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Tag 02FX

Chapter 28: Morphisms of Schemes > Section 28.27: Morphisms and dimensions of fibres

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 4901–4911 (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}

    Comments (2)

    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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