Degree of an element in a Field Extension

Question

If $\alpha \in \overline{F}$ s.t. $\alpha^2 = s$ for some $s \in F$ can we say that degree of $\alpha$ over $F$ equals $2$?

Answer 1

You can conclude that the degree of $\alpha$ is at most $2$. Indeed, $\alpha^{2}=s$ with $s\in F$ implies that $\alpha$ is root of $x^2-s\in F[x]$. Hence, the minimal polynomial of $\alpha$ over $F$ divides $x^2-s$, and so degree of $\alpha$ over $F$ is at most $2$.

Answer 2

Yes, the minimal polynomial that $\alpha$ will satisfy in $F \left[ x \right]$ is $x^2 - s$. Since $\alpha$ is algebraic over $F$ and the minimal polynomial it satisfies is of degree 2, we have $$ \left[ F ( \alpha ) : F \right] = 2 $$