33.19 Dimension of fibres

We have already seen that dimension of fibres of finite type morphisms typically jump up. In this section we discuss the phenomenon that in codimension $1$ this does not happen. More generally, we discuss how much the dimension of a fibre can jump. Here is a list of related results:

1. For a finite type morphism $X \to S$ the set of $x \in X$ with $\dim _ x(X_{f(x)}) \leq d$ is open, see Algebra, Lemma 10.125.6 and Morphisms, Lemma 29.28.4.

2. We have the dimension formula, see Algebra, Lemma 10.113.1 and Morphisms, Lemma 29.52.1.

3. Constant fibre dimension for an integral finite type scheme dominating a valuation ring, see Algebra, Lemma 10.125.9.

4. If $X \to S$ is of finite type and is quasi-finite at every generic point of $X$, then $X \to S$ is quasi-finite in codimension $1$, see Algebra, Lemma 10.113.2 and Lemma 33.17.1.

The last result mentioned above generalizes as follows.

Lemma 33.19.1. Let $f : X \to Y$ be locally of finite type. Let $x \in X$ be a point with image $y \in Y$ such that $\mathcal{O}_{Y, y}$ is Noetherian of dimension $\leq 1$. Let $d \geq 0$ be an integer such that for every generic point $\eta$ of an irreducible component of $X$ which contains $x$, we have $\dim _\eta (X_{f(\eta )}) = d$. Then $\dim _ x(X_ y) = d$.

Proof. Recall that $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ is the set of points of $Y$ specializing to $y$, see Schemes, Lemma 26.13.2. Thus we may replace $Y$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ and assume $Y = \mathop{\mathrm{Spec}}(B)$ where $B$ is a Noetherian local ring of dimension $\leq 1$ and $y$ is the closed point. We may also replace $X$ by an affine neighbourhood of $x$.

Let $X = \bigcup X_ i$ be the irreducible components of $X$ viewed as reduced closed subschemes. If we can show each fibre $X_{i, y}$ has dimension $d$, then $X_ y = \bigcup X_{i, y}$ has dimension $d$ as well. Thus we may assume $X$ is an integral scheme.

If $X \to Y$ maps the generic point $\eta$ of $X$ to $y$, then $X$ is a scheme over $\kappa (y)$ and the result is true by assumption. Assume that $X$ maps $\eta$ to a point $\xi \in Y$ corresponding to a minimal prime $\mathfrak q$ of $B$ different from $\mathfrak m_ B$. We obtain a factorization $X \to \mathop{\mathrm{Spec}}(B/\mathfrak q) \to \mathop{\mathrm{Spec}}(B)$. By the dimension formula (Morphisms, Lemma 29.52.1) we have

$\dim (\mathcal{O}_{X, x}) + \text{trdeg}_{\kappa (y)} \kappa (x) \leq \dim (B/\mathfrak q) + \text{trdeg}_{\kappa (\mathfrak q)}(R(X))$

We have $\dim (B/\mathfrak q) = 1$. We have $\text{trdeg}_{\kappa (\mathfrak q)}(R(X)) = d$ by our assumption that $\dim _\eta (X_\xi ) = d$, see Morphisms, Lemma 29.28.1. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_ s, x}$ has a kernel (as $\eta \mapsto \xi \not= y$) and since $\mathcal{O}_{X, x}$ is a Noetherian domain we see that $\dim (\mathcal{O}_{X, x}) > \dim (\mathcal{O}_{X_ y, x})$. We conclude that

$\dim _ x(X_ s) = \dim (\mathcal{O}_{X_ s, x}) + \text{trdeg}_{\kappa (y)} \kappa (x) \leq d$

(Morphisms, Lemma 29.28.1). On the other hand, we have $\dim _ x(X_ s) \geq \dim _\eta (X_{f(\eta )}) = d$ by Morphisms, Lemma 29.28.4. $\square$

Lemma 33.19.2. Let $f : X \to \mathop{\mathrm{Spec}}(R)$ be a morphism from an irreducible scheme to the spectrum of a valuation ring. If $f$ is locally of finite type and surjective, then the special fibre is equidimensional of dimension equal to the dimension of the generic fibre.

Proof. We may replace $X$ by its reduction because this does not change the dimension of $X$ or of the special fibre. Then $X$ is integral and the lemma follows from Algebra, Lemma 10.125.9. $\square$

The following lemma generalizes Lemma 33.19.1.

Lemma 33.19.3. Let $f : X \to Y$ be locally of finite type. Let $x \in X$ be a point with image $y \in Y$ such that $\mathcal{O}_{Y, y}$ is Noetherian. Let $d \geq 0$ be an integer such that for every generic point $\eta$ of an irreducible component of $X$ which contains $x$, we have $f(\eta ) \not= y$ and $\dim _\eta (X_{f(\eta )}) = d$. Then $\dim _ x(X_ y) \leq d + \dim (\mathcal{O}_{Y, y}) - 1$.

Proof. Exactly as in the proof of Lemma 33.19.1 we reduce to the case $X = \mathop{\mathrm{Spec}}(A)$ with $A$ a domain and $Y = \mathop{\mathrm{Spec}}(B)$ where $B$ is a Noetherian local ring whose maximal ideal corresponds to $y$. After replacing $B$ by $B/\mathop{\mathrm{Ker}}(B \to A)$ we may assume that $B$ is a domain and that $B \subset A$. Then we use the dimension formula (Morphisms, Lemma 29.52.1) to get

$\dim (\mathcal{O}_{X, x}) + \text{trdeg}_{\kappa (y)} \kappa (x) \leq \dim (B) + \text{trdeg}_ B(A)$

We have $\text{trdeg}_ B(A) = d$ by our assumption that $\dim _\eta (X_\xi ) = d$, see Morphisms, Lemma 29.28.1. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_ s, x}$ has a kernel (as $f(\eta ) \not= y$) and since $\mathcal{O}_{X, x}$ is a Noetherian domain we see that $\dim (\mathcal{O}_{X, x}) > \dim (\mathcal{O}_{X_ y, x})$. We conclude that

$\dim _ x(X_ s) = \dim (\mathcal{O}_{X_ s, x}) + \text{trdeg}_{\kappa (y)} \kappa (x) < \dim (B) + d$

(equality by Morphisms, Lemma 29.28.1) which proves what we want. $\square$

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