About transcendence degree of an affine $K$-domain Exercise 5.6 (G. Kemper)
If $A$ is an affine $K$-domain, then the transcendence degree of $A$ is the size
of a maximal algebraically independent subset of $A$.
Kemper's Proof:
Let $T\subseteq A$ be a maximal algebraically independent subset.
Then $T$ is a maximal algebraically independent subset of $Quot(A)$, so it
is a transcendence basis. Since any two transcendence basis has the same size, then $trdeg(Quot(A))=|T|=trdeg(A)$.
I'm trying to proof (unsuccessfully) the affirmation "$T$ is a maximal algebraically independent subset of $Quot(A)$" because it's not obvious to me,
but a more elementary question arrises (I'm learning):
when we say "maximal" we mean among all other algebraically independent subsets or it is related to the size of the set? For example, the set $\{\overline{x_1}\}$ is a maximal algebraically independent subset of
$K[x_1,x_2,x_3]/(x_1x_2,x_1x_3)$, but is not "maximal in size", since $\{\overline{x_2},\overline{x_3}\}$ is algebraically independent too.
If someone could give an explanation about these terms and proof of the affirmation I'm really glad. Thanks!
 A: Maximal in this context always means with respect to inclusion, not size. Formally, a maximal element of a poset $(P, \le)$ is an element $m$ such that if $m \le n$ then $m = n$ ("maximal with respect to $\le$"). A poset can have many maximal elements and maximal elements of a collection of subsets ordered with respect to inclusion can have many different sizes.
To show that $T$ is a maximal algebraically independent subset of $\text{Quot}(A)$ we can argue as follows. Let $t = \frac{f}{g} \in \text{Quot}(A)$ be arbitrary, where $f, g \in A$. By maximality $T \cup \{ f \}$ is algebraically dependent in $A$, so there exists some polynomial relation
$$\sum a_n f^n = 0$$
where $a_n \in k[T] \subseteq A$. Similarly $T \cup \{ g \}$ is algebraically dependent in $A$, so there exists some polynomial relation
$$\sum b_n g^n = 0$$
where $b_n \in k[T]$. So $\frac{f}{g}$ is a quotient of two elements algebraic over $\text{Quot}(k[T])$ and hence is itself algebraic over $\text{Quot}(k[T])$, meaning there exists some polynomial relation $\sum c_n t^n = 0$ where $c_n \in \text{Quot}(k[T])$. This means $T \cup \{ t \}$ is algebraically dependent as desired, for any $t$. So $T$ is maximal.
