Consider a following cartesian diagram of schemes.
$$\begin{array}
AX^{'} & \stackrel{v}{\longrightarrow} & X \\
\downarrow{f'} & & \downarrow{f} \\
Y^{'} & \stackrel{u}{\longrightarrow} & Y
\end{array}
$$
If $u$ is faithfully flat and quasi-compact and $f'$ is locally of finite type, then why is $f$ locally of finite type?
Wedhorn's book says that it follows from next lemma:
Lemma: Let $A \rightarrow A'$ be a faithfully flat ring homomorphism, let $B$ be an $A$-algebra, and write $B' = B \otimes_A A'$. If $B'$ is $A'$-algebra of finite type, then $B$ is $A$-algebra of finite type. ("algebra of finite type" means finitely generated algebra)
The property of being locally of finite type is local on the target and the source. Hence we may assume that $X$ and $Y$ is affine. However, can we assume that $Y'$ is affine? Does there exist some affine open subsets $U$ of $Y'$ such that $u|_U : U \rightarrow Y' \stackrel{u}{\rightarrow} Y$ is surjective?
Best Answer
Yes, one may assume that $Y'$ is affine. Every fpqc covering of an affine scheme admits a refinement to a covering of the same scheme by affine opens (see stacks 022E for a source). One may base change along this refinement to get a new cartesian square with $Y'$ affine preserving the property that $f'$ is locally of finite type.