## 65.34 Morphisms and dimensions of fibres

This section is the analogue of Morphisms, Section 29.28. The formulations in this section are a bit awkward since we do not have local rings of algebraic spaces at points.

Lemma 65.34.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$. Assume $f$ is locally of finite type. Then we have

$\begin{matrix} \text{relative dimension of }f\text{ at }x \\ = \\ \text{dimension of local ring of the fibre of }f\text{ at }x \\ + \\ \text{transcendence degree of }x/f(x) \end{matrix}$

where the notation is as in Definition 65.33.1.

Proof. This follows immediately from Morphisms, Lemma 29.28.1 applied to $h : U \to V$ and $u \in U$ as in Lemma 65.22.5. $\square$

Lemma 65.34.2. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$. Let $x \in |X|$ and set $y = f(x)$. Assume $f$ and $g$ locally of finite type. Then

1. $\begin{matrix} \text{relative dimension of }g \circ f\text{ at }x \\ \leq \\ \text{relative dimension of }f\text{ at }x \\ + \\ \text{relative dimension of }g\text{ at }y \end{matrix}$
2. equality holds in (1) if for some morphism $\mathop{\mathrm{Spec}}(k) \to Z$ from the spectrum of a field in the class of $g(f(x)) = g(y)$ the morphism $X_ k \to Y_ k$ is flat at $x$, for example if $f$ is flat at $x$,

3. $\begin{matrix} \text{transcendence degree of }x/g(f(x)) \\ = \\ \text{transcendence degree of }x/f(x) \\ + \\ \text{transcendence degree of }f(x)/g(f(x)) \end{matrix}$

Proof. Choose a diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \ar[r] & W \ar[d] \\ X \ar[r] & Y \ar[r] & Z }$

with $U, V, W$ schemes and vertical arrows étale and surjective. (See Spaces, Lemma 63.11.6.) Choose $u \in U$ mapping to $x$. Set $v, w$ equal to the images of $u$ in $V, W$. Apply Morphisms, Lemma 29.28.2 to the top row and the points $u, v, w$. Details omitted. $\square$

Lemma 65.34.3. Let $S$ be a scheme. Let

$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

be a fibre product diagram of algebraic spaces over $S$. Let $x' \in |X'|$. Set $x = g'(x')$. Assume $f$ locally of finite type. Then

1. $\begin{matrix} \text{relative dimension of }f\text{ at }x \\ = \\ \text{relative dimension of }f'\text{ at }x' \end{matrix}$
2. we have

$\begin{matrix} \text{dimension of local ring of the fibre of }f'\text{ at }x' \\ - \\ \text{dimension of local ring of the fibre of }f\text{ at }x \\ = \\ \text{transcendence degree of }x/f(x) \\ - \\ \text{transcendence degree of }x'/f'(x') \end{matrix}$

and the common value is $\geq 0$,

3. given $x$ and $y' \in |Y'|$ mapping to the same $y \in |Y|$ there exists a choice of $x'$ such that the integer in (2) is $0$.

Proof. Choose a surjective étale morphism $V \to Y$ with $V$ a scheme. Choose a surjective étale morphism $U \to V \times _ Y X$ with $U$ a scheme. Choose a surjective étale morphism $V' \to V \times _ Y Y'$ with $V'$ a scheme. Set $U' = V' \times _ V U$. Then the induced morphism $U' \to X'$ is also surjective and étale (argument omitted). Choose $u' \in U'$ mapping to $x'$. At this point parts (1) and (2) follow by applying Morphisms, Lemma 29.28.3 to the diagram of schemes involving $U', U, V', V$ and the point $u'$. To prove (3) first choose $v \in V$ mapping to $y$. Then using Properties of Spaces, Lemma 64.4.3 we can choose $v' \in V'$ mapping to $y'$ and $v$ and $u \in U$ mapping to $x$ and $v$. Finally, according to Morphisms, Lemma 29.28.3 we can choose $u' \in U'$ mapping to $v'$ and $u$ such that the integer is zero. Then taking $x' \in |X'|$ the image of $u'$ works. $\square$

Lemma 65.34.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $n \geq 0$. Assume $f$ is locally of finite type. The set

$W_ n = \{ x \in |X| \text{ such that the relative dimension of }f\text{ at } x \leq n\}$

is open in $|X|$.

