**Proof.**
To parse this one has to consult Morphisms of Spaces, Definition 66.33.1, Properties of Spaces, Definition 65.9.2, Properties of Spaces, Definition 65.9.1. 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 65.4.3) 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 65.22.5 and Morphisms of Spaces, Lemma 66.34.3. 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 66.34.3 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

Comment #8232 by Stacks Project on