If $\mathrm{char}(K)=p$ is prime, $L/K$ is separable if and only if $K(\alpha) =K(\alpha^p)$ for all $\alpha \in L$ I am trying to prove that if $L/K$ is an algebraic extension and if $\alpha \in L$, then 


*

*$\alpha$ is separable over $K$ if $\mathrm{char}(K)=0$. This is clear because $K$ is perfect which in turn implies that $L/K$ is seperable . 

*Now if $\mathrm{char}(K)=p$ is prime, then the statement is: $\alpha$ is separable if and only if $K(\alpha) =K(\alpha^p)$. 
This problem is from a past examination and looks very interesting, but somehow I am not able to connect what's happening with the extension of $K$ with $\alpha$ and $\alpha^p$ .
I need some help here.  
This somehow makes me think that the minimal polynomial of $\alpha$ and $\alpha^p$ have different roots and $\alpha$ is given by some root of minimal polynomial of $\alpha^p$ . 
Note that I am using the definition of separability as follows : 

An extension $L/K$ is called separable if for every $\alpha \in L$ the minimal polynomial of $\alpha$ has distinct roots in $L$ . 

Thanks for helping. 
 A: In characteristic $p$, the Frobenius map $\Phi:K\rightarrow K$ with $\Phi(\kappa)=\kappa^p$ is additive (that is, $\Phi(\kappa +\lambda)=\Phi(\kappa)+\Phi(\lambda)$ because all the binomial coefficients that arise in $(\kappa+\lambda)^p$ are divisible by $p$. Taking $p$ powers is always  multiplicative so $\Phi$ must be a field endomorphism  of the base field $K$. Field homomorphisms are always monomorphisms since the only ideals of a field $K$ are $0$ and $K$. 
Now let $Q(X)=X^m+\sum_{i=0}^{m-1} c_iX^i$ be the minimal polynomial for $\alpha$ over $K$ . Now use calculus to show that $Q$ has a multiple root in $K[\alpha]$ if and only if $Q$ is a polynomial in $X^p$. Also, because $Q$ is irreducible so not a $p^{th}$ power, some coefficient of $Q(X^p)$ must not be in $\Phi(K)$ to avoid $Q(X)=\left(\Phi^{-1}(Q)\right)^p$ over $K$. Now note that from $Q(\alpha)=0$ we get $\Phi(Q(X))$ has root $\alpha^p$ so eventually some power $\alpha^{p^k}$ will be a root of a separable factor of $\Phi^{k}(Q[X])$, and use this to show $K[\alpha]=K[\alpha^p]=\cdots =K[\alpha^{p^k}]$ if and only if $Q$ is separable. 
A: Hint: If $\alpha \in L$, then there is some $n \in \mathbb{N}$ such that $\alpha^{p^n}$ is separable.
