Proposition. some propositions [1902]

    1. Every field of characteristic $0$ is perfect.
    2. Every finite field is perfect.
    3. A field $F$ of characteristic $p$ is perfect if and only if $F = F^{p}$, i.e. every element of $F$ is a $p$-th power in $F$.

    Proof. sketch [1902A]

      1. Let $F$ be a field of characteristic $0$. Given finite field extension $K / F$ and $\alpha \in K$, let $m_{\alpha, F}(x) \in F[x]$ be the minimal polynomial of $\alpha$ over $F$. Then $m_{\alpha, F}(x)$ is irreducible. By (3) of [Proposition 1704], $m_{\alpha, F}(x)$ is separable. Hence $\alpha$ is separable over $F$. So $K / F$ is separable $\implies$ $F$ is perfect.
      2. Let $F$ be a finite field with $\operatorname{ch}(F) = p$ and $\lvert F \rvert = p^{n}$ for some integer $n$. By the proof of Theorem 1801 ,$\alpha = \alpha^{p^{n}} = (\alpha^{p^{n-1}})^{p}$ for every $\alpha \in F$. In particular, every element in $F$ is a $p$-th power in $F$. By (3), $F$ is perfect.
      3. Suppose that $F = F^{p}$. Let $K / F$ be a finite extension and $\alpha \in K$. Let $m_{\alpha, F}(x)$ be the minimal polynomial of $\alpha$ over $F$. We show that $m_{\alpha, F}(x)$ is separable by contradiction. Suppose the opposite. Then by (2) of Proposition 1704, $\gcd \left( m_{\alpha, F}(x), m_{\alpha, F}'(x) \right) \neq 1$, then $\gcd \left( m_{\alpha, F}(x), m_{\alpha, F}'(x) \right) = m_{\alpha, F}(x)$ because $m_{\alpha, F}(x)$ is irreducible. It follows that $m_{\alpha, F}(x) \mid m'_{\alpha, F}(x)$. We deduce that $m_{\alpha, F}'(x) = 0$. Otherwise, $\deg m_{\alpha, F}'(x) < \deg m_{\alpha, F}(x)$, so $m_{\alpha, F}(x)$ cannot divide $m_{\alpha, F}'(x)$. It follows from the definition of derivative of polynomials that$$m_{\alpha, F}'(x) = 0 \implies m_{\alpha, F}(x) = a_{m}x^{p^{m}} + a_{m-1}x^{p^{m-1}} + \cdots + a_{1}x^{p} + a_{0}.$$The condition $F = F^{p}$ implies that there exists $b_{0}, \dots, b_{m} \in F$ such that $a_{j} = b_{j}^{p}$. Then$$\begin{align*} m_{\alpha, F}(x) &= b_{m}^{p}x^{p^{m}} + b_{m-1}^{p}x^{p^{m-1}} + \cdots + b_{1}^{p}x^{p} + b_{0}^{p} \\ &= \left( b_{m}x^{p^{m-1}} + b_{m-1}x^{p^{m-1}} + \cdots + b_{1}x + b_{0} \right)^{p},\end{align*}$$and $m_{\alpha, F}(x)$ is not irreducible. We get a contradiction. Hence, $m_{\alpha, F}(x)$ is separable, and $\alpha$ is separable over $F$, so $K / F$ is separable and $F$ is perfect.