Transcendence degree of integral domain 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? 
 A: 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.
