## 10.125 Dimension of fibres

We study the behaviour of dimensions of fibres, using Zariski's main theorem. Recall that we defined the dimension $\dim _ x(X)$ of a topological space $X$ at a point $x$ in Topology, Definition 5.10.1.

Definition 10.125.1. Suppose that $R \to S$ is of finite type, and let $\mathfrak q \subset S$ be a prime lying over a prime $\mathfrak p$ of $R$. We define the relative dimension of $S/R$ at $\mathfrak q$, denoted $\dim _{\mathfrak q}(S/R)$, to be the dimension of $\mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p))$ at the point corresponding to $\mathfrak q$. We let $\dim (S/R)$ be the supremum of $\dim _{\mathfrak q}(S/R)$ over all $\mathfrak q$. This is called the relative dimension of $S/R$.

In particular, $R \to S$ is quasi-finite at $\mathfrak q$ if and only if $\dim _{\mathfrak q}(S/R) = 0$. The following lemma is more or less a reformulation of Zariski's Main Theorem.

Lemma 10.125.2. Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime. Suppose that $\dim _{\mathfrak q}(S/R) = n$. There exists a $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is quasi-finite over a polynomial algebra $R[t_1, \ldots , t_ n]$.

Proof. The ring $\overline{S} = S \otimes _ R \kappa (\mathfrak p)$ is of finite type over $\kappa (\mathfrak p)$. Let $\overline{\mathfrak q}$ be the prime of $\overline{S}$ corresponding to $\mathfrak q$. By definition of the dimension of a topological space at a point there exists an open $U \subset \mathop{\mathrm{Spec}}(\overline{S})$ with $\overline{q} \in U$ and $\dim (U) = n$. Since the topology on $\mathop{\mathrm{Spec}}(\overline{S})$ is induced from the topology on $\mathop{\mathrm{Spec}}(S)$ (see Remark 10.17.8), we can find a $g \in S$, $g \not\in \mathfrak q$ with image $\overline{g} \in \overline{S}$ such that $D(\overline{g}) \subset U$. Thus after replacing $S$ by $S_ g$ we see that $\dim (\overline{S}) = n$.

Next, choose generators $x_1, \ldots , x_ N$ for $S$ as an $R$-algebra. By Lemma 10.115.4 there exist elements $y_1, \ldots , y_ n$ in the $\mathbf{Z}$-subalgebra of $S$ generated by $x_1, \ldots , x_ N$ such that the map $R[t_1, \ldots , t_ n] \to S$, $t_ i \mapsto y_ i$ has the property that $\kappa (\mathfrak p)[t_1\ldots , t_ n] \to \overline{S}$ is finite. In particular, $S$ is quasi-finite over $R[t_1, \ldots , t_ n]$ at $\mathfrak q$. Hence, by Lemma 10.123.13 we may replace $S$ by $S_ g$ for some $g\in S$, $g \not\in \mathfrak q$ such that $R[t_1, \ldots , t_ n] \to S$ is quasi-finite. $\square$

Lemma 10.125.3. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. Assume

1. $R \to S$ is of finite type,

2. $\dim _{\mathfrak q}(S/R) = n$, and

3. $\text{trdeg}_{\kappa (\mathfrak p)}\kappa (\mathfrak q) = r$.

Then there exist $f \in R$, $f \not\in \mathfrak p$, $g \in S$, $g \not\in \mathfrak q$ and a quasi-finite ring map

$\varphi : R_ f[x_1, \ldots , x_ n] \longrightarrow S_ g$

such that $\varphi ^{-1}(\mathfrak qS_ g) = (\mathfrak p, x_{r + 1}, \ldots , x_ n)R_ f[x_{r + 1}, \ldots , x_ n]$

Proof. After replacing $S$ by a principal localization we may assume there exists a quasi-finite ring map $\varphi : R[t_1, \ldots , t_ n] \to S$, see Lemma 10.125.2. Set $\mathfrak q' = \varphi ^{-1}(\mathfrak q)$. Let $\overline{\mathfrak q}' \subset \kappa (\mathfrak p)[t_1, \ldots , t_ n]$ be the prime corresponding to $\mathfrak q'$. By Lemma 10.115.6 there exists a finite ring map $\kappa (\mathfrak p)[x_1, \ldots , x_ n] \to \kappa (\mathfrak p)[t_1, \ldots , t_ n]$ such that the inverse image of $\overline{\mathfrak q}'$ is $(x_{r + 1}, \ldots , x_ n)$. Let $\overline{h}_ i \in \kappa (\mathfrak p)[t_1, \ldots , t_ n]$ be the image of $x_ i$. We can find an element $f \in R$, $f \not\in \mathfrak p$ and $h_ i \in R_ f[t_1, \ldots , t_ n]$ which map to $\overline{h}_ i$ in $\kappa (\mathfrak p)[t_1, \ldots , t_ n]$. Then the ring map

$R_ f[x_1, \ldots , x_ n] \longrightarrow R_ f[t_1, \ldots , t_ n]$

becomes finite after tensoring with $\kappa (\mathfrak p)$. In particular, $R_ f[t_1, \ldots , t_ n]$ is quasi-finite over $R_ f[x_1, \ldots , x_ n]$ at the prime $\mathfrak q'R_ f[t_1, \ldots , t_ n]$. Hence, by Lemma 10.123.13 there exists a $g \in R_ f[t_1, \ldots , t_ n]$, $g \not\in \mathfrak q'R_ f[t_1, \ldots , t_ n]$ such that $R_ f[x_1, \ldots , x_ n] \to R_ f[t_1, \ldots , t_ n, 1/g]$ is quasi-finite. Thus we see that the composition

$R_ f[x_1, \ldots , x_ n] \longrightarrow R_ f[t_1, \ldots , t_ n, 1/g] \longrightarrow S_{\varphi (g)}$

is quasi-finite and we win. $\square$

Lemma 10.125.4. Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. If $R \to S$ is quasi-finite at $\mathfrak q$, then $\dim (S_{\mathfrak q}) \leq \dim (R_{\mathfrak p})$.

