TITLE: Finite Field Extension
QUESTION [7 upvotes]: Suppose $E/F$ is a field extension of degree $n$. Does it follow that $E = F(a_{1}, a_{2}, \ldots, a_{n})$ for some $a_{i} \in E$? I feel like this is true, but I'm getting confused with all the definitions.

REPLY [4 votes]: Yes, the statement that $E=F(a_1,\ldots,a_n)$ for some $a_i\in E$ is implied by the statement that $[E:F]=n$. Recall that $[E:F]=n$ is, by definition, the statement that $E$ is an $n$-dimensional vector space over $F$. Thus, there exists a basis for this vector space consisting of $n$ elements, that is, there exist $a_1,\ldots,a_n\in E$ such that any $b\in E$ has
$$b=c_1a_1+\cdots+c_na_n$$
where the $c_i\in F$. Thus, we will have $E=F(a_1,\ldots,a_n)$.
However, you should note that it might not be the case that all the $a_i$'s are necessary to generate $E$ as a field over $F$. For example, if $E=\mathbb{Q}(\sqrt[n]{2})$ and $F=\mathbb{Q}$, then $[E:F]=n$, and a basis for $E$ as a vector space over $F$ is 
$$1,\sqrt[n]{2},\ldots,(\sqrt[n]{2})^{n-1}$$
so that $E=F(1,\sqrt[n]{2},\ldots,(\sqrt[n]{2})^{n-1})$, but we actually already had $E=F(\sqrt[n]{2})$ - that is, $E$ can be generated over $F$ by a single element.