I want to prove that a field $F$ of characteristic $p$, is perfect if and only if every element in $F$ has a $p$th root in $F$.
We say that $F$ is perfect if every polynomial $f(x)\in F[x]$ is separable, where we say that $f(x)$ is separable if its irreducible factors have no repeated roots.
Every element of $F$ has a $p$th root in $F$ means that if we fix $\alpha\in F$, then the polynomial $x^p-\alpha$ has a root. But how can I factor $x^p-\alpha$?
Conversely, if $f(x)\in F[x]$ and every element has a $p$th root, how can I manage $f(x)$ in order to show it is separable?
Could anyone give me a hint?
Thank you.
Best Answer
Fill in the (usually small) details in the following:
It is probably easier to show the double negation: $\;\Bbb F\;$ is inseparable iff $\;\Bbb F^p\setminus\Bbb F\neq\emptyset\;$
== So take $\;a\in\Bbb F^p\setminus\Bbb F\;$ . If $\;\beta\;$ in some non-trivial extension of $\;\Bbb F\;$ is s.t. $\;\beta^p=a\;$ , then
$$f_a(x):=x^p-a=x^p-\beta^p=(x-\beta)^p\in\Bbb F[x]$$
If $\;f_a(x)\;$ isn't irreducible over $\;\Bbb F\;$, then the right hand above shows that it must have a non-trivial factor in $\;\Bbb F[x]\;$ of the form $\;(x-\beta)^n\;,\;\;1\le n\le p-1\;$ . This polynomial's coefficient of $\;x^{n-1}\;$ is $\;-n\beta\;$ and it belongs to $\;\Bbb F\;$ . Since $\; n(\neq0)\;$ belongs to the prime field $\;\Bbb F_p\le\Bbb F\;$ , we get that it is obviously invertible there and thus
$$-n\beta\in\Bbb F\implies \beta\in\Bbb F\;\implies\;a=\beta^p\in\Bbb F$$
which is a contradiction. Thus, $\;f_a(x)\;$ is irreducible.
== Suppose now there exists an inseparable irreducible $\;g(x)\in\Bbb F[x]\;$ . Then it must be that $\;g'(x)\equiv0\;$ , so that $\;g\;$ is a polynomial in $\;x^p\;$ :
$$g(x)=\sum_{k=0}^m a_kx^{pk}$$
If we had that $\;\Bbb F^p=\Bbb F\;$ , then for all $\;0\le k\le m\;$ there exist $\;b_k\;$ s.t. $\;a_k=b_k^p\;$ , but then
$$g(x)=\sum_{k=0}^m a_kx^{pk}=\sum_{k=0}^mb_k^p\left(x^k\right)^p=\left(\sum_{k=0}^m b_kx^k\right)^p$$
which contradicts the fact that $\;g\;$ is irreducible.