I was reading the following solution here Prove that $E=F[\alpha^2]$ and I was not sure how is this statement in a solution is correct? "since $\alpha$ is a root of $x^2-\alpha^2$ in$F[\alpha^2][x],\dots$ " as far as I know that $\alpha \notin F[\alpha^2],$ could someone clarify this to me please? Why can not we say the $x^2 - \alpha^2$ is a minimal polynomial of $\alpha$ in $F[\alpha^2]$?
Also, is there any way that the degree of an extension be zero? And why is the extension of a field by itself has degree one?
Any clarification will be greatly appreciated!
Edit:
Here is the full answer written there:
Since $\alpha^2\in F[\alpha]$, $F[\alpha^2]\subseteq F[\alpha]$. Thus $[F[\alpha]:F]=[F[\alpha]:F[\alpha^2]]*[F[\alpha^2]:F]$ since $\alpha$ is a root of $x^2-\alpha^2$ in$F[\alpha^2][x]$, the extension $[F[\alpha]:F[\alpha^2]]\leq 2$. Thus it must be 1 since the total extension is of odd degree showing that $F[\alpha]=F[\alpha^2]$.
To show that $\big[F[\alpha] : F[\alpha^2]\big] \leq 2$, it suffices to show that $\alpha$ is a root of a quadratic polynomial with coefficients in $F[\alpha^2]$. But now $T^2 - \alpha^2 \in F[\alpha^2][T]$ is appropriate.
For the second question, if you have a field extension $K \to L$ such that $[L : K] = 0$, it basically means that $L$ is the zero vector space over $K$, so $L = 0$ and $K = 0$ (because $L$ contains a copy of $K$ thanks to the extension). Now it depends on the definition of a field you choose, mine is that a field must have at least two elements (the neutral for addition and the neutral for multiplication) so it is not possible for me. If your definition is different and that $0$ is a field, then I have to think about it.