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If $a\in\mathbb{C}$ is algebraic, then $\mathbb{Q}(a)=\mathbb{Q}[a]$, and the converse holds too. I am having trouble proving that the same holds for several elements. Is it true that $a_1,\ldots, a_n$ are algebraic iff $\mathbb{Q}(a_1,\ldots, a_n)=\mathbb{Q}[a_1,\ldots, a_n]$?

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  • $\begingroup$ Yes. The key observation is that $F[a]=F(a)$ when $F$ is a field and $a$ is algebraic over $F$. For the converse, use a dimension argument. $\endgroup$ – Brett Frankel Mar 4 '13 at 0:59
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Try an inductive argument and observe that

$$\Bbb F(a_1,...,a_n)=\Bbb F(a_1,...,a_{n-1})(a_n)=\Bbb F(a_1,...,a_{n-1})[a_n]\Longleftrightarrow a_n$$

is algebraic over $\,\Bbb F(a_1,...,a_{n-1})\,$

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