Lemma 9.8.7. Let $E/k$ be a field extension. Then the elements of $E$ algebraic over $k$ form a subextension of $E/k$.

**Proof.**
Let $\alpha , \beta \in E$ be algebraic over $k$. Then $k(\alpha , \beta )/k$ is a finite extension by Lemma 9.8.6. It follows that $k(\alpha + \beta ) \subset k(\alpha , \beta )$ is a finite extension, which implies that $\alpha + \beta $ is algebraic by Lemma 9.8.5. Similarly for the difference, product and quotient of $\alpha $ and $\beta $.
$\square$

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