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The Stacks project

32.14 Universally closed morphisms

In this section we discuss when a quasi-compact (but not necessarily separated) morphism is universally closed. We first prove a lemma which will allow us to check universal closedness after a base change which is locally of finite presentation.

Lemma 32.14.1. Let f : X \to S be a quasi-compact morphism of schemes. Let g : T \to S be a morphism of schemes. Let t \in T be a point and Z \subset X_ T be a closed subscheme such that Z \cap X_ t = \emptyset . Then there exists an open neighbourhood V \subset T of t, a commutative diagram

\xymatrix{ V \ar[d] \ar[r]_ a & T' \ar[d]^ b \\ T \ar[r]^ g & S, }

and a closed subscheme Z' \subset X_{T'} such that

  1. the morphism b : T' \to S is locally of finite presentation,

  2. with t' = a(t) we have Z' \cap X_{t'} = \emptyset , and

  3. Z \cap X_ V maps into Z' via the morphism X_ V \to X_{T'}.

Moreover, we may assume V and T' are affine.

Proof. Let s = g(t). During the proof we may always replace T by an open neighbourhood of t. Hence we may also replace S by an open neighbourhood of s. Thus we may and do assume that T and S are affine. Say S = \mathop{\mathrm{Spec}}(A), T = \mathop{\mathrm{Spec}}(B), g is given by the ring map A \to B, and t correspond to the prime ideal \mathfrak q \subset B.

As X \to S is quasi-compact and S is affine we may write X = \bigcup _{i = 1, \ldots , n} U_ i as a finite union of affine opens. Write U_ i = \mathop{\mathrm{Spec}}(C_ i). In particular we have X_ T = \bigcup _{i = 1, \ldots , n} U_{i, T} = \bigcup _{i = 1, \ldots n} \mathop{\mathrm{Spec}}(C_ i \otimes _ A B). Let I_ i \subset C_ i \otimes _ A B be the ideal corresponding to the closed subscheme Z \cap U_{i, T}. The condition that Z \cap X_ t = \emptyset signifies that I_ i generates the unit ideal in the ring

C_ i \otimes _ A \kappa (\mathfrak q) = (B \setminus \mathfrak q)^{-1}\left( C_ i \otimes _ A B/\mathfrak q C_ i \otimes _ A B \right)

Since I_ i (B \setminus \mathfrak q)^{-1}(C_ i \otimes _ A B) = (B \setminus \mathfrak q)^{-1} I_ i this means that 1 = x_ i/g_ i for some x_ i \in I_ i and g_ i \in B, g_ i \not\in \mathfrak q. Thus, clearing denominators we can find a relation of the form

x_ i + \sum \nolimits _ j f_{i, j}c_{i, j} = g_ i

with x_ i \in I_ i, f_{i, j} \in \mathfrak q, c_{i, j} \in C_ i \otimes _ A B, and g_ i \in B, g_ i \not\in \mathfrak q. After replacing B by B_{g_1 \ldots g_ n}, i.e., after replacing T by a smaller affine neighbourhood of t, we may assume the equations read

x_ i + \sum \nolimits _ j f_{i, j}c_{i, j} = 1

with x_ i \in I_ i, f_{i, j} \in \mathfrak q, c_{i, j} \in C_ i \otimes _ A B.

To finish the argument write B as a colimit of finitely presented A-algebras B_\lambda over a directed set \Lambda . For each \lambda set \mathfrak q_\lambda = (B_\lambda \to B)^{-1}(\mathfrak q). For sufficiently large \lambda \in \Lambda we can find

  1. an element x_{i, \lambda } \in C_ i \otimes _ A B_\lambda which maps to x_ i,

  2. elements f_{i, j, \lambda } \in \mathfrak q_{i, \lambda } mapping to f_{i, j}, and

  3. elements c_{i, j, \lambda } \in C_ i \otimes _ A B_\lambda mapping to c_{i, j}.

