Can every element of an algebraic field extension, $E \subseteq F$ be represented as $f(s)$, where $s$ is another member and $f(x) \in F[x]$

Question

Specifically, could every member of $E$ be generated by one element $s$, by evaluating different polynomials of $s$ with coefficients in $F$?

Answer 1

No. If $F \to E$ is an infinite algebraic extension then any element $s \in E$ generates at best a finite subextension $F[s]$ of $E$, so $E$ will never be generated by one element. If $F \to E$ is a finite algebraic extension then a necessary and sufficient condition is given by the primitive element theorem: it is necessary and sufficient that there be finitely may intermediate extensions between $E$ and $F$.

This condition in particular always holds for finite separable extensions, which is the most commonly used special case of the primitive element theorem. For example, it follows that every finite extension of $\mathbb{Q}$ (that is, number fields) is generated by one element.

Example. Consider the extension $\mathbb{F}_p(x^p, y^p) \to \mathbb{F}_p(x, y)$. This is a purely inseparable extension of degree $p^2$. Any element $f \in \mathbb{F}_p(x, y)$ has the property that $f^p \in \mathbb{F}_p(x^p, y^p)$, and consequently generates an extension of $\mathbb{F}_p(x^p, y^p)$ of degree at most $p$; hence $f$ cannot generate the entire extension.