The Stacks project

Lemma 32.15.2. Let $S$ be a locally Noetherian scheme. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is locally of finite type. The following are equivalent:

  1. The morphism $f$ is separated.

  2. For any diagram ( there is at most one dotted arrow.

  3. For all diagrams ( with $A$ a discrete valuation ring there is at most one dotted arrow.

  4. For any irreducible component $X_0$ of $X$ with generic point $\eta \in X_0$, for any discrete valuation ring $A \subset K = \kappa (\eta )$ with fraction field $K$ and any diagram ( such that the morphism $\mathop{\mathrm{Spec}}(K) \to X$ is the canonical one (see Schemes, Section 26.13) there is at most one dotted arrow.

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
\begin{equation} \label{limits-equation-valuative-alpha} \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X_\alpha \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & S_\alpha } \end{equation}

would then give two distinct arrows in diagram ( 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 (, 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:
\begin{equation} \label{limits-equation-valuative-generic} \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X_0 \ar[d]^{\Delta } \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ur]& X_0 \times _ S X_0 \\ } \end{equation}

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:
\begin{equation} \label{limits-equation-valuative-nonexistent} \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X_0 \ar[d]\\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ur] & S \\ } \end{equation}

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). $\square$

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