
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 the cardinality of a countable product of copies of $F$ has cardinality at most $\max (\aleph _0, |F|)$. Thus so does $E$. $\square$

Comment #2950 by Herman Rohrbach on

... becaue the cardinality of a finite product of copies of $F$...

I think this should be

... because the cardinality of a countable product of copies of $F$...,

since the set of nonconstant polynomials with coefficients in $F$ is a subset of $F^{(\mathbb{N})}$

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