TITLE: Field extension of one element
QUESTION [0 upvotes]: Let's say $F$ is a field and $\alpha\notin F$. I'm slightly confused about field extensions, but then would $F(\alpha)$ be
$$
\{a_1 + a_2\alpha: a_1,a_2\in F\}
$$
I know that now we can consider $F(\alpha)$ as a vector space over $F$. So, then, if my interpretation of $F(\alpha)$ was correct, would $[F: F(\alpha)] = 2$ because $1_F, \alpha$ would be a basis of $F(\alpha)$?
Edit: Wait no, I just realized this must be wrong. Since $\mathbb{Q}(\sqrt[3]{2})$ isn't $\{a_1+ a_2\sqrt[3]{2}: a_1, a_2\in\mathbb{Q}\}$. I'm still confused on how to estimate the degree of a field extension, then, without knowing which field you're working in and what element you're extending it by.

REPLY [0 votes]: Firstly, $\alpha $ must be algebraic, that means that there exists a polynomial P with coefficients in F such that $\alpha$ is a root of P. Denote $n=degP$, and suppose that P is irreductible (since the polynomial ring $F[X]$ is UFD), then $F(\alpha)$ is an F-vector space of dimension $n$ and generated by {$1_{F}, \alpha, \alpha^{2},..., \alpha^{n-1}$}.
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