The Stacks project

26.19 Quasi-compact morphisms

A scheme is quasi-compact if its underlying topological space is quasi-compact. There is a relative notion which is defined as follows.

Definition 26.19.1. A morphism of schemes is called quasi-compact if the underlying map of topological spaces is quasi-compact, see Topology, Definition 5.12.1.

Lemma 26.19.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

  1. $f : X \to S$ is quasi-compact,

  2. the inverse image of every affine open is quasi-compact, and

  3. there exists some affine open covering $S = \bigcup _{i \in I} U_ i$ such that $f^{-1}(U_ i)$ is quasi-compact for all $i$.

Proof. Suppose we are given a covering $S = \bigcup _{i \in I} U_ i$ as in (3). First, let $U \subset S$ be any affine open. For any $u \in U$ we can find an index $i(u) \in I$ such that $u \in U_{i(u)}$. As standard opens form a basis for the topology on $U_{i(u)}$ we can find $W_ u \subset U \cap U_{i(u)}$ which is standard open in $U_{i(u)}$. By compactness we can find finitely many points $u_1, \ldots , u_ n \in U$ such that $U = \bigcup _{j = 1}^ n W_{u_ j}$. For each $j$ write $f^{-1}U_{i(u_ j)} = \bigcup _{k \in K_ j} V_{jk}$ as a finite union of affine opens. Since $W_{u_ j} \subset U_{i(u_ j)}$ is a standard open we see that $f^{-1}(W_{u_ j}) \cap V_{jk}$ is a standard open of $V_{jk}$, see Algebra, Lemma 10.17.4. Hence $f^{-1}(W_{u_ j}) \cap V_{jk}$ is affine, and so $f^{-1}(W_{u_ j})$ is a finite union of affines. This proves that the inverse image of any affine open is a finite union of affine opens.

Next, assume that the inverse image of every affine open is a finite union of affine opens. Let $K \subset S$ be any quasi-compact open. Since $S$ has a basis of the topology consisting of affine opens we see that $K$ is a finite union of affine opens. Hence the inverse image of $K$ is a finite union of affine opens. Hence $f$ is quasi-compact.

Finally, assume that $f$ is quasi-compact. In this case the argument of the previous paragraph shows that the inverse image of any affine is a finite union of affine opens. $\square$

Lemma 26.19.3. Being quasi-compact is a property of morphisms of schemes over a base which is preserved under arbitrary base change.

Proof. Omitted. $\square$

Lemma 26.19.4. The composition of quasi-compact morphisms is quasi-compact.

Proof. This follows from the definitions and Topology, Lemma 5.12.2. $\square$

Proof. Follows from the definitions and Topology, Lemma 5.12.3. $\square$

Example 26.19.6. An open immersion is in general not quasi-compact. The standard example of this is the open subspace $U \subset X$, where $X = \mathop{\mathrm{Spec}}(k[x_1, x_2, x_3, \ldots ])$, where $U$ is $X \setminus \{ 0\} $, and where $0$ is the point of $X$ corresponding to the maximal ideal $(x_1, x_2, x_3, \ldots )$.

Lemma 26.19.7. Let $f : X \to S$ be a quasi-compact morphism of schemes. The following are equivalent

  1. $f(X) \subset S$ is closed, and

  2. $f(X) \subset S$ is stable under specialization.

Proof. We have (1) $\Rightarrow $ (2) by Topology, Lemma 5.19.2. Assume (2). Let $U \subset S$ be an affine open. It suffices to prove that $f(X) \cap U$ is closed. Since $U \cap f(X)$ is stable under specializations in $U$, we have reduced to the case where $S$ is affine. Because $f$ is quasi-compact we deduce that $X = f^{-1}(S)$ is quasi-compact as $S$ is affine. Thus we may write $X = \bigcup _{i = 1}^ n U_ i$ with $U_ i \subset X$ open affine. Say $S = \mathop{\mathrm{Spec}}(R)$ and $U_ i = \mathop{\mathrm{Spec}}(A_ i)$ for some $R$-algebra $A_ i$. Then $f(X) = \mathop{\mathrm{Im}}(\mathop{\mathrm{Spec}}(A_1 \times \ldots \times A_ n) \to \mathop{\mathrm{Spec}}(R))$. Thus the lemma follows from Algebra, Lemma 10.41.5. $\square$

Lemma 26.19.8. Let $f : X \to S$ be a quasi-compact morphism of schemes. Then $f$ is closed if and only if specializations lift along $f$, see Topology, Definition 5.19.4.

Proof. According to Topology, Lemma 5.19.7 if $f$ is closed then specializations lift along $f$. Conversely, suppose that specializations lift along $f$. Let $Z \subset X$ be a closed subset. We may think of $Z$ as a scheme with the reduced induced scheme structure, see Definition 26.12.5. Since $Z \subset X$ is closed the restriction of $f$ to $Z$ is still quasi-compact. Moreover specializations lift along $Z \to S$ as well, see Topology, Lemma 5.19.5. Hence it suffices to prove $f(X)$ is closed if specializations lift along $f$. In particular $f(X)$ is stable under specializations, see Topology, Lemma 5.19.6. Thus $f(X)$ is closed by Lemma 26.19.7. $\square$


Comments (8)

Comment #278 by Keenan Kidwell on

In the proof of Lemma 01K4, the assumption that is standard open in is never used. The only part that matters is that is standard open in , so the result about the existence of simultaneous standard opens in intersections of affines isn't needed (only that standard opens form a base for the Zariski topology of an affine scheme).

Comment #1321 by Erik Visse on

There is a typo in the first line of the proof of Lemma 25.19.2. X should be S.

Comment #1474 by sdf on

The two instances of in the second paragraph of Lemma 25.19.2. should be S, since .

Comment #2462 by Andrea on

There is a typo in the sixth line of the proof of lemma 25.19.2: instead of there should be , I think.


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