Lemma 9.8.8. Let $E/k$ and $F/E$ be algebraic extensions of fields. Then $F/k$ is an algebraic extension of fields.

Proof. Choose $\alpha \in F$. Then $\alpha$ is algebraic over $E$. The key observation is that $\alpha$ is algebraic over a finitely generated subextension of $k$. That is, there is a finite set $S \subset E$ such that $\alpha$ is algebraic over $k(S)$: this is clear because being algebraic means that a certain polynomial in $E[x]$ that $\alpha$ satisfies exists, and as $S$ we can take the coefficients of this polynomial. It follows that $\alpha$ is algebraic over $k(S)$. In particular, the extension $k(S, \alpha )/ k(S)$ is finite. Since $S$ is a finite set, and $k(S)/k$ is algebraic, Lemma 9.8.6 shows that $k(S)/k$ is finite. Using multiplicativity (Lemma 9.7.7) we find that $k(S,\alpha )/k$ is finite, so $\alpha$ is algebraic over $k$. $\square$

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