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!
Best Answer
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.