The Stacks project

Proof. If $U \to X$ is separated, then this is More on Morphisms, Lemma 37.45.4. In general, we may assume $X$ is affine. Choose a finite affine open covering $U = \bigcup U_ j$. Apply the previous case to all the morphisms $U_ j \to X$ and $U_ j \cap U_{j'} \to X$ and choose a common refinement $X = \coprod X_ i$ of the resulting partitions. After refining the partition further we may assume $X_ i$ affine as well. Fix $i$ and set $V = U \times _ X X_ i$. The morphisms $V_ j = U_ j \times _ X X_ i \to X_ i$ and $V_{jj'} = (U_ j \cap U_{j'}) \times _ X X_ i \to X_ i$ are finite étale. Hence $V_ j$ and $V_{jj'}$ are affine schemes and $V_{jj'} \subset V_ j$ is closed as well as open (since $V_{jj'} \to X_ i$ is proper, so Morphisms, Lemma 29.41.7 applies). Then $V = \bigcup V_ j$ is separated because $\mathcal{O}(V_ j) \to \mathcal{O}(V_{jj'})$ is surjective, see Schemes, Lemma 26.21.7. Thus the previous case applies to $V \to X_ i$ and we can further refine the partition if needed (it actually isn't but we don't need this). $\square$

Proof. This is Morphisms, Lemma 29.51.1. $\square$

Proof. Setting $X' = \coprod X'_ i$ we see that (2) implies (1). Write $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. Write $A$ as a filtered colimit of finite type $\mathbf{Z}$-algebras $A_ i$. Since $B$ is an $A$-algebra of finite presentation, we see that there exists $0 \in I$ and a finite type ring map $A_0 \to B_0$ such that $B = \mathop{\mathrm{colim}}\nolimits B_ i$ with $B_ i = A_ i \otimes _{A_0} B_0$, see Algebra, Lemma 10.127.8. For $i$ sufficiently large we see that $A_ i \to B_ i$ is quasi-finite, see Limits, Lemma 32.18.2. Thus we reduce to the case of finite type algebras over $\mathbf{Z}$, in particular we reduce to the Noetherian case. (Details omitted.)

Assume $X$ and $Y$ Noetherian. In this case any locally closed subset of $X$ is constructible. By Lemma 59.72.2 and Noetherian induction we see that there is a finite partition $X = \coprod X_ i$ of $X$ by locally closed strata such that $Y \times _ X X_ i \to X_ i$ is finite. We can refine this partition to get affine strata. Thus after replacing $X$ by $X' = \coprod X_ i$ we may assume $Y \to X$ is finite.

Assume $X$ and $Y$ Noetherian and $Y \to X$ finite. Suppose that we can prove (2) after base change by a surjective, flat, quasi-finite morphism $U \to X$. Thus we have a partition $U = \coprod U_ i$ and finite locally free morphisms $U'_ i \to U_ i$ such that $U'_ i \times _ X Y \to U'_ i$ is isomorphic to $\coprod _{j = 1}^{n_ i} (U'_ i)_{red} \to (U'_ i)_{red}$ for some $n_ i$. Then, by the argument in the previous paragraph, we can find a partition $X = \coprod X_ j$ with locally closed affine strata such that $X_ j \times _ X U_ i \to X_ j$ is finite for all $i, j$. By Morphisms, Lemma 29.48.2 each $X_ j \times _ X U_ i \to X_ j$ is finite locally free. Hence $X_ j \times _ X U'_ i \to X_ j$ is finite locally free (Morphisms, Lemma 29.48.3). It follows that $X = \coprod X_ j$ and $X_ j' = \coprod _ i X_ j \times _ X U'_ i$ is a solution for $Y \to X$. Thus it suffices to prove the result (in the Noetherian case) after a surjective flat quasi-finite base change.

Applying Morphisms, Lemma 29.48.6 we see we may assume that $Y$ is a closed subscheme of an affine scheme $Z$ which is (set theoretically) a finite union $Z = \bigcup _{i \in I} Z_ i$ of closed subschemes mapping isomorphically to $X$. In this case we will find a finite partition of $X = \coprod X_ j$ with affine locally closed strata that works (in other words $X'_ j = X_ j$). Set $T_ i = Y \cap Z_ i$. This is a closed subscheme of $X$. As $X$ is Noetherian we can find a finite partition of $X = \coprod X_ j$ by affine locally closed subschemes, such that each $X_ j \times _ X T_ i$ is (set theoretically) a union of strata $X_ j \times _ X Z_ i$. Replacing $X$ by $X_ j$ we see that we may assume $I = I_1 \amalg I_2$ with $Z_ i \subset Y$ for $i \in I_1$ and $Z_ i \cap Y = \emptyset $ for $i \in I_2$. Replacing $Z$ by $\bigcup _{i \in I_1} Z_ i$ we see that we may assume $Y = Z$. Finally, we can replace $X$ again by the members of a partition as above such that for every $i, i' \subset I$ the intersection $Z_ i \cap Z_{i'}$ is either empty or (set theoretically) equal to $Z_ i$ and $Z_{i'}$. This clearly means that $Y$ is (set theoretically) equal to a disjoint union of the $Z_ i$ which is what we wanted to show. $\square$