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}Comments (0)
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