## 66.35 The dimension formula

The analog of the dimension formula (Morphisms, Lemma 29.52.1) is a bit tricky to formulate, because we would have to define integral algebraic spaces (we do this later) as well as universally catenary algebraic spaces. However, the following version is straightforward.

Lemma 66.35.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $Y$ is locally Noetherian and $f$ locally of finite type. Let $x \in |X|$ with image $y \in |Y|$. Then we have

\begin{align*} & \text{the dimension of the local ring of }X\text{ at }x \leq \\ & \text{the dimension of the local ring of }Y\text{ at }y + E - \\ & \text{ the transcendence degree of }x/y \end{align*}

Here $E$ is the maximum of the transcendence degrees of $\xi /f(\xi )$ where $\xi \in |X|$ runs over the points specializing to $x$ at which the local ring of $X$ has dimension $0$.

Proof. Choose an affine scheme $V$, an étale morphism $V \to Y$, and a point $v \in V$ mapping to $y$. Choose an affine scheme $U$ , an étale morphism $U \to X \times _ Y V$ and a point $u \in U$ mapping to $v$ in $V$ and $x$ in $X$. Unwinding Definition 66.33.1 and Properties of Spaces, Definition 65.10.2 we have to show that

$\dim (\mathcal{O}_{U, u}) \leq \dim (\mathcal{O}_{V, v}) + E - \text{trdeg}_{\kappa (v)}(\kappa (u))$

Let $\xi _ U \in U$ be a generic point of an irreducible component of $U$ which contains $u$. Then $\xi _ U$ maps to a point $\xi \in |X|$ which is in the list used to define the quantity $E$ and in fact every $\xi$ used in the definition of $E$ occurs in this manner (small detail omitted). In particular, there are only a finite number of these $\xi$ and we can take the maximum (i.e., it really is a maximum and not a supremum). The transcendence degree of $\xi$ over $f(\xi )$ is $\text{trdeg}_{\kappa (\xi _ V)}(\kappa (\xi _ U))$ where $\xi _ V \in V$ is the image of $\xi _ U$. Thus the lemma follows from Morphisms, Lemma 29.52.2. $\square$

Lemma 66.35.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $Y$ is locally Noetherian and $f$ is locally of finite type. Then

$\dim (X) \leq \dim (Y) + E$

where $E$ is the supremum of the transcendence degrees of $\xi /f(\xi )$ where $\xi$ runs through the points at which the local ring of $X$ has dimension $0$.

Proof. Immediate consequence of Lemma 66.35.1 and Properties of Spaces, Lemma 65.10.3. $\square$

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