Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{Q}(a_1,\ldots, a_n)=\mathbb{Q}[a_1,\ldots, a_n]$?

share|cite|improve this question
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. – Brett Frankel Mar 4 '13 at 0:59

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.