Lemma 72.15.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that

$Y$ is locally Noetherian,

$X$ and $Y$ are integral algebraic spaces,

$f$ is dominant, and

$f$ is locally of finite type.

If $x \in |X|$ and $y \in |Y|$ are the generic points, then

\[ \dim (X) \leq \dim (Y) + \text{transcendence degree of }x/y. \]

If $f$ is proper, then equality holds.

**Proof.**
Recall that $|X|$ and $|Y|$ are irreducible sober topological spaces, see discussion following Definition 72.4.1. Thus the fact that $f$ is dominant means that $|f|$ maps $x$ to $y$. Moreover, $x \in |X|$ is the unique point at which the local ring of $X$ has dimension $0$, see Decent Spaces, Lemma 68.20.1. By Morphisms of Spaces, Lemma 67.35.1 we see that the dimension of the local ring of $X$ at any point $x' \in |X|$ is at most the dimension of the local ring of $Y$ at $y' = f(x')$ plus the transcendence degree of $x/y$. Since the dimension of $X$, resp. dimension of $Y$ is the supremum of the dimensions of the local rings at $x'$, resp. $y'$ (Properties of Spaces, Lemma 66.10.3) we conclude the inequality holds.

Assume $f$ is proper. Let $V \subset Y$ be a nonempty quasi-compact open subspace. If we can prove the equality for the morphism $f^{-1}(V) \to V$, then we get the equality for $X \to Y$. Thus we may assume that $X$ and $Y$ are quasi-compact. Observe that $X$ is quasi-separated as a locally Noetherian decent algebraic space, see Decent Spaces, Lemma 68.14.1. Thus we may choose $Y' \to Y$ finite surjective where $Y'$ is a scheme, see Limits of Spaces, Proposition 70.16.1. After replacing $Y'$ by a suitable closed subscheme, we may assume $Y'$ is integral, see for example the more general Lemma 72.8.5. By the same lemma, we may choose a closed subspace $X' \subset X \times _ Y Y'$ such that $X'$ is integral and $X' \to X$ is finite surjective. Now $X'$ is also locally Noetherian (Morphisms of Spaces, Lemma 67.23.5) and we can use Limits of Spaces, Proposition 70.16.1 once more to choose a finite surjective morphism $X'' \to X'$ with $X''$ a scheme. As before we may assume that $X''$ is integral. Picture

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

By Lemma 72.15.1 we have $\dim (X'') = \dim (X)$ and $\dim (Y') = \dim (Y)$. Since $X$ and $Y$ have open neighbourhoods of $x$, resp. $y$ which are schemes, we readily see that the generic points $x'' \in X''$, resp. $y' \in Y'$ are the unique points mapping to $x$, resp. $y$ and that the residue field extensions $\kappa (x'')/\kappa (x)$ and $\kappa (y')/\kappa (y)$ are finite. This implies that the transcendence degree of $x''/y'$ is the same as the transcendence degree of $x/y$. Thus the equality follows from the case of schemes whicn is Morphisms, Lemma 29.52.4.
$\square$

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