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This question really has me stumped.

We define $\bar{\mathbb{Q}}$ to be the set of elements in $\mathbb{C}$ that are algebraic over $\mathbb{Q}$. I have shown that this is a field.

Now, I'm assuming that we have some element $b \in \mathbb{C}$ that is algebraic over $\bar{\mathbb{Q}}$. I want to show $b \in \bar{\mathbb{Q}}$.

My attempt:

Since $b$ is algebraic over $\bar{\mathbb{Q}}$, we know the extension $[\bar{\mathbb{Q}}(b) : \bar{\mathbb{Q}}] = n < \infty$. Also, there's a polynomial $f \in \bar{\mathbb{Q}}[x]$ s.t $f(b) = 0$. So, I've let $f = x^n + a_{n-1} x^{n-1} + ... + a_0$, where the $a_i \in \bar{\mathbb{Q}}$. In particular, we know that these $a_i$ are algebraic over $\mathbb{Q}$.

Next, we consider the extension $[\mathbb{Q}(a_0, a_1, ... , a_{n-1}) : \mathbb{Q}]$. Since all the $a_i$ are algebraic over $\mathbb{Q}$, this extension is finite.

Now, we consider $[\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}]$. By Tower Law, this equals $[\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}(b)] \times [\mathbb{Q}(b) : \mathbb{Q}]$. Both of these are finite extensions, so $[\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}]$ is finite.

And this is where I get stuck. I'm not quite sure where to go from here and I just feel kinda confused. I'm not sure if this is even taking me towards the answer, but I'm not sure what to do. Any nudges would be great!

EDIT: Corrected typos listed below.

Mystery_Jay
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    You have shown that $\beta$ (later called $b$) is algebraic over $\Bbb Q$ (because contained in a finite degree extension). Now look at what you wanted to prove in the first place. – Marc van Leeuwen Oct 29 '14 at 20:46
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    If an extension is finite, then every element of the extension is algebraic on the base field. So $b$ is algebraic over $\mathbb{Q}$. You're missing a bar where you say $[\bar{\mathbb{Q}}(b):\mathbb{Q}]<n$ (it should be $[\bar{\mathbb{Q}}(b):\bar{\mathbb{Q}}]<n$). – egreg Oct 29 '14 at 20:46
  • Thanks all; that seems to have helped. And thanks @HenningMakholm - didn't realise about the other thread. :) – Mystery_Jay Oct 29 '14 at 20:50

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