## 29.29 Morphisms of given relative dimension

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.

1. We say $f$ is of relative dimension $\leq d$ at $x$ if $\dim _ x(X_{f(x)}) \leq d$.

2. We say $f$ is of relative dimension $\leq d$ if $\dim _ x(X_{f(x)}) \leq d$ for all $x \in X$.

3. 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

1. $f$ has relative dimension $d$,

2. $g$ has relative dimension $e$, and

3. $f$ 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

1. $f$ is flat,

2. $f$ is locally of finite presentation, and

3. 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,

$\dim (\mathcal{O}_{X,a}) = \dim (\mathcal{O}_{Y,b}) + \dim (\mathcal{O}_{X_ b, a}).$

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. $\square$

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