# The Stacks Project

## Tag 02FW

### 28.27. Morphisms and dimensions of fibres

Let $X$ be a topological space, and $x \in X$. Recall that we have defined $\dim_x(X)$ as the minimum of the dimensions of the open neighbourhoods of $x$ in $X$. See Topology, Definition 5.10.1.

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. $\square$

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. $\square$

Lemma 28.27.3. Let $$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^f \\ S' \ar[r]^g & S }$$ be a fibre product diagram of schemes. Assume $f$ locally of finite type. Suppose that $x' \in X'$, $x = g'(x')$, $s' = f'(x')$ and $s = g(s') = f(x)$. Then

1. $\dim_x(X_s) = \dim_{x'}(X'_{s'})$,
2. if $F$ is the fibre of the morphism $X'_{s'} \to X_s$ over $x$, then $$\dim(\mathcal{O}_{F, x'}) = \dim(\mathcal{O}_{X'_{s'}, x'}) - \dim(\mathcal{O}_{X_s, x}) = \text{trdeg}_{\kappa(s)}(\kappa(x)) - \text{trdeg}_{\kappa(s')}(\kappa(x'))$$ In particular $\dim(\mathcal{O}_{X'_{s'}, x'}) \geq \dim(\mathcal{O}_{X_s, x})$ and $\text{trdeg}_{\kappa(s)}(\kappa(x)) \geq \text{trdeg}_{\kappa(s')}(\kappa(x'))$.
3. given $s', s, x$ there exists a choice of $x'$ such that $\dim(\mathcal{O}_{X'_{s'}, x'}) = \dim(\mathcal{O}_{X_s, x})$ and $\text{trdeg}_{\kappa(s)}(\kappa(x)) = \text{trdeg}_{\kappa(s')}(\kappa(x'))$.

Proof. Part (1) follows immediately from Algebra, Lemma 10.115.6. Parts (2) and (3) from Algebra, Lemma 10.115.7. $\square$

The following lemma follows from a nontrivial algebraic result. Namely, the algebraic version of Zariski's main theorem.

Lemma 28.27.4. Let $f : X \to S$ be a morphism of schemes. Let $n \geq 0$. Assume $f$ is locally of finite type. The set $$U_n = \{x \in X \mid \dim_x X_{f(x)} \leq n\}$$ is open in $X$.

Proof. This is immediate from Algebra, Lemma 10.124.6 $\square$

Lemma 28.27.5. Let $f : X \to S$ be a morphism of schemes. Let $n \geq 0$. Assume $f$ is locally of finite presentation. The open $$U_n = \{x \in X \mid \dim_x X_{f(x)} \leq n\}$$ of Lemma 28.27.4 is retrocompact in $X$. (See Topology, Definition 5.12.1.)

Proof. The topological space $X$ has a basis for its topology consisting of affine opens $U \subset X$ such that the induced morphism $f|_U : U \to S$ factors through an affine open $V \subset S$. Hence it is enough to show that $U \cap U_n$ is quasi-compact for such a $U$. Note that $U_n \cap U$ is the same as the open $\{x \in U \mid \dim_x U_{f(x)} \leq n\}$. This reduces us to the case where $X$ and $S$ are affine. In this case the lemma follows from Algebra, Lemma 10.124.8 (and Lemma 28.20.2). $\square$

Lemma 28.27.6. Let $f : X \to S$ be a morphism of schemes. Let $x \leadsto x'$ be a nontrivial specialization of points in $X$ lying over the same point $s \in S$. Assume $f$ is locally of finite type. Then

1. $\dim_x(X_s) \leq \dim_{x'}(X_s)$,
2. $\dim(\mathcal{O}_{X_s, x}) < \dim(\mathcal{O}_{X_s, x'})$, and
3. $\text{trdeg}_{\kappa(s)}(\kappa(x)) > \text{trdeg}_{\kappa(s)}(\kappa(x'))$.

Proof. Part (1) follows from the fact that any open of $X_s$ containing $x'$ also contains $x$. Part (2) follows since $\mathcal{O}_{X_s, x}$ is a localization of $\mathcal{O}_{X_s, x'}$ at a prime ideal, hence any chain of prime ideals in $\mathcal{O}_{X_s, x}$ is part of a strictly longer chain of primes in $\mathcal{O}_{X_s, x'}$. The last inequality follows from Algebra, Lemma 10.115.2. $\square$

The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4889–5057 (see updates for more information).

\section{Morphisms and dimensions of fibres}
\label{section-dimension-fibres}

\noindent
Let $X$ be a topological space, and $x \in X$.
Recall that we have defined $\dim_x(X)$ as the minimum of the
dimensions of the open neighbourhoods of $x$ in $X$.
See Topology, Definition \ref{topology-definition-Krull}.

