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Let $L$ and $K$ two fields such that $K \subset L$. Let $a,b \in L$ be algebraic over $K$.

Show that $K[a,b]$ (the smallest ring that contains $K$, $a$ and $b$) is a field.

I have shown that $\{x \in L \mid x \text{ is algebraic over } K\}$ is a field, but now I'm stuck.

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Hint: First prove $K[\alpha]=K(\alpha)$, then note that $K[\alpha,\beta]=(K[\alpha])[\beta]$ and that $\beta$ is also algebraic over $K[\alpha]=K(\alpha)$.

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Thanks for the hint. $K[a]=K(a)$ because a in algebraic on $k$... Then because $b$ is algebraic on $K$ it is also on $K(a)$ and then $(K(a))[b]=(K(a))(b)$. Hence: $K[a,b]=(K[a])([b])=(K(a))[b]=(K(a))(b)=K(a,b)$ It's right? – Madara Jan 4 '13 at 10:34
@Madara - Yes, exactly what I had in mind – Belgi Jan 4 '13 at 10:35

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