# $\alpha$ is algebraic if and only if $\alpha^2$ is algebraic

We have a field extension $K/k$ and we're asked to prove that $\alpha \in K$ is algebraic over $k$ $\iff$ $\alpha^2$ is algebraic over $k$.

I've got an argument for the left arrow $\implies$, but I'm not sure how to proceed with $\impliedby$. Do we need to manipulate a certain polynomial or is it a linear algebra argument, perhaps?

Any help would be appreciated!

If $\alpha^2$ is a root of $P(x)$ then $\alpha$ is root of $P(x^2)$.
We have $\alpha^2 \in k[\alpha]$ where $k[\alpha]$ is the field where $\alpha$ is adjoined to $k$. Hence $k[\alpha^2] \subset k[\alpha]$ and we must have
$$[\space k[\alpha^2]:k] \leq [k[\alpha]:k \space].$$
Therefore, if the extension $k[\alpha]/k$ has finite degree, also the extension $k[\alpha^2]/k$ has finite degree. And finite degree means algebraic.