\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}

\begin{lemma}
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}

\begin{lemma}
\label{lemma-dimension-fibre-after-base-change}
Let
$$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^f \\ S' \ar[r]^g & S }$$
be a fibre product diagram of schemes. Assume $f$ locally of finite type.
Suppose that $x' \in X'$, $x = g'(x')$, $s' = f'(x')$ and
$s = g(s') = f(x)$. Then
\begin{enumerate}
\item $\dim_x(X_s) = \dim_{x'}(X'_{s'})$,
\item if $F$ is the fibre of the morphism $X'_{s'} \to X_s$
over $x$, then
$$\dim(\mathcal{O}_{F, x'}) = \dim(\mathcal{O}_{X'_{s'}, x'}) - \dim(\mathcal{O}_{X_s, x}) = \text{trdeg}_{\kappa(s)}(\kappa(x)) - \text{trdeg}_{\kappa(s')}(\kappa(x'))$$
In particular $\dim(\mathcal{O}_{X'_{s'}, x'}) \geq \dim(\mathcal{O}_{X_s, x})$
and $\text{trdeg}_{\kappa(s)}(\kappa(x)) \geq \text{trdeg}_{\kappa(s')}(\kappa(x'))$.
\item given $s', s, x$ there exists a choice of $x'$ such that
$\dim(\mathcal{O}_{X'_{s'}, x'}) = \dim(\mathcal{O}_{X_s, x})$ and
$\text{trdeg}_{\kappa(s)}(\kappa(x)) = \text{trdeg}_{\kappa(s')}(\kappa(x'))$.
\end{enumerate}
\end{lemma}

\begin{proof}
Part (1) follows immediately from
Algebra,
Lemma \ref{algebra-lemma-dimension-at-a-point-preserved-field-extension}.
Parts (2) and (3) from
Algebra, Lemma \ref{algebra-lemma-inequalities-under-field-extension}.
\end{proof}

\noindent
The following lemma follows from a nontrivial algebraic result.
Namely, the algebraic version of Zariski's main theorem.

\begin{lemma}
\label{lemma-openness-bounded-dimension-fibres}
Let $f : X \to S$ be a morphism of schemes.
Let $n \geq 0$. Assume $f$ is locally of finite type.
The set
$$U_n = \{x \in X \mid \dim_x X_{f(x)} \leq n\}$$
is open in $X$.
\end{lemma}

\begin{proof}
This is immediate from
Algebra,
Lemma \ref{algebra-lemma-dimension-fibres-bounded-open-upstairs}
\end{proof}

\begin{lemma}
\label{lemma-openness-bounded-dimension-fibres-finite-presentation}
Let $f : X \to S$ be a morphism of schemes.
Let $n \geq 0$. Assume $f$ is locally of finite presentation.
The open
$$U_n = \{x \in X \mid \dim_x X_{f(x)} \leq n\}$$
of Lemma \ref{lemma-openness-bounded-dimension-fibres} is retrocompact
in $X$. (See Topology, Definition \ref{topology-definition-quasi-compact}.)
\end{lemma}

\begin{proof}
The topological space $X$ has a basis for its topology consisting of
affine opens $U \subset X$ such that the induced morphism
$f|_U : U \to S$ factors through an affine open $V \subset S$. Hence
it is enough to show that $U \cap U_n$ is quasi-compact for such a $U$.
Note that $U_n \cap U$ is the same as the open
$\{x \in U \mid \dim_x U_{f(x)} \leq n\}$. This reduces us to the case
where $X$ and $S$ are affine. In this case the lemma follows from
Algebra,
Lemma \ref{algebra-lemma-dimension-fibres-bounded-quasi-compact-open-upstairs}
(and Lemma \ref{lemma-locally-finite-presentation-characterize}).
\end{proof}

\begin{lemma}
\label{lemma-dimension-fibre-specialization}
Let $f : X \to S$ be a morphism of schemes.
Let $x \leadsto x'$ be a nontrivial specialization of points in $X$
lying over the same point $s \in S$. Assume $f$ is locally of finite type.
Then
\begin{enumerate}
\item $\dim_x(X_s) \leq \dim_{x'}(X_s)$,
\item $\dim(\mathcal{O}_{X_s, x}) < \dim(\mathcal{O}_{X_s, x'})$, and
\item $\text{trdeg}_{\kappa(s)}(\kappa(x)) > \text{trdeg}_{\kappa(s)}(\kappa(x'))$.
\end{enumerate}
\end{lemma}

\begin{proof}
Part (1) follows from the fact that any open of $X_s$ containing $x'$
also contains $x$. Part (2) follows since $\mathcal{O}_{X_s, x}$ is a
localization of $\mathcal{O}_{X_s, x'}$ at a prime ideal, hence any chain
of prime ideals in $\mathcal{O}_{X_s, x}$ is part of a strictly longer
chain of primes in $\mathcal{O}_{X_s, x'}$. The last inequality follows from
Algebra, Lemma \ref{algebra-lemma-tr-deg-specialization}.
\end{proof}

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