## 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)

## Add a comment on tag `09GK`

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.