After increasing \lambda a bit more the equation

x_{i, \lambda } + \sum \nolimits _ j f_{i, j, \lambda }c_{i, j, \lambda } = 1

will hold. Fix such a \lambda and set T' = \mathop{\mathrm{Spec}}(B_\lambda ). Then t' \in T' is the point corresponding to the prime \mathfrak q_\lambda . Finally, let Z' \subset X_{T'} be the scheme theoretic image of Z \to X_ T \to X_{T'}. As X_ T \to X_{T'} is affine, we can compute Z' on the affine open pieces U_{i, T'} as the closed subscheme associated to \mathop{\mathrm{Ker}}(C_ i \otimes _ A B_\lambda \to C_ i \otimes _ A B/I_ i), see Morphisms, Example 29.6.4. Hence x_{i, \lambda } is in the ideal defining Z'. Thus the last displayed equation shows that Z' \cap X_{t'} is empty. \square

Lemma 32.14.2. Let f : X \to S be a quasi-compact morphism of schemes. The following are equivalent

  1. f is universally closed,

  2. for every morphism S' \to S which is locally of finite presentation the base change X_{S'} \to S' is closed, and

  3. for every n the morphism \mathbf{A}^ n \times X \to \mathbf{A}^ n \times S is closed.

Proof. It is clear that (1) implies (2). Let us prove that (2) implies (1). Suppose that the base change X_ T \to T is not closed for some scheme T over S. By Schemes, Lemma 26.19.8 this means that there exists some specialization t_1 \leadsto t in T and a point \xi \in X_ T mapping to t_1 such that \xi does not specialize to a point in the fibre over t. Set Z = \overline{\{ \xi \} } \subset X_ T. Then Z \cap X_ t = \emptyset . Apply Lemma 32.14.1. We find an open neighbourhood V \subset T of t, a commutative diagram

\xymatrix{ V \ar[d] \ar[r]_ a & T' \ar[d]^ b \\ T \ar[r]^ g & S, }

and a closed subscheme Z' \subset X_{T'} such that

  1. the morphism b : T' \to S is locally of finite presentation,

  2. with t' = a(t) we have Z' \cap X_{t'} = \emptyset , and

  3. Z \cap X_ V maps into Z' via the morphism X_ V \to X_{T'}.

Clearly this means that X_{T'} \to T' maps the closed subset Z' to a subset of T' which contains a(t_1) but not t' = a(t). Since a(t_1) \leadsto a(t) = t' we conclude that X_{T'} \to T' is not closed. Hence we have shown that X \to S not universally closed implies that X_{T'} \to T' is not closed for some T' \to S which is locally of finite presentation. In order words (2) implies (1).

Assume that \mathbf{A}^ n \times X \to \mathbf{A}^ n \times S is closed for every integer n. We want to prove that X_ T \to T is closed for every scheme T which is locally of finite presentation over S. We may of course assume that T is affine and maps into an affine open V of S (since X_ T \to T being a closed is local on T). In this case there exists a closed immersion T \to \mathbf{A}^ n \times V because \mathcal{O}_ T(T) is a finitely presented \mathcal{O}_ S(V)-algebra, see Morphisms, Lemma 29.21.2. Then T \to \mathbf{A}^ n \times S is a locally closed immersion. Hence we get a cartesian diagram

\xymatrix{ X_ T \ar[d]_{f_ T} \ar[r] & \mathbf{A}^ n \times X \ar[d]^{f_ n} \\ T \ar[r] & \mathbf{A}^ n \times S }

of schemes where the horizontal arrows are locally closed immersions. Hence any closed subset Z \subset X_ T can be written as X_ T \cap Z' for some closed subset Z' \subset \mathbf{A}^ n \times X. Then f_ T(Z) = T \cap f_ n(Z') and we see that if f_ n is closed, then also f_ T is closed. \square

Lemma 32.14.3. Let S be a scheme. Let f : X \to S be a separated morphism of finite type. The following are equivalent:

  1. The morphism f is proper.

  2. For any morphism S' \to S which is locally of finite type the base change X_{S'} \to S' is closed.

  3. For every n \geq 0 the morphism \mathbf{A}^ n \times X \to \mathbf{A}^ n \times S is closed.

First proof. In view of the fact that a proper morphism is the same thing as a separated, finite type, and universally closed morphism, this lemma is a special case of Lemma 32.14.2. \square

