My friend's assignment has the following question:
Let $F$ be a field and $E=F(\alpha)$, where $\alpha\notin F$ and $\alpha^2 \in F$. Find the set of all $\beta \in E$ such that $\beta^2 \in F$.
This seems to be really easy:
Solution. Write $\beta = a+b\alpha$, obtain $\beta^2 = (a^2+b^2\alpha^2) + 2ab\alpha,$ deduce $2ab = 0$, hence $a = 0$ or $b = 0$. So the only elements with this property are $F \cup \alpha F$.
This seems a bit too easy though, considering the difficulty of the other questions on the assignment. Are there any mistakes in the above argument that make it incorrect?
Your reasoning is correct; indeed any element $\beta$ in $F(\alpha)$ can be written as $\beta=a+b\alpha $ with $a,b \in F$. As you have already noticed $\beta^2=(a^2+b^2\alpha^2)+2ab\alpha$ and since $\alpha^2\in F$ you may deduce that $\beta^2\in F$ if and only if $2ab\alpha=0$. But now there is a little casistic.
Suppose $\textsf{char}F=2$, then $2ab\alpha=0$ for any $a,b\in F$; hence any $\beta\in E$ is such that $\beta^2\in F$.
Otherwise $\textsf{char}F\neq2$, in such cases $\beta^2\in E$ if and only if $ab=0$, hence $a=0$ or $b=0$.