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 $E$. $\square$

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 $F^{(\mathbb{N})}$

Comment #9027 by João Candeias on

I think this is still not quite correct, as a countable product of countables is not countable (already a countable product of finite sets isn't countable), so in particular if we take $F=\mathbb{Q}$, we see that a countable product of copies of $F$ has bigger cadinality than that of $F$ (which is equal to $\aleph_0$). Hence this is not a valid argument.

However, we are only interested in a subset of $F^{\mathbb{N}}$, which can be obtained as a countable union (all possible degrees $n$) of finite products (all possible $n+1$ coefficients of the polynomial) of $F$, and this is indeed bounded by $\max{(\aleph_0,|F|)}$.

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