Lemma 29.28.1. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ and set $s = f(x)$. Assume $f$ is locally of finite type. Then

$\dim _ x(X_ s) = \dim (\mathcal{O}_{X_ s, x}) + \text{trdeg}_{\kappa (s)}(\kappa (x)).$

Proof. This immediately reduces to the case $S = s$, and $X$ affine. In this case the result follows from Algebra, Lemma 10.116.3. $\square$

## Comments (2)

Comment #2448 by Andy on

What is $S=s$ supposed to mean?

Comment #2490 by on

When $s \in S$ is a point of a scheme, then we think of $s$ as a scheme too, namely we think of $s$ as the spectrum of its residue field. So what is meant is that the base change to $s$ reduces the lemma to the case where the base scheme is the spectrum of a field.

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