Mathematics Asked on January 1, 2022

Let $X,Y$ be schemes of finite type over a field $k$. In particular, they are quasi-compact. Let $f: X to Y$ be a morphism of finite type (for all open affines $U subset Y$, $f^{-1}(U)$ is quasi-compact and $Gamma(V, O_X)$ is finitely generated over $Gamma(U, O_Y)$ for all open affine $V subset f^{-1}(U)$).

I would like to deduce that the fibre $f^{-1}(y)$ as a scheme over $operatorname{Spec}kappa(y)$ is quasi-compact, for any $y in Y$. How can I prove this? Any comments are appreciated. Thank you!

We know that morphisms of finite type are stable under base extension (see Hartshorne exercise 3.13 (d)). Then $f:X rightarrow Y$ being of finite type implies that $f^{-1}(y) rightarrow $ Spec $k(y)$ is of finite type. Note that morphisms of finite type are necessarily quasi-compact (see Hartshorne exercise 3.3(a)), since the preimage of Spec $k(y)$ is $f^{-1}(y)$, we must have it is quasi-compact space.

Answered by Rikka on January 1, 2022

A scheme is quasicompact iff it can be covered by finitely many open affine subschemes.

When $X, Y$ are affine, it’s easy to see that the fiber is the spectrum of $mathcal{O}_X(X) otimes_{mathcal{O}_Y(Y)} kappa(y)$ so is quasicompact.

In general, let $y in V subset Y$ be an affine open subset, let $f^{-1}(V)=bigcup_i{U_i}$ be a finite reunion of affine open subsets of $X$ (as $X$ is a Noetherian scheme, all its affine open subsets are quasicompact). Then $X_y$ is the (finite) reunion of the $(U_i)_y$ which are open affine (they are fibers of $U_i rightarrow V$ affine of finite type), so is quasi-compact.

Answered by Mindlack on January 1, 2022

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