows are the same. Therefore, we have reduced to $X \to S$ is finite type.

Assume $X \to S$ of finite type and assume (4). Since $X \to S$ is finite type, and $S$ is an affine Noetherian scheme, $X$ is also Noetherian (see Morphisms, Lemma 29.15.6). Therefore, $X \to X \times _ S X$ will be a quasi-compact immersion of Noetherian schemes. We proceed by contradiction. Assume that $X \to X \times _ S X$ is not closed. Then, there is some $y \in X \times _ S X$ in the closure of the image that is not in the image. As $X$ is Noetherian it has finitely many irreducible components. Therefore, $y$ is in the closure of the image of one of the irreducible components $X_0 \subset X$. Give $X_0$ the reduced induced structure. The composition $X_0 \to X \to X \times _ S X$ factors through the closed subscheme $X_0 \times _ S X_0 \subset X \times _ S X$. Denote the closure of $\Delta (X_0)$ in $X_0 \times _ S X_0$ by $\bar X_0$ (again as a reduced closed subscheme). Thus $y \in \bar X_0$. Since $X_0 \to X_0 \times _ S X_0$ is an immersion, the image of $X_0$ will be open in $\bar X_0$. Hence $X_0$ and $\bar X_0$ are birational. Since $\bar{X}_0$ is a closed subscheme of a Noetherian scheme, it is Noetherian. Thus, the local ring $\mathcal O_{{\bar X_0, y}}$ is a local Noetherian domain with fraction field $K$ equal to the function field of $X_0$. By the Krull-Akizuki theorem (see Algebra, Lemma 10.119.13), there exists a discrete valuation ring $A$ dominating $\mathcal O_{{\bar X_0, y}}$ with fraction field $K$. This allows to construct a diagram:

which sends $\mathop{\mathrm{Spec}}K$ to the generic point of $\Delta (X_0)$ and the closed point of $A$ to $y \in X_0 \times _ S X_0$ (use the material in Schemes, Section 26.13 to construct the arrows). There cannot even exist a set theoretic dotted arrow, since $y$ is not in the image of $\Delta$ by our choice of $y$. By categorical means, the existence of the dotted arrow in the above diagram is equivalent to the uniqueness of the dotted arrow in the following diagram:

Therefore, we have non-uniqueness in this latter diagram by the nonexistence in the first. Therefore, $X_0$ does not satisfy uniqueness for discrete valuation rings, and since $X_0$ is an irreducible component of $X$, we have that $X \to S$ does not satisfy (4). Therefore, we have shown (4) implies (1). $\square$