Proposition 5

— If K is a field of characteristic 0, or is finite, \ or algebraically closed \*, it is perfect. In particular every prime field is perfect.* Suppose that $K$ has characteristic $p \neq 0$. If $K$ is finite, then the subfield $K^p$ of $K$ has the same cardinal as $K$, whence $K^p = K$. \* If $K$ is algebraically closed, the polynomial $X^p - a$ has a root $x$ in $K$ for each $a \in K$ (V, p. 20, Def. 1) whence $x^p = a$ and so $K^p = K$. \* Finally a prime field is of characteristic 0 or finite. Let $K_0$ be a field of characteristic $p \neq 0$ and $K = K_0(X)$ the field of rational fractions in an indeterminate $X$ over $K_0$. Then $K$ is imperfect, for there exists no element $u(X)/v(X)$ of $K$ (u, v polynomials in $K_0[X]$) such that $(u(X)/v(X))^p = X$. This may be seen by writing this relation in the form $u(X)^p = Xv(X)^p$ and comparing the degrees of the two sides.