A couple of weeks ago, we were given the following theorem in class:
Let $\alpha \in \bar{\mathbb{Q}}$ and $K = \mathbb{Q}(\alpha)$. Then $\bar{K} = \bar{\mathbb{Q}}$.
I can see that $\bar{\mathbb{Q}} \subseteq \bar{K}$ since $\mathbb{Q}\subseteq K$. But I am having trouble proving $\bar{K} \subseteq \bar{\mathbb{Q}}$. So if $\beta \in \bar{K}$, then there is a $f \in K[t]$ such that $f(\beta) = 0$. Must $m(t)$ (the minimal polynomial of $\alpha$ in $\mathbb{Q}$) divide $f$? If so, can we say the quotient $\frac{f(t)}{m(t)}$ is the polynomial over $\mathbb{Q}$ that we are looking for? I.e. $\frac{f(\beta)}{m(\beta)} = 0$. (Must this quotient be in $\mathbb{Q}[t]$?)
Another characterisation of algebraic elements over field $K$ is that $K(\beta)$ has finite dimension as a $K$-vector space. If $\dim_K K(\beta)=n$ and $\dim_{\mathbf Q}K=p$, then $$\dim_{\mathbf Q}(K(\beta))=np$$ Now $\mathbf Q(\beta)$ is a $\mathbf Q$-subspace of $K(\beta)$, hence it has finite dimension, which proves $\beta $ is algebraic over $\mathbf Q$.