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A bit more on dimension of spaces

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aisejohan committed Dec 6, 2017
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@@ -1775,6 +1775,142 @@ \section{Geometrically integral algebraic spaces}
In this section we continue the discussion about dimension.
Here is a list of previous material:
\item dimension is defined in
Properties of Spaces, Section
\item dimension of local ring is defined in
Properties of Spaces, Section
\item a couple of results in Properties of Spaces, Lemmas
\ref{spaces-properties-lemma-dimension-local-ring} and
\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
\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
In More on Morphisms of Spaces, Section
we will discuss jumping of
dimension in fibres of a finite type morphism.
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 $\dim(X) = \dim(Y)$.
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
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}.
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. Assume that
\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.
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.
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
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.
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
Thus we may choose $Y' \to Y$ finite surjective where $Y'$
is a scheme, see Limits of Spaces, Proposition
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
and we can use Limits of Spaces, Proposition
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
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}.
\section{Spaces smooth over fields}

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