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 k, 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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