Surjective ring maps always induce isomorphic residue field? Context. I am trying to understand why 10.115.4 follows from 10.115.3. I believe it boils down to the following.

Let $(S',m') \rightarrow (S,m)$ be a surjection of finite type local $k$ algebras. Then we have induced isomoprhism of residue fields
$$ \kappa(m') \rightarrow \kappa(m) $$

But this seems to be true more general: that we do note need the condition of finite type local $k$ algebras. I believe that

If $S' \rightarrow S$ is a surjective ring map, then for all prime $p$ of $S$ with preimage $p'$, we have induced isomorphism of residue fields.
$$ \kappa(p')\rightarrow \kappa(p)$$

The argument is simply because the induced map is surjective, and any field homomorphism is either $0$ or injective.
Am I correct?
 A: You are correct when you claim that the following is true.
Proposition: Let $S'\to S$ be a surjective ring homomorphism, let $\mathfrak{p}$ be a prime ideal of $S,$ and let $\mathfrak{p}'$ be its preimage in $S'.$ Then the induced map $\kappa(\mathfrak{p}')\to\kappa(\mathfrak{p})$ is an isomorphism.
And you are also correct that the implication 10.115.3$\implies$10.115.4 boils down to this claim, as 10.115.3 implies that
\begin{align*}
\dim_{x'}X' - \dim_x X &= (\dim(S'_{\mathfrak{p}'}) + \operatorname{trdeg}_k(\kappa(\mathfrak{p}'))) - (\dim(S_{\mathfrak{p}}) + \operatorname{trdeg}_k(\kappa(\mathfrak{p})))\\
&= \operatorname{height}(\mathfrak{p}') - \operatorname{height}(\mathfrak{p}) + \operatorname{trdeg}_k(\kappa(\mathfrak{p}')) - \operatorname{trdeg}_k(\kappa(\mathfrak{p})),
\end{align*}
so that all you need to prove is that $\kappa(\mathfrak{p}')\to\kappa(\mathfrak{p})$ is an isomorphism.
However, the requirement that $S'$ and $S$ are finite type $k$-algebras is hidden in the proof of 10.115.3. To prove 10.115.3, they invoke 10.113.5, which in turn relies on 10.113.4:
Lemma 10.113.4: Let $k$ be a field. Let $S$ be a finite type $k$-algebra which is an integral domain. Then $\dim(S)=\dim(S_\mathfrak{m})$ for any maximal ideal $\mathfrak{m}$ of $S$. In words: every maximal chain of primes has length equal to the dimension of $S$.
The proof of this lemma requires us to write $S\cong k[x_1,\dots, x_n]/\mathfrak{p}.$ While I do not have a counterexample to the statement of lemma 10.113.4 if we drop the assumption that $S$ is finite type at the moment, I would imagine that odd things can happen and this could fail.
