Past exam of the logic and set theory course of the University of Oxford / 2011 / B / 6 / c

For the standard definition of algebraic independence, finite tuples are required to have distinct entries. Let $\mathcal{P}$ be the set of all subsets of $L$ with algebraic independence over $K$, ordered by inclusion. It is nonempty because $\varnothing \in \mathcal{P}$.

If $\mathcal{C}$ is a chain in $\mathcal{P}$, put $U=\bigcup \mathcal{C}$. A finite tuple of distinct elements of $U$ lies in one member of $\mathcal{C}$: choose a member containing each element and take the largest of those finitely many members. A polynomial relation on that tuple would then be a polynomial relation in that member, contradicting its algebraic independence. Thus $U\in \mathcal{P}$ and is an upper bound. For the empty chain, $\varnothing$ is an upper bound. The Zorn's lemma now gives a maximal member $B$, which is a transcendence basis by the specified maximality criterion. To also recover the usual field-theoretic definition, if $a\in L$ were transcendental over $K(B)$, clearing denominators in a polynomial relation would show that $B\cup \{a\}$ still has algebraic independence over $K$, contradicting maximality. Thus $L$ is algebraic over $K(B)$. Hence every field extension has a transcendence basis.

There is a literal issue in the printed definition: it does not require the tuple entries to be distinct. With that wording, any nonempty $X$ fails the test: for $a\in X$, the nonzero polynomial $X_1-X_2$ vanishes on $(a,a)$, in every characteristic. Thus under the literal printed test the family is exactly $\{\varnothing \}$, and $\varnothing$ is its maximal member. The same chain-and-Zorn's lemma argument still proves existence for that literal notion, but it is not the usual transcendence basis notion. The conventional result above uses distinct tuple entries.