Transcendence basis as subset of generators

Question

Let $K \subset L = K(a_1,...,a_n)$ be a field extension finitely generated as $K$-algebra with transcendence degree $\operatorname{Trdeg}_K(L):= m \le n$. It is well known that the choice of a transcendence basis is hightly not unique. The question is why it is always possible to find a transcendence basis which is a subset of the set of generators $\{a_1,...,a_n\}$ of $L$?

Answer 1

Note that if the set $a_1,\dots,a_n$ was algebraically independent that set already forms a transcendence basis for $L/K$. If not then there is some non-trivial polynomial $p$ over $K$ with $$p(a_1,\dots,a_n)=0$$ At least one of the $a_i$ has to appear on the left side. Wlog assume that $a_n$ appears there. Then $a_n$, and therefore also $L$, is algebraic over $K(a_1,\dots,a_{n-1})$. Now continue in the same way until you are left with algebraically independent elements $a_1,\dots,a_{n-i}$, so that the other elements $a_{n-i+1},\dots,a_n$ are algebraic over $K(a_1,\dots,a_{n-i})$. Then $\{a_1,\dots,a_{n-i}\}$ will be your transcendence base.
Note that this argument heavily relied on the fact that we have a finite set (otherwise the process might not stop). Here is another argument (which is actually not that different) which also works in the infinite case:
Let $S$ be the set of all algebraically independent subsets of $A:=\{a_1,\dots,a_n\}$. Note that this set is not empty as $\emptyset\in S$. As $S$ is finite $S$ has a maximal element $B$ with respect to the inclusion (in the infinite case we would need to invoke Zorn's lemma here to get the existence of a maximal element). I claim that this $B$ is a transcendence base for $L/K$. By definition $B$ is algebraically independent. Now for any $a\in A\setminus B$ by maximality of $B$ the set $B\cup \{a\}$ is algebraically independent. It follows that there is a non-trivial polynomial $p$ over $K$ with $$p(B,a)=0$$ As $B$ is algebraically independent $a$ has to appear on the left side, so $a$ is algebraic over $K(B)$. It follows that $L=K(A)$ is algebraic over $K(B)$, so $B$ is indeed a transcendence base.

You might see that the reasoning above is very similar to that in linear algebra when dealing with bases of vector spaces. The principle that is behind these two things is the notion of a Dependence relation.