Lemma 9.8.5. A finite extension is algebraic. In fact, an extension $E/k$ is algebraic if and only if every subextension $k(\alpha )/k$ generated by some $\alpha \in E$ is finite.

**Proof.**
Let $E/k$ be finite, say of degree $n$. Choose $\alpha \in E$. Then the elements $\{ 1, \alpha , \ldots , \alpha ^ n\} $ are linearly dependent over $E$, or we would necessarily have $[E : k] > n$. A relation of linear dependence now gives the desired polynomial that $\alpha $ must satisfy.

For the last assertion, note that a monogenic extension $k(\alpha )/k$ is finite if and only if $\alpha $ is algebraic over $k$, by Examples 9.7.4 and 9.7.6. So if $E/k$ is algebraic, then each $k(\alpha )/k$, $\alpha \in E$, is a finite extension, and conversely. $\square$

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