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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.

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    What can you say about $F(\alpha):F$? And about $F(\alpha^2):F(\alpha)$? – 2017-02-08
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    They are finite. but, how do you show the second extension is finite? – 2017-02-08
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    $F(\alpha)$ is a field, right?... – 2017-02-08
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    but $F(\alpha^2)$ is not a extension field of $F(\alpha)$ right? – 2017-02-08
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    Right, but $F(\alpha)$ is an extension of $F(\alpha^2)$. – 2017-02-08
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    Indeed, and the same idea can show the converse ($\alpha^2$ algebraic implies $\alpha$ algebraic). – 2017-02-08

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