Lemma 70.15.1. Let $S$ be a scheme. Let $f : X \to Y$ be an integral morphism of algebraic spaces. Then $\dim (X) \leq \dim (Y)$. If $f$ is surjective then $\dim (X) = \dim (Y)$.

## 70.15 Dimension

In this section we continue the discussion about dimension. Here is a list of previous material:

dimension is defined in Properties of Spaces, Section 64.9,

dimension of local ring is defined in Properties of Spaces, Section 64.10,

a couple of results in Properties of Spaces, Lemmas 64.22.4 and 64.22.5,

relative dimension is defined in Morphisms of Spaces, Section 65.33,

results on dimension of fibres in Morphisms of Spaces, Section 65.34,

a weak form of the dimension formula Morphisms of Spaces, Section 65.35,

a result on smoothness and dimension Morphisms of Spaces, Lemma 65.37.10,

dimension is $\dim (|X|)$ for decent spaces Decent Spaces, Lemma 66.12.5,

quasi-finite maps and dimension Decent Spaces, Lemmas 66.12.6 and 66.12.7.

In More on Morphisms of Spaces, Section 74.31 we will discuss jumping of dimension in fibres of a finite type morphism.

**Proof.**
Choose $V \to Y$ surjective étale with $V$ a scheme. Then $U = X \times _ Y V$ is a scheme and $U \to V$ is integral (and surjective if $f$ is surjective). By Properties of Spaces, Lemma 64.22.5 we have $\dim (X) = \dim (U)$ and $\dim (Y) = \dim (V)$. Thus the result follows from the case of schemes which is Morphisms, Lemma 29.43.9.
$\square$

Lemma 70.15.2. 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

If $f$ is proper, then equality holds.

**Proof.**
Recall that $|X|$ and $|Y|$ are irreducible sober topological spaces, see discussion following Definition 70.4.1. 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 66.20.1. By Morphisms of Spaces, Lemma 65.35.1 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 64.10.3) 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 66.14.1. Thus we may choose $Y' \to Y$ finite surjective where $Y'$ is a scheme, see Limits of Spaces, Proposition 68.16.1. After replacing $Y'$ by a suitable closed subscheme, we may assume $Y'$ is integral, see for example the more general Lemma 70.8.5. 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 65.23.5) and we can use Limits of Spaces, Proposition 68.16.1 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

By Lemma 70.15.1 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 29.51.4. $\square$

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