Proof. If $R_{\mathfrak p}$ is Noetherian (and hence $S_{\mathfrak q}$ Noetherian since it is essentially of finite type over $R_{\mathfrak p}$) then this follows immediately from Lemma 10.112.6 and the definitions. In the general case, let $S'$ be the integral closure of $R_\mathfrak p$ in $S_\mathfrak p$. By Zariski's Main Theorem 10.123.12 we have $S_{\mathfrak q} = S'_{\mathfrak q'}$ for some $\mathfrak q' \subset S'$ lying over $\mathfrak q$. By Lemma 10.112.3 we have $\dim (S') \leq \dim (R_\mathfrak p)$ and hence a fortiori $\dim (S_\mathfrak q) = \dim (S'_{\mathfrak q'}) \leq \dim (R_\mathfrak p)$. $\square$

Lemma 10.125.5. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Suppose there is a quasi-finite $k$-algebra map $k[t_1, \ldots , t_ n] \subset S$. Then $\dim (S) \leq n$.

Proof. By Lemma 10.114.1 the dimension of any local ring of $k[t_1, \ldots , t_ n]$ is at most $n$. Thus the result follows from Lemma 10.125.4. $\square$

Lemma 10.125.6. Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime. Suppose that $\dim _{\mathfrak q}(S/R) = n$. There exists an open neighbourhood $V$ of $\mathfrak q$ in $\mathop{\mathrm{Spec}}(S)$ such that $\dim _{\mathfrak q'}(S/R) \leq n$ for all $\mathfrak q' \in V$.

Proof. By Lemma 10.125.2 we see that we may assume that $S$ is quasi-finite over a polynomial algebra $R[t_1, \ldots , t_ n]$. Considering the fibres, we reduce to Lemma 10.125.5. $\square$

In other words, the lemma says that the set of points where the fibre has dimension $\leq n$ is open in $\mathop{\mathrm{Spec}}(S)$. The next lemma says that formation of this open commutes with base change. If the ring map is of finite presentation then this set is quasi-compact open (see below).

Lemma 10.125.7. Let $R \to S$ be a finite type ring map. Let $R \to R'$ be any ring map. Set $S' = R' \otimes _ R S$ and denote $f : \mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$ the associated map on spectra. Let $n \geq 0$. The inverse image $f^{-1}(\{ \mathfrak q \in \mathop{\mathrm{Spec}}(S) \mid \dim _{\mathfrak q}(S/R) \leq n\} )$ is equal to $\{ \mathfrak q' \in \mathop{\mathrm{Spec}}(S') \mid \dim _{\mathfrak q'}(S'/R') \leq n\}$.

Proof. The condition is formulated in terms of dimensions of fibre rings which are of finite type over a field. Combined with Lemma 10.116.6 this yields the lemma. $\square$

Lemma 10.125.8. Let $R \to S$ be a ring homomorphism of finite presentation. Let $n \geq 0$. The set

$V_ n = \{ \mathfrak q \in \mathop{\mathrm{Spec}}(S) \mid \dim _{\mathfrak q}(S/R) \leq n\}$

is a quasi-compact open subset of $\mathop{\mathrm{Spec}}(S)$.

Proof. It is open by Lemma 10.125.6. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ be a presentation of $S$. Let $R_0$ be the $\mathbf{Z}$-subalgebra of $R$ generated by the coefficients of the polynomials $f_ i$. Let $S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Then $S = R \otimes _{R_0} S_0$. By Lemma 10.125.7 $V_ n$ is the inverse image of an open $V_{0, n}$ under the quasi-compact continuous map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S_0)$. Since $S_0$ is Noetherian we see that $V_{0, n}$ is quasi-compact. $\square$

Lemma 10.125.9. Let $R$ be a valuation ring with residue field $k$ and field of fractions $K$. Let $S$ be a domain containing $R$ such that $S$ is of finite type over $R$. If $S \otimes _ R k$ is not the zero ring then

$\dim (S \otimes _ R k) = \dim (S \otimes _ R K)$

In fact, $\mathop{\mathrm{Spec}}(S \otimes _ R k)$ is equidimensional.

Proof. It suffices to show that $\dim _{\mathfrak q}(S/k)$ is equal to $\dim (S \otimes _ R K)$ for every prime $\mathfrak q$ of $S$ containing $\mathfrak m_ RS$. Pick such a prime. By Lemma 10.125.6 the inequality $\dim _{\mathfrak q}(S/k) \geq \dim (S \otimes _ R K)$ holds. Set $n = \dim _{\mathfrak q}(S/k)$. By Lemma 10.125.2 after replacing $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ there exists a quasi-finite ring map $R[t_1, \ldots , t_ n] \to S$. If $\dim (S \otimes _ R K) < n$, then $K[t_1, \ldots , t_ n] \to S \otimes _ R K$ has a nonzero kernel. Say $f = \sum a_ I t_1^{i_1}\ldots t_ n^{i_ n}$. After dividing $f$ by a nonzero coefficient of $f$ with minimal valuation, we may assume $f\in R[t_1, \ldots , t_ n]$ and some $a_ I$ does not map to zero in $k$. Hence the ring map $k[t_1, \ldots , t_ n] \to S \otimes _ R k$ has a nonzero kernel which implies that $\dim (S \otimes _ R k) < n$. Contradiction. $\square$

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