Lemma 89.2.3. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local domain of dimension $\geq 1$. Let $f : X \to \mathop{\mathrm{Spec}}(A)$ be a morphism of algebraic spaces. Assume at least one of the following conditions is satisfied
$f$ is a modification (Spaces over Fields, Definition 72.8.1),
$f$ is an alteration (Spaces over Fields, Definition 72.8.3),
$f$ is locally of finite type, quasi-separated, $X$ is integral, and there is exactly one point of $|X|$ mapping to the generic point of $\mathop{\mathrm{Spec}}(A)$,
$f$ is locally of finite type, $X$ is decent, and the points of $|X|$ mapping to the generic point of $\mathop{\mathrm{Spec}}(A)$ are the generic points of irreducible components of $|X|$,
add more here.
Then $\dim (X_\kappa ) \leq \dim (A) - 1$.
Proof.
Cases (1), (2), and (3) are special cases of (4). Choose an affine scheme $U = \mathop{\mathrm{Spec}}(B)$ and an étale morphism $U \to X$. The ring map $A \to B$ is of finite type. We have to show that $\dim (U_\kappa ) \leq \dim (A) - 1$. Since $X$ is decent, the generic points of irreducible components of $U$ are the points lying over generic points of irreducible components of $|X|$, see Decent Spaces, Lemma 68.20.1. Hence the fibre of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ over $(0)$ is the (finite) set of minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ r$ of $B$. Thus $A_ f \to B_ f$ is finite for some nonzero $f \in A$ (Algebra, Lemma 10.122.10). We conclude $\kappa (\mathfrak q_ i)$ is a finite extension of the fraction field of $A$. Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak m$. Then
\[ \dim (B_\mathfrak q) = \max \dim ((B/\mathfrak q_ i)_{\mathfrak q}) \leq \dim (A) \]
the inequality by the dimension formula for $A \subset B/\mathfrak q_ i$, see Algebra, Lemma 10.113.1. However, the dimension of $B_\mathfrak q/\mathfrak m B_\mathfrak q$ (which is the local ring of $U_\kappa $ at the corresponding point) is at least one less because the minimal primes $\mathfrak q_ i$ are not in $V(\mathfrak m)$. We conclude by Properties, Lemma 28.10.2.
$\square$
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