$\alpha$ algebraic over field $F$ implies $[F(\alpha):F]=\text{degree of minimal polynomial over $\alpha$}$ Let $\alpha$ be some element algebraic over a field $F$. Then $F(\alpha)$ is isomorphic to $F[x]/\langle m_{\alpha,F}\rangle$, where $m_{\alpha,F}$ is the minimal polynomial with root $\alpha$ over $F$. Moreover, $[F(\alpha) : F] = \deg(m_{\alpha,F})$.
The first part of this theorem appears to be clear, but what about the second part (after the word "moreover")? Why does the index of $F(\alpha)$ over $F$ equal the degree of the minimal polynomial in $F[x]$ with root $\alpha$? I think this should be almost obvious, but perhaps I'm lacking some theory.

It seems the other answer and comment are about $F[\alpha]$ being a $F$-vector space of dimension $deg(p)$ ($F[\alpha]$ is the smallest ring containing $F$ and $\alpha$, i.e. $F[\alpha] = \{ \sum_{n =0}^d c_n \alpha^n, c_n \in F\}$).
For showing that $F[\alpha] = F(\alpha)$ you need to prove that $\varphi : F[x]/(p(x)) \to F[\alpha],\ \ \varphi(f(x)) = f(\alpha)$ is an isomorphism of rings ($\varphi$ is clearly surjective, and it is injective by definition of the minimal polynomial)
Hence $F[\alpha]$ is a field, so $F[\alpha] = F(\alpha)$ and $[F(\alpha):F]= deg(p)$
