# stacks/stacks-project

A bit more on dimension of spaces

 @@ -1775,6 +1775,142 @@ \section{Geometrically integral algebraic spaces} \section{Dimension} \label{section-dimension} \noindent In this section we continue the discussion about dimension. Here is a list of previous material: \begin{enumerate} \item dimension is defined in Properties of Spaces, Section \ref{spaces-properties-section-dimension}, \item dimension of local ring is defined in Properties of Spaces, Section \ref{spaces-properties-section-dimension-local-ring}, \item a couple of results in Properties of Spaces, Lemmas \ref{spaces-properties-lemma-dimension-local-ring} and \ref{spaces-properties-lemma-dimension-decent-invariant-under-etale}, \item relative dimension is defined in Morphisms of Spaces, Section \ref{spaces-morphisms-section-relative-dimension}, \item results on dimension of fibres in Morphisms of Spaces, Section \ref{spaces-morphisms-section-dimension-fibres}, \item a weak form of the dimension formula Morphisms of Spaces, Section \ref{spaces-morphisms-section-dimension-formula}, \item a result on smoothness and dimension Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-smoothness-dimension-spaces}, \item dimension is $\dim(|X|)$ for decent spaces Decent Spaces, Lemma \ref{decent-spaces-lemma-dimension-decent-space}, \item quasi-finite maps and dimension Decent Spaces, Lemmas \ref{decent-spaces-lemma-dimension-local-ring-quasi-finite} and \ref{decent-spaces-lemma-dimension-quasi-finite}. \end{enumerate} In More on Morphisms of Spaces, Section \ref{spaces-more-morphisms-section-dimension} we will discuss jumping of dimension in fibres of a finite type morphism. \begin{lemma} \label{lemma-integral-dimension} 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 $\dim(X) = \dim(Y)$. \end{lemma} \begin{proof} Choose $V \to Y$ surjective \'etale 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 \ref{spaces-properties-lemma-dimension-decent-invariant-under-etale} we have $\dim(X) = \dim(U)$ and $\dim(Y) = \dim(V)$. Thus the result follows from the case of schemes which is Morphisms, Lemma \ref{morphisms-lemma-integral-dimension}. \end{proof} \begin{lemma} \label{lemma-alteration-dimension} Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that \begin{enumerate} \item $Y$ is locally Noetherian, \item $X$ and $Y$ are integral algebraic spaces, \item $f$ is dominant, and \item $f$ is locally of finite type. \end{enumerate} 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. \end{lemma} \begin{proof} Recall that $|X|$ and $|Y|$ are irreducible sober topological spaces, see discussion following Definition \ref{definition-integral-algebraic-space}. 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 \ref{decent-spaces-lemma-decent-generic-points}. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-dimension-formula-general} 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 \ref{spaces-properties-lemma-dimension}) we conclude the inequality holds. \medskip\noindent 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 \ref{decent-spaces-lemma-locally-Noetherian-decent-quasi-separated}. Thus we may choose $Y' \to Y$ finite surjective where $Y'$ is a scheme, see Limits of Spaces, Proposition \ref{spaces-limits-proposition-there-is-a-scheme-finite-over}. After replacing $Y'$ by a suitable closed subscheme, we may assume $Y'$ is integral, see for example the more general Lemma \ref{lemma-alteration-contained-in}. 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 \ref{spaces-morphisms-lemma-locally-finite-type-locally-noetherian}) and we can use Limits of Spaces, Proposition \ref{spaces-limits-proposition-there-is-a-scheme-finite-over} 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 \ref{lemma-integral-dimension} 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 \ref{morphisms-lemma-alteration-dimension}. \end{proof} \section{Spaces smooth over fields} \label{section-smooth}