Field Extensions as $F$ adjoin some element Let $F$ be a field and $E$ an extension of $F$. Is it always possible to write $E=F(\alpha_1,\alpha_2,\ldots)$?
If $E$ is a finite extension then I think it is possible to write $E=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$. My reason is that if we take $\alpha\in E$ then as $[E:F]<\infty$  for some $n$ we must have $\alpha^n\in\text{Span}\{\alpha, \ldots,\alpha^{n-1}\}$. Meaning that $\alpha$ satisfies an (irreducible) polynomial in $F[x]$. If we keep doing this for each element in $E$ then we get $E=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$. Is this correct?
What about the case when $E$ is not a finite extension?
Thanks
 A: An example  is  $\mathbb R$ as  extension of  $\mathbb Q$.
A: When $E$ is a finite extension of $F$ we can always do this. For suppose that $[E:F] = n$. Then this means that we can choose $\alpha_1 \in E \setminus F$. Now consider $F(\alpha_1)$ as a subspace of $E$. As a subspace of a finite dimensional vector space it is finite dimensional so that $F(\alpha_1)/F$ is a finite extension. Now if $[F(\alpha_1):F] = n$, then we are done because $E = F(\alpha_1)$. Otherwise if it is less than $n$, we repeat this procedure again and find $\alpha_2 \in E \setminus F(\alpha_1)$ and look at $[F(\alpha_1,\alpha_2):F] = [\big(F(\alpha_1)\big)(\alpha_2):F]$. Eventually we will stop because $\dim_F E$ is finite, so that 
$$E = F(\alpha_1,\ldots, \alpha_n)$$ for some $\alpha_1,\ldots \alpha_n \in E$.
Edit: Qiaochu has posted an example here on MO to show that it is not true that every algebraic extension is obtained by adjoining a countable number of elements.
A: If $E/F$ is a finite extension, then $E$ is a finite dimensional vector space over $F$ with some basis $\{\alpha_1, \ldots, \alpha_n\}$ and thus $E = F(\alpha_1, \ldots, \alpha_n)$.
