The Stacks project

Proof. By Morphisms of Spaces, Lemma 67.11.2 the sheaves $f_*\mathcal{O}_ X$, $(X_ y \to Y)_*\mathcal{O}_{X_ y}$, and $y_*\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}$ are quasi-coherent sheaves of $\mathcal{O}_ Y$-algebras. Consider the maps

The fibre product is a quasi-coherent sheaf of $\mathcal{O}_ Y$-algebras $\mathcal{A}'$ and we can define $X' \to Y$ as the relative spectrum of $\mathcal{A}'$ over $Y$, see Morphisms, Lemma 29.11.5. This construction commutes with arbitrary change of base. In particular, it is clear that over the open subspace $|Y| \setminus \{ y\} $ the morphism $X \to X'$ is an isomorphism and over $|Y| \setminus \{ y\} $ the morphism $X' \to Y$ is integral. It remains to prove the statements in a small neighbourhood of $y$. Choose an affine scheme $V = \mathop{\mathrm{Spec}}(R)$ and an étale morphism $\varphi : V \to Y$ such that $y$ is in the image of $\varphi $. Then $V_ y$ is a closed subscheme of $V$ étale over $k$, whence consists of finitely many points each with residue field separable over $k$ (see Decent Spaces, Remark 68.4.1). After shrinking $V$ we may assume there is a unique closed point $v = \mathop{\mathrm{Spec}}(l) \to V$ mapping to $y$ with $l/k$ finite separable. We may write $V \times _ Y X = \mathop{\mathrm{Spec}}(C)$ with $R \to C$ an integral ring map. The stated compatibility with base change gives us that $U \times _ X Y' = \mathop{\mathrm{Spec}}(C')$ where

Since $R \to l$ is surjective, also $C \to C \otimes _ R l$ is surjective and we see that this is a fibre product of the kind studied in More on Algebra, Situation 15.6.1 (with $A, A', B, B'$ corresponding to $C \otimes _ R l, C, l, C'$). Observe that $C'$ is an $R$-subalgebra of $C$ and hence is integral over $R$; this proves (1). Finally, More on Algebra, Lemma 15.6.2 shows that $V \times _ X Y' = \mathop{\mathrm{Spec}}(C')$ has a unique point $y''$ lying over $v$ with residue $l$ (this corresponds with the obvious surjective map $C' \to l$). Thus $X_ y \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(l)$ has a unique point with residue field $l$. Since $l/k$ is finite separable, this implies $X'_ y$ has a unique point with residue field $k$, i.e., (3) holds.

To prove the final statement, observe that if $Y$ is locally Noetherian, then $R$ is a Noetherian ring and if $f$ is finite, then $R \to C$ is finite. Then $C'$ is a finite type $R$-algebra by More on Algebra, Lemma 15.5.1. This proves that $X' \to Y$ is finite. $\square$

Proof. The problem is local on $B$. Thus we may assume $B$ is quasi-compact. By Decent Spaces, Lemma 68.14.1 we see $B$ is quasi-separated. By Limits of Spaces, Proposition 70.16.1 we can choose a finite surjective morphism $\pi : Y \to X$ where $Y$ is a scheme. Claim: $\delta _ Y$ is a dimension function.

The claim implies the lemma. With $X \to B$ as in the lemma set $Z = Y \times _ B X$ with projections $p : Z \to Y$ and $q : Z \to X$. Then we have

and $\delta _ Z(z) = \delta _ X(q(z))$. This follows from Morphisms of Spaces, Lemma 67.34.2 and the fact that these transcendence degrees are zero for finite morphisms. By Decent Spaces, Lemma 68.25.2 and the claim we find that $\delta _ Z$ is a dimension function. Then we find that $\delta _ X$ is a dimension function by Decent Spaces, Lemma 68.25.6.

Proof of the claim. Consider a specialization $y \leadsto y'$, $y \not= y'$ of points of the Noetherian scheme $Y$. Then $\delta _ Y(y) > \delta _ Y(y')$ because there are no specializations between points in fibres of $Y$ (see Decent Spaces, Lemma 68.18.10). Using this for a chain of specializations we find

Our task is to show equality. By Properties, Lemma 28.5.9 we can choose a specialization $y' \leadsto y_0$. It suffices to show $\delta _ Y(y) - \delta _ Y(y_0) = \text{codim}(\overline{\{ y_0\} }, \overline{\{ y\} })$ because this will imply the equality for both $y \leadsto y'$ and $y' \leadsto y_0$.

Choose a maximal chain $y = y_ c \leadsto y_{c - 1} \leadsto \ldots \leadsto y_0$ of specializations in $Y$. Set $b = \pi (y)$ and $b_0 = \pi (y_0)$. Choose a maximal chain $b = b_ e \leadsto b_{e - 1} \leadsto \ldots \leadsto b_0$ of specializations in $|B|$. We have to show $e = c$. Since $\pi $ is closed (Morphisms of Spaces, Lemma 67.45.9) we can find a sequence of specializations $y = y'_ e \leadsto y'_{e - 1} \leadsto \ldots \leadsto y'_0$ mapping to $b = b_ e \leadsto b_{e - 1} \leadsto \ldots \leadsto b_0$. Observe that $y'_ e \leadsto y'_{e - 1} \leadsto \ldots \leadsto y'_0$ is a maximal chain as well. If $y_0 = y'_0$, then because $Y$ is catenary, we conclude that $e = c$ as desired. In the next paragraph we reduce to this case by sleight of hand and we conclude in the same manner.

Since $\pi $ is closed we see that $b_0$ is a closed point of $|B|$. By Decent Spaces, Lemma 68.14.6 we can represent $b_0$ by a closed immersion $b_0 : \mathop{\mathrm{Spec}}(k) \to B$. By Lemma 76.32.1 we can find a factorization

with $\pi ' : Y' \to X$ finite and $Y \to Y'$ a morphism which map $y_0$ and $y'_0$ to the same point and is an isomorphism away from the inverse image of $b_0$. (Of course $Y'$ won't be a scheme but this doesn't matter for the argument that follows.) Clearly the maximal chains of specializations $y_ c \leadsto y_{c - 1} \leadsto \ldots \leadsto y_0$ and $y'_ e \leadsto y'_{e - 1} \leadsto \ldots \leadsto y'_0$ map to maximal chains of specializations in $Y'$ having the same start and end. Since $B$ is universally catenary, we see that $|Y'|$ is catenary and we conclude as before. $\square$