Proposition 3

— Let K be a field, E an extension field of K which is p-radical over K and $u$ a homomorphism of K into a perfect field F. Then there exists a unique homomorphism $v$ of E into F extending $u$.

Let $E_n$ be the set of elements of E which are p-radical of height $\leq n$ over K. By Prop. 2 the field E is the union of the ascending sequence $(E_n)_{n \geq 0}$ of subfields. Let $v$ be a homomorphism of E into F extending $u$; for any $x \in E_n$ we have $x^{p^n} \in K$, whence $v(x)^{p^n} = v(x^{p^n}) = u(x^{p^n})$; we thus have $v(x) = u(x^{p^n})^{p^{-n}}$ for all $n \geq 0$ and all $x \in E_n$, whence the uniqueness of the extension of $u$ to E.

Let $n$ be a positive integer; for any $x \in E_n$ we have $x^{p^n} \in K$ and since F is perfect, we can define an element $v_n(x)$ of F by $v_n(x) = u(x^{p^n})^{p^{-n}}$. It is clear that $v_n$ is a homomorphism of $E_n$ into F, that $v_0 = u$ and that $v_{n+1}$ induces $v_n$ on $E_n$. Hence there exists a homomorphism $v$ of E into F inducing $v_n$ on $E_n$ for all $n \geq 0$ and, in particular, coinciding with $v_0 = u$ on $E_0 = K$.