I was reading the proof of Prop. 2.54
Let $A$ be an integral domain, $k \subseteq A$ a field contained in $A$. Then Milne claims,
$$\deg tr_k A = \deg tr_{\bar{k}} A $$ here $\bar{k}$ denotes the algebraic closure of $k$ in $A$.
I have two confusions:
(i) How does $\bar{k}$ exist?
(ii) Why are the degrees still the same?
The proof of (1) is the same as the proof that an algebraic closure exists for fields in general, but actually easier since we can avoid some set-theoretic issues. Consider the collection of all algebraic field extensions $F_\alpha$ of $k$ inside $A$, that is $k\subset F_\alpha\subset A$. Then for any chain of such extensions, their union is again an algebraic field extension of $k$ inside $A$, and therefore we may apply Zorn's lemma to find a maximal element $F$. If this maximal element is not algebraically closed, then we may find $\beta\in A$ such that $\beta\in A$ is algebraic over $k$ but $\beta\notin F$. But then $F(\beta)$ is a proper extension of $F$, contradicting it's status as a maximal element. So $F=\overline{k}$.
The proof of (2) is that we have an inclusion of fields $k\subset \overline{k}\subset Frac(A)$ and we know that transcendence degree adds over compositions of field extensions. As $k\subset \overline{k}$ is algebraic, it has transcendence degree zero, so the transcendence degrees of the extensions $k\subset Frac(A)$ and $\overline{k}\subset Frac(A)$ are the same.