In my notes, I have written that the field extension of $k$ generated by an element $\alpha \in K$, where $K$ a larger field, is defined to be $$k(\alpha) = \bigcap_{ \alpha \in E} E$$ where $E \subset K$ are subfields of $K$, and $k\subset E$.
A $\textbf{separable}$ field extension is a field extension $K$ over a smaller field $k$ such that every element of $K$ is a root of a separable polynomial over $k$, let $k$ be a finite field.
Let us consider $f(x) \in k[x]$, irreducible, then $f(x)$ is separable. Now let $\alpha \in K$ be a root of $f(x)=0$ and let us generate the finite field extension $k(\alpha)$.
In my notes I also have that all finite extensions of finite fields are separable. So in this case, $k(\alpha)$ is finite extension of a finite field and thus should be separable.
I know that, of course, for $\alpha$, there exists a separable polynomial over $k$, namely the irreducible $f(x)$. However, how do we know that ALL of the elements of $k(\alpha)$ are the roots of separable polynomials in $k[x]$ ?
Overall, how are the elements of $k(\alpha)$ related to $\alpha$?