Let $K$ a field of characteristic $p>0$. We say that a field is perfect if all algebraic extension is separable. Show that the following assertions are equivalents:
1) $K$ perfect,
2) Every irreducible polynomial is separable,
3) $K^{p^n}=K$ for all $n\geq 1$.
Proof
$1)\implies 2)$. Let $f\in K[X]$ irreducible and $K_f=K[X]/(f)$ it splitting field. Since $K_f$ is algebraic, it's separable. Let $\alpha\in K_f$ a root of $f$ and $p_\alpha $ it's minimal polynomial. Then $p_\alpha $ has only simple roots, and $p_\alpha \mid f$.
Q1) Can I conclude that $p_\alpha =f$ ? and thus $f$ has only simple roots and thus separable ?
$2)\implies 3)$. $K^{p^n}=\{x^{p^n}\mid x\in K\}$. Therefore $K^{p^n}\subset K$. For the other inclusion, let $\alpha \in K$. Since $car(K)=p$, $$\alpha ^{p^n}\equiv \alpha \pmod p$$ and thus $\alpha \in K^{p^n}$.
Q2) Does it work ? But to be honest, I don't really understand what I said, any explanation ?
For $3)\implies 1)$ I have no problem.