Lemma 9.8.9. Let E/F be an algebraic extension of fields. Then the cardinality |E| of E is at most \max (\aleph _0, |F|).
Proof. Let S be the set of nonconstant polynomials with coefficients in F. For every P \in S the set of roots r(P, E) = \{ \alpha \in E \mid P(\alpha ) = 0\} is finite (details omitted). Moreover, the fact that E is algebraic over F implies that E = \bigcup _{P \in S} r(P, E). It is clear that S has cardinality bounded by \max (\aleph _0, |F|) because it is a countable union of finite products of copies of F. Thus so does E. \square
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