Perfect field of characteristic $p$

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.

• If $\beta\in F$ is a root of the polynomial $x^p-\alpha$, then we have the factorization $x^p-\alpha=(x-\beta)^p$. – Zev Chonoles Jun 6 '15 at 7:00
• How it is related to an arbitrary $f(x)\in F[x]$? – Sandor Jun 6 '15 at 7:29
• @timbuc the former, he included that in the question already – Christopher Jun 6 '15 at 10:24
• @user73985 Right. Thank you. – Timbuc Jun 6 '15 at 10:32

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.

• I loved this answer. Thank you!! – Sandor Jun 6 '15 at 18:10