Second proof. Clearly (1) implies (2), and (2) implies (3), so we just need to show (3) implies (1). First we reduce to the case when S is affine. Assume that (3) implies (1) when the base is affine. Now let f: X \to S be a separated morphism of finite type. Being proper is local on the base (see Morphisms, Lemma 29.41.3), so if S = \bigcup _\alpha S_\alpha is an open affine cover, and if we denote X_\alpha := f^{-1}(S_\alpha ), then it is enough to show that f|_{X_\alpha }: X_\alpha \to S_\alpha is proper for all \alpha . Since S_\alpha is affine, if the map f|_{X_\alpha } satisfies (3), then it will satisfy (1) by assumption, and will be proper. To finish the reduction to the case S is affine, we must show that if f: X \to S is separated of finite type satisfying (3), then f|_{X_\alpha } : X_\alpha \to S_\alpha is separated of finite type satisfying (3). Separatedness and finite type are clear. To see (3), notice that \mathbf{A}^ n \times X_\alpha is the open preimage of \mathbf{A}^ n \times S_\alpha under the map 1 \times f. Fix a closed set Z \subset \mathbf A^ n \times X_\alpha . Let \bar Z denote the closure of Z in \mathbf{A}^ n \times X. Then for topological reasons,

1 \times f(\bar Z) \cap \mathbf{A}^ n \times S_\alpha = 1 \times f(Z).

Hence 1 \times f(Z) is closed, and we have reduced the proof of (3) \Rightarrow (1) to the affine case.

Assume S affine, and f : X \to S separated of finite type. We can apply Chow's Lemma 32.12.1 to get \pi : X' \to X proper surjective and X' \to \mathbf{P}^ n_ S an immersion. If X is proper over S, then X' \to S is proper (Morphisms, Lemma 29.41.4). Since \mathbf{P}^ n_ S \to S is separated, we conclude that X' \to \mathbf{P}^ n_ S is proper (Morphisms, Lemma 29.41.7) and hence a closed immersion (Schemes, Lemma 26.10.4). Conversely, assume X' \to \mathbf{P}^ n_ S is a closed immersion. Consider the diagram:

32.14.3.1
\begin{equation} \label{limits-equation-check-proper} \xymatrix{ X' \ar[r] \ar@{->>}[d]_{\pi } & \mathbf{P}^ n_ S \ar[d] \\ X \ar[r]^ f & S } \end{equation}

All maps are a priori proper except for X \to S. Hence we conclude that X \to S is proper by Morphisms, Lemma 29.41.9. Therefore, we have shown that X \to S is proper if and only if X' \to \mathbf{P}^ n_ S is a closed immersion.

Assume S is affine and (3) holds, and let n, X', \pi be as above. Since being a closed morphism is local on the base, the map X \times \mathbf{P}^ n \to S \times \mathbf{P}^ n is closed since by (3) X \times \mathbf{A}^ n \to S \times \mathbf{A}^ n is closed and since projective space is covered by copies of affine n-space, see Constructions, Lemma 27.13.3. By Morphisms, Lemma 29.41.5 the morphism

X' \times _ S \mathbf{P}^ n_ S \to X \times _ S \mathbf{P}^ n_ S = X \times \mathbf{P}^ n

is proper. Since \mathbf{P}^ n is separated, the projection

X' \times _ S \mathbf{P}^ n_ S = \mathbf{P}^ n_{X'} \to X'

will be separated as it is just a base change of a separated morphism. Therefore, the map X' \to X' \times _ S \mathbf{P}^ n_ S is proper, since it is a section to a separated map (see Schemes, Lemma 26.21.11). Composing these morphisms

X' \to X' \times _ S \mathbf{P}^ n_ S \to X \times _ S \mathbf{P}^ n_ S = X \times \mathbf{P}^ n \to S \times \mathbf{P}^ n = \mathbf{P}^ n_ S

we find that the immersion X' \to \mathbf{P}^ n_ S is closed, and hence a closed immersion. \square


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