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Tag 09GK

Chapter 9: Fields > Section 9.8: Algebraic extensions

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 finite product of copies of $F$ has cardinality at most $\max(\aleph_0, |F|)$. Thus so does $E$. $\square$

    The code snippet corresponding to this tag is a part of the file fields.tex and is located in lines 737–741 (see updates for more information).

    \begin{lemma}
    \label{lemma-size-algebraic-extension}
    Let $E/F$ be an algebraic extension of fields. Then the cardinality $|E|$
    of $E$ is at most $\max(\aleph_0, |F|)$.
    \end{lemma}
    
    \begin{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 finite product of copies of $F$ has
    cardinality at most $\max(\aleph_0, |F|)$.
    Thus so does $E$.
    \end{proof}

    Comments (1)

    Comment #2950 by Herman Rohrbach on October 11, 2017 a 11:53 am UTC

    The penultimate sentence reads

    ... 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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