The Stacks project

Lemma 76.31.1. Let $S$ be a scheme. Let $f : X \to Y$ be a finite type morphism of algebraic spaces over $S$. Let $y \in |Y|$. The following quantities are the same

  1. $d = -\infty $ if $y$ is not in the image of $|f|$ and otherwise the minimal integer $d$ such that $f$ has relative dimension $\leq d$ at every $x \in |X|$ mapping to $y$,

  2. the dimension of the algebraic space $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$ for any morphism $\mathop{\mathrm{Spec}}(k) \to Y$ in the equivalence class defining $y$.

Proof. To parse this one has to consult Morphisms of Spaces, Definition 67.33.1, Properties of Spaces, Definition 66.9.2, Properties of Spaces, Definition 66.9.1. We will show that the numbers in (1) and (2) are equal for a fixed morphism $\mathop{\mathrm{Spec}}(k) \to Y$. Choose an étale morphism $V \to Y$ where $V$ is an affine scheme and a point $v \in V$ mapping to $y$. Since $V \times _ Y \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ is surjective étale (by Properties of Spaces, Lemma 66.4.3) we can find a finite separable extension $k'/k$ (by Morphisms, Lemma 29.36.7) and a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(k') \ar[r] \ar[d] & V \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & Y } \]

We may replace $X \to Y$ by $V \times _ Y X \to V$ and $X_ k$ by $X_{k'} = \mathop{\mathrm{Spec}}(k') \times _ V (V \times _ Y X)$ because this does not change the dimensions in question by Properties of Spaces, Lemma 66.22.5 and Morphisms of Spaces, Lemma 67.34.3. Thus we may assume that $Y$ is an affine scheme. In this case we may assume that $k = \kappa (y)$ because the dimension of $X_{\kappa (y)}$ and $X_ k$ are the same by the aforementioned Morphisms of Spaces, Lemma 67.34.3 and the fact that for an algebraic space $Z$ over a field $K$ the relative dimension of $Z$ at a point $z \in |Z|$ is the same as $\dim _ z(Z)$ by definition. Assume $Y$ is affine and $k = \kappa (y)$. Then $X$ is quasi-compact we can choose an affine scheme $U$ and an surjective étale morphism $U \to X$. Then $\dim (X_ k) = \dim (U_ k) = \max \dim _ u(U_ k)$ is equal to the number given in (1) by definition. $\square$

Comments (2)

Comment #8140 by Laurent Moret-Bailly on

If the fiber is empty, (2) gives the value . Does this count as an integer? If not, (1) gives the value .

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