Quasi-Compact Morphisms

by iarnab_kundu, Dec 9, 2018, 6:07 PM

Def- Let $X,Y$ be schemes. A morphism $f:Y\to X$ is said to be quasi-compact if the pre-image of every open quasi-compact subset of $X$ is an open quasi-compact subset of $Y$.
At this juncture it seems to me to be just a topological condition.

Proposition- $f:Y\to X$ is quasi compact iff the pre-image of every open affine sub-scheme of $X$ can be written as a finite union of open affine sub-schemes of $Y$.
Proof- Every open quasi-compact subset can be written as a finite union of open affine subsets. Also note that any affine scheme is quasi-compact.

Lemma- Let $f:\text{Spec}(B)\to\text{Spec}(A)$ be a morphism of schemes, then $f^{-1}(D(r))=D(f(r))$.
Corollary- A map of affine schemes is quasi-compact.

Proposition- $f:Y\to X$ is quasi-compact iff there is a finite open affine cover $U_i=\text{Spec}(A_i)$ of $X$ such that for each $i$ the pre-image $f^{-1}(U_i)$ can be written as a finite union of open affines of $X$.
Proof- Suppose there is an open affine cover like the above and we want to show that $f$ is quasi-compact. Thus we take an open affine $U=\text{Spec}(A)$ of $X$. Then we cover $U$ by open affines $W_j=\text{Spec}(C_j)$ such that for each $j$ there exists an $i$ such that $W_j=D(r_j)\subset U_i$ is a basic open set. Since $U$ is quasi-compact we extract a finite affine sub-cover from that. Let $f^{-1}(U_i)=\cup_k V_{ik}=\text{Spec}(B_{ik})$, and $f_ik:A_i\to B_{ik}$. Then $f^{-1}(V_j)=\cup_k D(f_{ik}(r_j))$ is a finite union of affine opens. Since there is a finite sub-collection of open affines $V_j$ we are done.

Proposition-
(a) Closed immersions are quasi-compact.
(b) Composition of quasi-compact morphisms is quasi-compact.
(b) Base change of quasi-compact is quasi-compact.
Proof- (a) OMITTED

Warning! Open immersions are NOT, in general, quasi-compact.
This post has been edited 4 times. Last edited by iarnab_kundu, Dec 9, 2018, 10:06 PM

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