That's it, I'm trying to prove that if $f:X\to S$ is a quasi-compact morphism of schemes and $g:T\to S$ is any morphism, then the base change $f_T:X\times_ST\to T$ is also quasi-compact.
The proof in 01K5 is simply "ommited".
This is what I've tried so far: I know that if $U\subset T$ is any open set, then $f_T^{-1}(U)=X\times_SU$ (for example, by 01JR). By 01K4 it suffices to show that $U$ affine and quasi-compact implies $f_T^{-1}(U)$ quasi-compact. On this case, we have that $g(U)$ is quasi-compact (as continuity preserves quasi-compactness). Therefore, there is a finite cover of $g(U)$ by affine open subsets of $S$. Call $W$ to the union of the sets of this cover, so $W$ is quasi-compact as quasi-compactness is stable under finite unions. Since $g(U)\subset W$, again by 01JR, we deduce $f_T^{-1}(U)=X\times_SU=f^{-1}(W)\times_WU$, where $f^{-1}(W)$ is quasi-compact. Thus, we've reduced the problem to showing
Exercise. If $X$ and $Y$ are quasi-compact schemes over a quasi-compact scheme $S$, then $X\times_SY$ is quasi-compact. (We may assume if necessary that at least one of $X$ or $Y$ is affine.)
But I don't know how to show this (in case it's true). Any ideas?
Best Answer
We change your notation slightly so that we use $\phi:S'\to S$ instead of $g:T\to S$. Let $X'=X\times_S S'$ and let $f':X'\to X$ be the obvious map.
Suppose $U\subset S$ is an affine open subscheme, and cover $f^{-1}(U)$ by finitely many affine opens $V_1,\cdots,V_n\subset X$. Let $U'\subset\phi^{-1}(U)$ be an affine open. Then $$f'^{-1}(U') = f^{-1}(U)\times_U U' = \left(\bigcup^n V_i\right) \times_U U' = \bigcup^n \left(V_i\times_U U'\right)$$ and $V_i\times_U U'$ is again affine by the construction of the fiber product. Since a scheme is quasi-compact iff it admits a cover by finitely many affine opens, we've shown that $f'^{-1}(U')$ is quasi-compact; by varying $U$ and $U'$ we get an open cover of $S'$ by affine opens with quasi-compact preimage and therefore by 01K4 for instance we've shown that $f'$ is quasi-compact.