Lemma 9.26.3. Let E/F be a field extension. A transcendence basis of E over F exists. Any two transcendence bases have the same cardinality.
Proof. Let A be an algebraically independent subset of E. Let G be a subset of E containing A that generates E/F. We claim we can find a transcendence basis B such that A \subset B \subset G. To prove this, consider the collection \mathcal{B} of algebraically independent subsets whose members are subsets of G that contain A. Define a partial ordering on \mathcal{B} using inclusion. Then \mathcal{B} contains at least one element A. The union of the elements of a totally ordered subset T of \mathcal{B} is an algebraically independent subset of E over F since any algebraic dependence relation would have occurred in one of the elements of T (since polynomials only involve finitely many variables). The union also contains A and is contained in G. By Zorn's lemma, there is a maximal element B \in \mathcal{B}. Now we claim E is algebraic over F(B). This is because if it wasn't then there would be an element f \in G transcendental over F(B) since F(G) = E. Then B \cup \{ f\} would be algebraically independent contradicting the maximality of B. Thus B is our transcendence basis.
Let B and B' be two transcendence bases. Without loss of generality, we can assume that |B'| \leq |B|. Now we divide the proof into two cases: the first case is that B is an infinite set. Then for each \alpha \in B', there is a finite set B_{\alpha } \subset B such that \alpha is algebraic over F(B_{\alpha }) since any algebraic dependence relation only uses finitely many indeterminates. Then we define B^* = \bigcup _{\alpha \in B'} B_{\alpha }. By construction, B^* \subset B, but we claim that in fact the two sets are equal. To see this, suppose that they are not equal, say there is an element \beta \in B \setminus B^*. We know \beta is algebraic over F(B') which is algebraic over F(B^*). Therefore \beta is algebraic over F(B^*), a contradiction. So |B| \leq |\bigcup _{\alpha \in B'} B_{\alpha }|. Now if B' is finite, then so is B so we can assume B' is infinite; this means
because each B_\alpha is finite and B' is infinite. Therefore in the infinite case, |B| = |B'|.
Now we need to look at the case where B is finite. In this case, B' is also finite, so suppose B = \{ \alpha _1, \ldots , \alpha _ n\} and B' = \{ \beta _1, \ldots , \beta _ m\} with m \leq n. We perform induction on m: if m = 0 then E/F is algebraic so B = \emptyset so n = 0. If m > 0, there is an irreducible polynomial f \in F[x, y_1, \ldots , y_ n] such that f(\beta _1, \alpha _1, \ldots , \alpha _ n) = 0 and such that x occurs in f. Since \beta _1 is not algebraic over F, f must involve some y_ i so without loss of generality, assume f uses y_1. Let B^* = \{ \beta _1, \alpha _2, \ldots , \alpha _ n\} . We claim that B^* is a basis for E/F. To prove this claim, we see that we have a tower of algebraic extensions
since \alpha _1 is algebraic over F(B^*). Now we claim that B^* (counting multiplicity of elements) is algebraically independent over F because if it weren't, then there would be an irreducible g\in F[x, y_2, \ldots , y_ n] such that g(\beta _1, \alpha _2, \ldots , \alpha _ n) = 0 which must involve x making \beta _1 algebraic over F(\alpha _2, \ldots , \alpha _ n) which would make \alpha _1 algebraic over F(\alpha _2, \ldots , \alpha _ n) which is impossible. So this means that \{ \alpha _2, \ldots , \alpha _ n\} and \{ \beta _2, \ldots , \beta _ m\} are bases for E over F(\beta _1) which means by induction, m = n. \square
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