## 29.50 The dimension formula

For morphisms between Noetherian schemes we can say a little more about dimensions of local rings. Here is an important (and not so hard to prove) result. Recall that $R(X)$ denotes the function field of an integral scheme $X$.

Lemma 29.50.1. Let $S$ be a scheme. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$, and set $s = f(x)$. Assume

1. $S$ is locally Noetherian,

2. $f$ is locally of finite type,

3. $X$ and $S$ integral, and

4. $f$ dominant.

We have

29.50.1.1
$$\label{morphisms-equation-dimension-formula} \dim (\mathcal{O}_{X, x}) \leq \dim (\mathcal{O}_{S, s}) + \text{trdeg}_{R(S)}R(X) - \text{trdeg}_{\kappa (s)} \kappa (x).$$

Moreover, equality holds if $S$ is universally catenary.

Proof. The corresponding algebra statement is Algebra, Lemma 10.112.1. $\square$

Lemma 29.50.2. Let $S$ be a scheme. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$, and set $s = f(x)$. Assume $S$ is locally Noetherian and $f$ is locally of finite type, We have

29.50.2.1
$$\label{morphisms-equation-dimension-formula-general} \dim (\mathcal{O}_{X, x}) \leq \dim (\mathcal{O}_{S, s}) + E - \text{trdeg}_{\kappa (s)} \kappa (x).$$

where $E$ is the maximum of $\text{trdeg}_{\kappa (f(\xi ))}(\kappa (\xi ))$ where $\xi$ runs over the generic points of irreducible components of $X$ containing $x$.

Proof. Let $X_1, \ldots , X_ n$ be the irreducible components of $X$ containing $x$ endowed with their reduced induced scheme structure. These correspond to the minimal primes $\mathfrak q_ i$ of $\mathcal{O}_{X, x}$ and hence there are finitely many of them (Schemes, Lemma 26.13.2 and Algebra, Lemma 10.30.6). Then $\dim (\mathcal{O}_{X, x}) = \max \dim (\mathcal{O}_{X, x}/\mathfrak q_ i) = \max \dim (\mathcal{O}_{X_ i, x})$. The $\xi$'s occurring in the definition of $E$ are exactly the generic points $\xi _ i \in X_ i$. Let $Z_ i = \overline{\{ f(\xi _ i)\} } \subset S$ endowed with the reduced induced scheme structure. The composition $X_ i \to X \to S$ factors through $Z_ i$ (Schemes, Lemma 26.12.7). Thus we may apply the dimension formula (Lemma 29.50.1) to see that $\dim (\mathcal{O}_{X_ i, x}) \leq \dim (\mathcal{O}_{Z_ i, x}) + \text{trdeg}_{\kappa (f(\xi ))}(\kappa (\xi )) - \text{trdeg}_{\kappa (s)} \kappa (x)$. Putting everything together we obtain the lemma. $\square$

An application is the construction of a dimension function on any scheme of finite type over a universally catenary scheme endowed with a dimension function. For the definition of dimension functions, see Topology, Definition 5.20.1.

Lemma 29.50.3. Let $S$ be a locally Noetherian and universally catenary scheme. Let $\delta : S \to \mathbf{Z}$ be a dimension function. Let $f : X \to S$ be a morphism of schemes. Assume $f$ locally of finite type. Then the map

\begin{align*} \delta = \delta _{X/S} : X & \longrightarrow \mathbf{Z} \\ x & \longmapsto \delta (f(x)) + \text{trdeg}_{\kappa (f(x))} \kappa (x) \end{align*}

is a dimension function on $X$.

Proof. Let $f : X \to S$ be locally of finite type. Let $x \leadsto y$, $x \not= y$ be a specialization in $X$. We have to show that $\delta _{X/S}(x) > \delta _{X/S}(y)$ and that $\delta _{X/S}(x) = \delta _{X/S}(y) + 1$ if $y$ is an immediate specialization of $x$.