Proof. Choose a diagram

$\xymatrix{ U \ar[r]_ h \ar[d]_ a & V \ar[d] \\ X \ar[r] & Y }$

where $U$ and $V$ are schemes and the vertical arrows are surjective and étale, see Spaces, Lemma 63.11.6. By Morphisms, Lemma 29.28.4 the set $U_ n$ of points where $h$ has relative dimension $\leq n$ is open in $U$. By our definition of relative dimension for morphisms of algebraic spaces at points we see that $U_ n = a^{-1}(W_ n)$. The lemma follows by definition of the topology on $|X|$. $\square$

Lemma 65.34.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ Let $n \geq 0$. Assume $f$ is locally of finite presentation. The open

$W_ n = \{ x \in |X| \text{ such that the relative dimension of }f\text{ at } x \leq n\}$

of Lemma 65.34.4 is retrocompact in $|X|$. (See Topology, Definition 5.12.1.)

Proof. Choose a diagram

$\xymatrix{ U \ar[r]_ h \ar[d]_ a & V \ar[d] \\ X \ar[r] & Y }$

where $U$ and $V$ are schemes and the vertical arrows are surjective and étale, see Spaces, Lemma 63.11.6. In the proof of Lemma 65.34.4 we have seen that $a^{-1}(W_ n) = U_ n$ is the corresponding set for the morphism $h$. By Morphisms, Lemma 29.28.6 we see that $U_ n$ is retrocompact in $U$. The lemma follows by definition of the topology on $|X|$, compare with Properties of Spaces, Lemma 64.5.5 and its proof. $\square$

Lemma 65.34.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type. Then $f$ is locally quasi-finite if and only if $f$ has relative dimension $0$ at each $x \in |X|$.

Proof. Choose a diagram

$\xymatrix{ U \ar[r]_ h \ar[d]_ a & V \ar[d] \\ X \ar[r] & Y }$

where $U$ and $V$ are schemes and the vertical arrows are surjective and étale, see Spaces, Lemma 63.11.6. The definitions imply that $h$ is locally quasi-finite if and only if $f$ is locally quasi-finite, and that $f$ has relative dimension $0$ at all $x \in |X|$ if and only if $h$ has relative dimension $0$ at all $u \in U$. Hence the result follows from the result for $h$ which is Morphisms, Lemma 29.29.5. $\square$

Lemma 65.34.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type. Then there exists a canonical open subspace $X' \subset X$ such that $f|_{X'} : X' \to Y$ is locally quasi-finite, and such that the relative dimension of $f$ at any $x \in |X|$, $x \not\in |X'|$ is $\geq 1$. Formation of $X'$ commutes with arbitrary base change.

Lemma 65.34.8. Let $S$ be a scheme. Consider a cartesian diagram

$\xymatrix{ X \ar[d] & F \ar[l]^ p \ar[d] \\ Y & \mathop{\mathrm{Spec}}(k) \ar[l] }$

where $X \to Y$ is a morphism of algebraic spaces over $S$ which is locally of finite type and where $k$ is a field over $S$. Let $z \in |F|$ be such that $\dim _ z(F) = 0$. Then, after replacing $X$ by an open subspace containing $p(z)$, the morphism

$X \longrightarrow Y$

is locally quasi-finite.

Proof. Let $X' \subset X$ be the open subspace over which $f$ is locally quasi-finite found in Lemma 65.34.7. Since the formation of $X'$ commutes with arbitrary base change we see that $z \in X' \times _ Y \mathop{\mathrm{Spec}}(k)$. Hence the lemma is clear. $\square$

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