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 4882–5050 (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}
    \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}
    
    \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}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 02FW

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?