Choose an affine open $V \subset S$ containing the image of $y$ and choose an affine open $U \subset X$ mapping into $V$ and containing $y$. We may clearly replace $X$ by $U$ and $S$ by $V$. Thus we may assume that $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$ and that $f$ is given by a ring map $R \to A$. The ring $R$ is universally catenary (Lemma 29.16.2) and the map $R \to A$ is of finite type (Lemma 29.14.2).

Let $\mathfrak q \subset A$ be the prime ideal corresponding to the point $x$ and let $\mathfrak p \subset R$ be the prime ideal corresponding to $f(x)$. The restriction $\delta '$ of $\delta$ to $S' = \mathop{\mathrm{Spec}}(R/\mathfrak p) \subset S$ is a dimension function. The ring $R/\mathfrak p$ is universally catenary. The restriction of $\delta _{X/S}$ to $X' = \mathop{\mathrm{Spec}}(A/\mathfrak q)$ is clearly equal to the function $\delta _{X'/S'}$ constructed using the dimension function $\delta '$. Hence we may assume in addition to the above that $R \subset A$ are domains, in other words that $X$ and $S$ are integral schemes, and that $x$ is the generic point of $X$ and $f(x)$ is the generic point of $S$.

Note that $\mathcal{O}_{X, x} = R(X)$ and that since $x \leadsto y$, $x \not= y$, the spectrum of $\mathcal{O}_{X, y}$ has at least two points (Schemes, Lemma 26.13.2) hence $\dim (\mathcal{O}_{X, y}) > 0$ . If $y$ is an immediate specialization of $x$, then $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, y}) = \{ x, y\}$ and $\dim (\mathcal{O}_{X, y}) = 1$.

Write $s = f(x)$ and $t = f(y)$. We compute

\begin{align*} \delta _{X/S}(x) - \delta _{X/S}(y) & = \delta (s) + \text{trdeg}_{\kappa (s)} \kappa (x) - \delta (t) - \text{trdeg}_{\kappa (t)} \kappa (y) \\ & = \delta (s) - \delta (t) + \text{trdeg}_{R(S)} R(X) - \text{trdeg}_{\kappa (t)} \kappa (y) \\ & = \delta (s) - \delta (t) + \dim (\mathcal{O}_{X, y}) - \dim (\mathcal{O}_{S, t}) \end{align*}

where we use equality in (29.50.1.1) in the last step. Since $\delta$ is a dimension function on the scheme $S$ and $s \in S$ is the generic point, the difference $\delta (s) - \delta (t)$ is equal to $\text{codim}(\overline{\{ t\} }, S)$ by Topology, Lemma 5.20.2. This is equal to $\dim (\mathcal{O}_{S, t})$ by Properties, Lemma 28.10.3. Hence we conclude that

$\delta _{X/S}(x) - \delta _{X/S}(y) = \dim (\mathcal{O}_{X, y})$

and the lemma follows from what we said above about $\dim (\mathcal{O}_{X, y})$. $\square$

Another application of the dimension formula is that the dimension does not change under “alterations” (to be defined later).

Lemma 29.50.4. Let $f : X \to Y$ be a morphism of schemes. Assume that

1. $Y$ is locally Noetherian,

2. $X$ and $Y$ are integral schemes,

3. $f$ is dominant, and

4. $f$ is locally of finite type.

Then we have

$\dim (X) \leq \dim (Y) + \text{trdeg}_{R(Y)} R(X).$

If $f$ is closed1 then equality holds.

Proof. Let $f : X \to Y$ be as in the lemma. Let $\xi _0 \leadsto \xi _1 \leadsto \ldots \leadsto \xi _ e$ be a sequence of specializations in $X$. Set $x = \xi _ e$ and $y = f(x)$. Observe that $e \leq \dim (\mathcal{O}_{X, x})$ as the given specializations occur in the spectrum of $\mathcal{O}_{X, x}$, see Schemes, Lemma 26.13.2. By the dimension formula, Lemma 29.50.1, we see that

\begin{align*} e & \leq \dim (\mathcal{O}_{X, x}) \\ & \leq \dim (\mathcal{O}_{Y, y}) + \text{trdeg}_{R(Y)} R(X) - \text{trdeg}_{\kappa (y)} \kappa (x) \\ & \leq \dim (\mathcal{O}_{Y, y}) + \text{trdeg}_{R(Y)} R(X) \end{align*}

Hence we conclude that $e \leq \dim (Y) + \text{trdeg}_{R(Y)} R(X)$ as desired.

