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
\dim _ r(R) = \dim _ v(V) = \dim (\mathcal{O}_{V, v}) + \text{trdeg}_{s(k)}(\kappa (v)) = \dim (\mathcal{O}_{R, \overline{r}}) + \text{trdeg}(r/_{\! \! s}k)
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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