Lemma 29.28.1. 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

## 29.28 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.

**Proof.**
This immediately reduces to the case $S = s$, and $X$ affine. In this case the result follows from Algebra, Lemma 10.116.3.
$\square$

Lemma 29.28.2. 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

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 29.28.1 the statement is equivalent to

For this see Algebra, Lemma 10.112.6. For the flat case see Algebra, Lemma 10.112.7. $\square$

Lemma 29.28.3. Let

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

$\dim _ x(X_ s) = \dim _{x'}(X'_{s'})$,

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 $\text{trdeg}_{\kappa (s)}(\kappa (x)) \geq \text{trdeg}_{\kappa (s')}(\kappa (x'))$.

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 10.116.6. Parts (2) and (3) from Algebra, Lemma 10.116.7.
$\square$

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

Lemma 29.28.4. Let $f : X \to S$ be a morphism of schemes. Let $n \geq 0$. Assume $f$ is locally of finite type. The set

is open in $X$.

**Proof.**
This is immediate from Algebra, Lemma 10.125.6
$\square$

Lemma 29.28.5. Let $f : X \to Y$ be a morphism of finite type with $Y$ quasi-compact. Then the dimension of the fibres of $f$ is bounded.

**Proof.**
By Lemma 29.28.4 the set $U_ n \subset X$ of points where the dimension of the fibre is $\leq n$ is open. Since $f$ is of finite type, every point is contained in some $U_ n$ (because the dimension of a finite type algebra over a field is finite). Since $Y$ is quasi-compact and $f$ is of finite type, we see that $X$ is quasi-compact. Hence $X = U_ n$ for some $n$.
$\square$

Lemma 29.28.6. Let $f : X \to S$ be a morphism of schemes. Let $n \geq 0$. Assume $f$ is locally of finite presentation. The open

of Lemma 29.28.4 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 10.125.8 (and Lemma 29.21.2).
$\square$

Lemma 29.28.7. 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

$\dim _ x(X_ s) \leq \dim _{x'}(X_ s)$,

$\dim (\mathcal{O}_{X_ s, x}) < \dim (\mathcal{O}_{X_ s, x'})$, and

$\text{trdeg}_{\kappa (s)}(\kappa (x)) > \text{trdeg}_{\kappa (s)}(\kappa (x'))$.

**Proof.**
Part (1) follows 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 10.116.2.
$\square$

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