Next, assume $f$ is also closed. Say $\overline{\xi }_0 \leadsto \overline{\xi }_1 \leadsto \ldots \leadsto \overline{\xi }_ d$ is a sequence of specializations in $Y$. We want to show that $\dim (X) \geq d + r$. We may assume that $\overline{\xi }_0 = \eta$ is the generic point of $Y$. The generic fibre $X_\eta$ is a scheme locally of finite type over $\kappa (\eta ) = R(Y)$. It is nonempty as $f$ is dominant. Hence by Lemma 29.15.10 it is a Jacobson scheme. Thus by Lemma 29.15.8 we can find a closed point $\xi _0 \in X_\eta$ and the extension $\kappa (\eta ) \subset \kappa (\xi _0)$ is a finite extension. Note that $\mathcal{O}_{X, \xi _0} = \mathcal{O}_{X_\eta , \xi _0}$ because $\eta$ is the generic point of $Y$. Hence we see that $\dim (\mathcal{O}_{X, \xi _0}) = r$ by Lemma 29.50.1 applied to the scheme $X_\eta$ over the universally catenary scheme $\mathop{\mathrm{Spec}}(\kappa (\eta ))$ (see Lemma 29.16.4) and the point $\xi _0$. This means that we can find $\xi _{-r} \leadsto \ldots \leadsto \xi _{-1} \leadsto \xi _0$ in $X$. On the other hand, as $f$ is closed specializations lift along $f$, see Topology, Lemma 5.19.7. Thus, as $\xi _0$ lies over $\eta = \overline{\xi }_0$ we can find specializations $\xi _0 \leadsto \xi _1 \leadsto \ldots \leadsto \xi _ d$ lying over $\overline{\xi }_0 \leadsto \overline{\xi }_1 \leadsto \ldots \leadsto \overline{\xi }_ d$. In other words we have

$\xi _{-r} \leadsto \ldots \leadsto \xi _{-1} \leadsto \xi _0 \leadsto \xi _1 \leadsto \ldots \leadsto \xi _ d$

which means that $\dim (X) \geq d + r$ as desired. $\square$

Lemma 29.50.5. Let $f : X \to Y$ be a morphism of schemes. Assume that $Y$ is locally Noetherian and $f$ is locally of finite type. Then

$\dim (X) \leq \dim (Y) + E$

where $E$ is the supremum of $\text{trdeg}_{\kappa (f(\xi ))}(\kappa (\xi ))$ where $\xi$ runs through the generic points of the irreducible components of $X$.

Proof. Immediate consequence of Lemma 29.50.2 and Properties, Lemma 28.10.2. $\square$

[1] For example if $f$ is proper, see Definition 29.39.1.

Comment #4939 by Antoine Chambert-Loir on

I wonder whether these results do extend when the assumption on the base (locally Noetherian) is replaced by the assumption on the morphism (finite presentation). It looks automatic to me.

I have one example in mind where such a reduction is used — for Lemma 2.4.12 of Maclagan/Sturmfels book, the base is a valuation ring (but they would need some slightly different result, though, because it is not clear a priori that if $I$ is an ideal of $K[T]$ (K, field of fractions of a valuation ring R), then $I \cap R[T]$ is finitely generated ; but they consider a finitely generated subideal of $I\cap R[T]$ which does the job.

Comment #4994 by alexis bouthier on

They do not, take any ring $A$ of finite Krull dimension such that $dim(A[t])\geq dimA +2$

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