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Tag 02JS

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

Lemma 28.27.2. Let $f : X \to Y$ and $g : Y \to S$ be morphisms of schemes. Let $x \in X$ and set $y = f(x)$, $s = g(y)$. Assume $f$ and $g$ locally of finite type. Then $$ \dim_x(X_s) \leq \dim_x(X_y) + \dim_y(Y_s). $$ Moreover, equality holds if $\mathcal{O}_{X_s, x}$ is flat over $\mathcal{O}_{Y_s, y}$, which holds for example if $\mathcal{O}_{X, x}$ is flat over $\mathcal{O}_{Y, y}$.

Proof. Note that $\text{trdeg}_{\kappa(s)}(\kappa(x)) = \text{trdeg}_{\kappa(y)}(\kappa(x)) + \text{trdeg}_{\kappa(s)}(\kappa(y))$. Thus by Lemma 28.27.1 the statement is equivalent to $$ \dim(\mathcal{O}_{X_s, x}) \leq \dim(\mathcal{O}_{X_y, x}) + \dim(\mathcal{O}_{Y_s, y}). $$ For this see Algebra, Lemma 10.111.6. For the flat case see Algebra, Lemma 10.111.7. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4919–4931 (see updates for more information).

    \begin{lemma}
    \label{lemma-dimension-fibre-at-a-point-additive}
    Let $f : X \to Y$ and $g : Y \to S$ be morphisms of schemes.
    Let $x \in X$ and set $y = f(x)$, $s = g(y)$.
    Assume $f$ and $g$ locally of finite type.
    Then
    $$
    \dim_x(X_s) \leq \dim_x(X_y) + \dim_y(Y_s).
    $$
    Moreover, equality holds if $\mathcal{O}_{X_s, x}$ is flat
    over $\mathcal{O}_{Y_s, y}$, which holds for example if $\mathcal{O}_{X, x}$
    is flat over $\mathcal{O}_{Y, y}$.
    \end{lemma}
    
    \begin{proof}
    Note that $\text{trdeg}_{\kappa(s)}(\kappa(x)) =
    \text{trdeg}_{\kappa(y)}(\kappa(x)) + \text{trdeg}_{\kappa(s)}(\kappa(y))$.
    Thus by Lemma \ref{lemma-dimension-fibre-at-a-point} the statement
    is equivalent to
    $$
    \dim(\mathcal{O}_{X_s, x})
    \leq
    \dim(\mathcal{O}_{X_y, x}) + \dim(\mathcal{O}_{Y_s, y}).
    $$
    For this see Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-total}.
    For the flat case see
    Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}.
    \end{proof}

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