Separable extension

Question

Let $\alpha$ algebraic over $k$ of characteristic $p>0$

Prove that $\alpha$ is separable over $k$ if and only if $k(\alpha)=k(\alpha^p)$.

Any suggestion, please.

Answer 1

Let $\alpha$ belongs to K($\alpha$)Note that $\alpha$ is separable over K imply $\alpha$ is separable over K($\alpha^p$).But $\alpha$ satisfies polynomial $x^p$- $\alpha^p$ over K($\alpha^p$). So minimal polynomial g(x)of $\alpha$ over K($\alpha^p)$ divides $x^p-\alpha^p$. Since g(x) is separable we see that degree of g(x)=1.So $\alpha$ belongs to K($\alpha^p)$.That imply K($\alpha)$ $\subset$K($\alpha^p)$.So we have K($\alpha)$ =K($\alpha^p)$

Answer 2

Suppose $k(\alpha)=k(\alpha^p)$, then we show $\alpha$ is separable over $k$. Since $\alpha$ is algebraic, then $\alpha^p$ is algebraic, and so $[k(\alpha^p):k]$ is finite. So $k(\alpha^p)$ has a $k$-basis $1, \alpha^p, (\alpha^p)^2, \ldots, (\alpha^p)^n$. Since $\alpha \in k(\alpha^p)$ by assumption, we can write $\alpha = \sum_{i=0}^n c_i(\alpha^p)^i $ where $c_i \in k$. Then consider the polynomial $f(x)=x-\sum c_ix^{pi}$. Its derivative is $f'(x)=1-\sum c_ipix^{pi-1} =1 -0$ and so $f(x)$ is separable (as $gcd(f(x), f'(x)=1$), and $f(x)$ has $\alpha$ as a root. Hence $\alpha$ is separable.

Saha has given the proof of the other direction.