# The Stacks Project

## Tag 02FX

Lemma 28.27.1. 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 10.115.3. $\square$

The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4898–4908 (see updates for more information).

\begin{lemma}
\label{lemma-dimension-fibre-at-a-point}
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
$$\dim_x(X_s) = \dim(\mathcal{O}_{X_s, x}) + \text{trdeg}_{\kappa(s)}(\kappa(x)).$$
\end{lemma}

\begin{proof}
This immediately reduces to the case $S = s$, and $X$ affine.
In this case the result follows from
Algebra, Lemma \ref{algebra-lemma-dimension-at-a-point-finite-type-field}.
\end{proof}

Comment #2448 by Andy on March 9, 2017 a 3:48 pm UTC

What is $S=s$ supposed to mean?

Comment #2490 by Johan (site) on April 13, 2017 a 11:00 pm UTC

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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