Lemma 79.9.9. In Situation 79.9.2 assume $s, t$ are locally of finite type. For all $r \in |R|$
$\dim (R) = \dim _ r(R)$,
the transcendence degree of $r$ over $\mathop{\mathrm{Spec}}(k)$ via $s$ equals the transcendence degree of $r$ over $\mathop{\mathrm{Spec}}(k)$ via $t$, and
if the transcendence degree mentioned in (2) is $0$, then $\dim (R) = \dim (\mathcal{O}_{R, \overline{r}})$.
Proof. Let $r \in |R|$. Denote $\text{trdeg}(r/_{\! \! s}k)$ the transcendence degree of $r$ over $\mathop{\mathrm{Spec}}(k)$ via $s$. Choose an étale morphism $\varphi : V \to R$ where $V$ is a scheme and $v \in V$ mapping to $r$. Using the definitions mentioned above the lemma we see that
and similarly for $t$ (the second equality by Morphisms, Lemma 29.28.1). Hence we see that $\text{trdeg}(r/_{\! \! s}k) = \text{trdeg}(r/_{\! \! t}k)$, i.e., (2) holds.
Let $k'/k$ be a field extension. Note that the restriction $R'$ of $R$ to $\mathop{\mathrm{Spec}}(k')$ (see Lemma 79.9.5) is obtained from $R$ by two base changes by morphisms of fields. Thus Morphisms of Spaces, Lemma 67.34.3 shows the dimension of $R$ at a point is unchanged by this operation. Hence in order to prove (1) we may assume, by Lemma 79.9.6, that $r$ is represented by a morphism $r : \mathop{\mathrm{Spec}}(k) \to R$ such that both $s \circ r$ and $t \circ r$ are automorphisms of $\mathop{\mathrm{Spec}}(k)$. In this case there exists an automorphism $R \to R$ which maps $r$ to $e$ (Lemma 79.9.7). Hence we see that $\dim _ r(R) = \dim _ e(R)$ for any $r$. By definition this means that $\dim _ r(R) = \dim (R)$.
Part (3) is a formal consequence of the results obtained in the discussion above. $\square$
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