# Field extension generated by $\alpha$ and separability

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 $\alpha$?

• To better understand your question, let us take the example of, say, $k(\alpha)=\mathbb{Q}(\sqrt{2})$. "How do we know that ALL of the elements are the roots of separable extensions?". Well, $a\in \mathbb{Q}$ is a root of $f(x)=x-a$, and $\alpha=\sqrt{2}$ is a root of $f(x)=x^2-2$. – Dietrich Burde Jul 15 '15 at 19:28
• How about the elements of $k(\alpha)$ which are not $\alpha$ or $a \in \mathbb{Q}$ ? I am having a hard time understanding what the elements of $k(\alpha)$ are. – Yuugi Jul 15 '15 at 21:47

## 1 Answer

All the elements of $$k(\alpha)$$ are separable, because all irreducible polynomials over a finite field are separable. Indeed if $$f(x)$$ is not separable, $$f(x)$$ and $$f'(x)$$ have a common root in some extension of $$k$$, hence $$\deg\gcd(f(x),f'(x))\ge 1$$, which is impossible if $$f$$ is irreducible, unless $$f(x)=g(x^p)$$, for some polynomial $$g$$ (in which case $$f'(x)=p x^{p-1}g'(x)=0$$).

But the Frobenius morphism $$x\mapsto x^p$$ is injective, hence surjective, since we're in a finite field, so that $$f(x)=g(x^p)=\bigl(h(x)\bigr)^p$$ for some $$h(x)\in k[x]$$, and $$f(x)$$ would not be irreducible.