Show if $\alpha$ is algebraic over $F$, then $\alpha ^2$ is algebraic over $F$

Question

I am asked to show that if $\alpha$ is algebraic over $F$, then $\alpha ^2$ is algebraic over $F$. I was considering the $b_0\alpha_0 +...+ b_n\alpha_n^n = 0$, but I don't know how to connect this with $\alpha^2$ being a root of some polynomial.

Answer 1

$\alpha \in F(\alpha) \Rightarrow \alpha^2 \in F(\alpha) \because F(\alpha)$ is a field.

Hence $F \subset F(\alpha)$ and $\alpha^2 \in F(\alpha).$

This implies $F(\alpha^2) \subset F(\alpha) \because F(\alpha^2)$ is the minimal field which contains both $F$ and $\alpha^2$.

Since $\alpha$ is algebraic over $F$, $[F(\alpha):F]$ is finite.

We have $[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F].$

Now if $[F(\alpha^2):F]$ were infinite, then $[F(\alpha):F]$ would also be infinite which would give us contradiction. Hence $[F(\alpha^2):F]$ is finite $\Rightarrow \alpha^2$ is algebraic over $F$.