Lemma 75.32.2. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $\delta : |B| \to \mathbf{Z}$ be a function. Assume $B$ is decent, locally Noetherian, and universally catenary and $\delta$ is a dimension function. If $X$ is a decent algebraic space over $B$ whose structure morphism $f : X \to B$ is locally of finite type we define $\delta _ X : |X| \to \mathbf{Z}$ by the rule

$\delta _ X(x) = \delta (f(x)) + \text{transcendence degreeof }x/f(x)$

(Morphisms of Spaces, Definition 66.33.1). Then $\delta _ X$ is a dimension function.

Proof. The problem is local on $B$. Thus we may assume $B$ is quasi-compact. By Decent Spaces, Lemma 67.14.1 we see $B$ is quasi-separated. By Limits of Spaces, Proposition 69.16.1 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

$\delta _ Z(z) = \delta _ Y(p(z)) + \text{transcendence degreeof }z/p(z)$

and $\delta _ Z(z) = \delta _ X(q(z))$. This follows from Morphisms of Spaces, Lemma 66.34.2 and the fact that these transcendence degrees are zero for finite morphisms. By Decent Spaces, Lemma 67.25.2 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 67.25.6.

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 67.18.10). Using this for a chain of specializations we find

$\delta _ Y(y) - \delta _ Y(y') \geq \text{codim}(\overline{\{ y'\} }, \overline{\{ y\} })$

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 66.45.9) 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 67.14.6 we can represent $b_0$ by a closed immersion $b_0 : \mathop{\mathrm{Spec}}(k) \to B$. By Lemma 75.32.1 we can find a factorization

$Y \to Y' \to X$

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. $\square$

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