In order to be able to speak comfortably about morphisms of a given relative dimension we introduce the following notion.
Definition 29.29.1. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is locally of finite type.
We say $f$ is of relative dimension $\leq d$ at $x$ if $\dim _ x(X_{f(x)}) \leq d$.
We say $f$ is of relative dimension $\leq d$ if $\dim _ x(X_{f(x)}) \leq d$ for all $x \in X$.
We say $f$ is of relative dimension $d$ if all nonempty fibres $X_ s$ are equidimensional of dimension $d$.
This is not a particularly well behaved notion, but it works well in a number of situations.
Lemma 29.29.2. Let $f : X \to S$ be a morphism of schemes which is locally of finite type. If $f$ has relative dimension $d$, then so does any base change of $f$. Same for relative dimension $\leq d$.
Proof. This is immediate from Lemma 29.28.3. $\square$
Lemma 29.29.3. Let $f : X \to Y$, $g : Y \to Z$ be locally of finite type. If $f$ has relative dimension $\leq d$ and $g$ has relative dimension $\leq e$ then $g \circ f$ has relative dimension $\leq d + e$. If
$f$ has relative dimension $d$,
$g$ has relative dimension $e$, and
$f$ is flat,
then $g \circ f$ has relative dimension $d + e$.
Proof. This is immediate from Lemma 29.28.2. $\square$
In general it is not possible to decompose a morphism into its pieces where the relative dimension is a given one. However, it is possible if the morphism has Cohen-Macaulay fibres and is flat of finite presentation.
Lemma 29.29.4. Let $f : X \to S$ be a morphism of schemes. Assume that
$f$ is flat,
$f$ is locally of finite presentation, and
for all $s \in S$ the fibre $X_ s$ is Cohen-Macaulay (Properties, Definition 28.8.1)
Then there exist open and closed subschemes $X_ d \subset X$ such that $X = \coprod _{d \geq 0} X_ d$ and $f|_{X_ d} : X_ d \to S$ has relative dimension $d$.
Proof. This is immediate from Algebra, Lemma 10.130.8. $\square$
Lemma 29.29.5. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is locally of finite type. Let $x \in X$ with $s = f(x)$. Then $f$ is quasi-finite at $x$ if and only if $\dim _ x(X_ s) = 0$. In particular, $f$ is locally quasi-finite if and only if $f$ has relative dimension $0$.
Proof. First proof. If $f$ is quasi-finite at $x$ then $\kappa (x)$ is a finite extension of $\kappa (s)$ (by Lemma 29.20.5) and $x$ is isolated in $X_ s$ (by Lemma 29.20.6), hence $\dim _ x(X_ s) = 0$ by Lemma 29.28.1. Conversely, if $\dim _ x(X_ s) = 0$ then by Lemma 29.28.1 we see $\kappa (s) \subset \kappa (x)$ is algebraic and there are no other points of $X_ s$ specializing to $x$. Hence $x$ is closed in its fibre by Lemma 29.20.2 and by Lemma 29.20.6 (3) we conclude that $f$ is quasi-finite at $x$.
Second proof. The fibre $X_ s$ is a scheme locally of finite type over a field, hence locally Noetherian (Lemma 29.15.6). The result now follows from Lemma 29.20.6 and Properties, Lemma 28.10.7. $\square$
Lemma 29.29.6. Let $f : X \to Y$ be a morphism of locally Noetherian schemes which is flat, locally of finite type and of relative dimension $d$. For every point $x$ in $X$ with image $y$ in $Y$ we have $\dim _ x(X) = \dim _ y(Y) + d$.
Proof. After shrinking $X$ and $Y$ to open neighborhoods of $x$ and $y$, we can assume that $\dim (X) = \dim _ x(X)$ and $\dim (Y) = \dim _ y(Y)$, by definition of the dimension of a scheme at a point (Properties, Definition 28.10.1). The morphism $f$ is open by Lemmas 29.21.9 and 29.25.10. Hence we can shrink $Y$ to arrange that $f$ is surjective. It remains to show that $\dim (X) = \dim (Y) + d$.
Let $a$ be a point in $X$ with image $b$ in $Y$. By Algebra, Lemma 10.112.7,
Taking the supremum over all points $a$ in $X$, it follows that $\dim (X) = \dim (Y) + d$, as we want, see Properties, Lemma 28.10.2. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)