The Stacks project

66.33 Relative dimension

In this section we define the relative dimension of a morphism of algebraic spaces at a point, and some closely related properties.

Definition 66.33.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$. Let $d, r \in \{ 0, 1, 2, \ldots , \infty \} $.

  1. We say the dimension of the local ring of the fibre of $f$ at $x$ is $d$ if the equivalent conditions of Lemma 66.22.5 hold for the property $\mathcal{P}_ d$ described in Descent, Lemma 35.33.6.

  2. We say the transcendence degree of $x/f(x)$ is $r$ if the equivalent conditions of Lemma 66.22.5 hold for the property $\mathcal{P}_ r$ described in Descent, Lemma 35.33.7.

  3. We say $f$ has relative dimension $d$ at $x$ if the equivalent conditions of Lemma 66.22.5 hold for the property $\mathcal{P}_ d$ described in Descent, Lemma 35.33.8.

Let us spell out what this means. Namely, choose some diagrams

\[ \xymatrix{ U \ar[d]_ a \ar[r]_ h & V \ar[d]^ b \\ X \ar[r]^ f & Y } \quad \quad \xymatrix{ u \ar[d] \ar[r] & v \ar[d] \\ x \ar[r] & y } \]

as in Lemma 66.22.5. Then we have

\[ \begin{matrix} \text{relative dimension of }f\text{ at }x & = & \dim _ u (U_ v) \\ \text{dimension of local ring of the fibre of }f\text{ at }x & = & \dim (\mathcal{O}_{U_ v, u}) \\ \text{transcendence degree of }x/f(x) & = & \text{trdeg}_{\kappa (v)}(\kappa (u)) \end{matrix} \]

Note that if $Y = \mathop{\mathrm{Spec}}(k)$ is the spectrum of a field, then the relative dimension of $X/Y$ at $x$ is the same as $\dim _ x(X)$, the transcendence degree of $x/f(x)$ is the transcendence degree over $k$, and the dimension of the local ring of the fibre of $f$ at $x$ is just the dimension of the local ring at $x$, i.e., the relative notions become absolute notions in that case.

Definition 66.33.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $d \in \{ 0, 1, 2, \ldots \} $.

  1. We say $f$ has relative dimension $\leq d$ if $f$ has relative dimension $\leq d$ at all $x \in |X|$.

  2. We say $f$ has relative dimension $d$ if $f$ has relative dimension $d$ at all $x \in |X|$.

Having relative dimension equal to $d$ means roughly speaking that all nonempty fibres are equidimensional of dimension $d$.

Lemma 66.33.3. Let $S$ be a scheme. Let $X \to Y \to Z$ be morphisms of algebraic spaces over $S$. Let $x \in |X|$ and let $y \in |Y|$, $z \in |Z|$ be the images. Assume $X \to Y$ is locally quasi-finite and $Y \to Z$ locally of finite type. Then the transcendence degree of $x/z$ is equal to the transcendence degree of $y/z$.

Proof. We can choose commutative diagrams

\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \ar[r] & W \ar[d] \\ X \ar[r] & Y \ar[r] & Z } \quad \quad \xymatrix{ u \ar[d] \ar[r] & v \ar[d] \ar[r] & w \ar[d] \\ x \ar[r] & y \ar[r] & z } \]

where $U, V, W$ are schemes and the vertical arrows are étale. By definition the morphism $U \to V$ is locally quasi-finite which implies that $\kappa (v) \subset \kappa (u)$ is finite, see Morphisms, Lemma 29.20.5. Hence the result is clear. $\square$

Lemma 66.33.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is locally of finite type, $Y$ is Jacobson (Properties of Spaces, Remark 65.7.3), and $x \in |X|$ is a finite type point of $X$, then the transcendence degree of $x/f(x)$ is $0$.

Proof. Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Choose a scheme $U$ and a surjective étale morphism $U \to X \times _ Y V$. By Lemma 66.25.5 we can find a finite type point $u \in U$ mapping to $x$. After shrinking $U$ we may assume $u \in U$ is closed (Morphisms, Lemma 29.16.4). Let $v \in V$ be the image of $u$. By Morphisms, Lemma 29.16.8 the extension $\kappa (u)/\kappa (v)$ is finite. This finishes the proof. $\square$

Lemma 66.33.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of locally Noetherian algebraic spaces over $S$ which is flat, locally of finite type and of relative dimension $d$. For every point $x$ in $|X|$ with image $y$ in $|Y|$ we have $\dim _ x(X) = \dim _ y(Y) + d$.

Proof. By definition of the dimension of an algebraic space at a point (Properties of Spaces, Definition 65.9.1) and by definition of having relative dimension $d$, this reduces to the corresponding statement for schemes (Morphisms, Lemma 29.29.6). $\square$


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