As an example:
Let p be an odd prime. When α, β ∈ $\mathbb{Z}$p are not squares, then α / β is a square. Let L, K both be fields of size p2.
The rest of the exercise is irrelevant to what I am trying to learn here. I am reading the solution to the exercise and it says that:
[L : $\mathbb{Z}$p] = 2
Why is this exactly? I am not sure if I am missing something obvious, or if there was some detail that I missed in class.
Given any field extension $F\subseteq F^\prime$ the degree $[F^\prime:F]$ is defined to be the dimension of $F^\prime$ as vector space over $F$, i.e. $$ [F^\prime:F]=\dim_F(F^\prime). $$ Thus, in the case of finite fields where, say, $F=\Bbb F_p$ (the field with $p$ elements) there is an equality $$ |L|=p^{\dim_{\Bbb F_p}(L)}=p^{[F^\prime:F]} $$ because it is true for every vector field $V$ that the choice of a basis ${e_1,...,e_d}$ defines an identification $V\simeq\Bbb F_p^d$ and thus $|V|=p^d$.