It suffices to prove that the polynomial $X^{p^e} - a$ is irreducible in $K[X]$. By the definition of height of x we have $a \notin K^p$, so that the proposition follows from the lemma:

Lemma 1. — Let K be a field and a an element of K such that $a \notin K^p$. Then for every integer $e \geq 0$, the polynomial $f(X) = X^{p^e} - a$ is irreducible in $K[X]$.

Let $\Omega$ be an algebraic closure of $K$ and let b be the element $a^{p^{-e}}$ of $\Omega$; we denote by g the minimal polynomial of b over $K$. We have $f(X) = (X - b)^{p^e}$ and hence every irreducible polynomial in $K[X]$ which divides $f$ admits b as root, and hence is equal to g. There exists thus (IV, p. 13, Prop. 13) an integer $q \geq 1$ such that $f = g^q$; since q divides the degree $p^e$ of $f$, there exists an integer $e'$ such that $0 \leq e' \leq e$ and $q = p^{e'}$. If c is the constant term of g, we have $-a = c^q$; since we assumed that a does not lie in $K^p$, we must have $q = 1$, that is, $f = g$, so the lemma follows.

For each integer $n \geq 0$ the set $E_n$ consists of all elements $x$ of E such that $x^{p^n} \in K$; since the mapping $x \mapsto x^{p^n}$ is an endomorphism of the field E, we deduce that $E_n$ is a subextension of E. The sequence $(E_n)_{n \geq 0}$ is ascending, with union $E_r$ and hence $E_r$ is a subextension of E (V, p. 11, Prop. 3). It is clear that $E_r$ is a p-radical extension of K and that $E_r$ contains every subextension of E which is p-radical over K.

For $E_r$ is a subextension of E and by hypothesis $E = K(E_\infty)$ whence $E = E_r$; so E is a p-radical extension of K.

Under the conditions of Proposition 2, $E_r$ is said to be the relative $p\text{-radical}$ closure of K in E.

We shall apply Prop. 2 particularly to the case where E is an algebraically closed extension of K ; then E is perfect and we have $E_n = K^{p^{-n}}$ for all $n \geq 0$. In this case we denote by $K^{p^{-\infty}}$ the set of elements of E that are p-radical over K ; it is the subfield of E which is the union of the ascending sequence $(K^{p^{-n}})_{n > 0}$ of subfields of E. By Prop. 2, $K^{p^{-\infty}}$ is the largest subextension of E which is p-radical over K ; by Prop. 3 of V, p. 6, $K^{p^{-\infty}}$ is a perfect closure of K and it is also the smallest perfect subfield of E containing K. When K is perfect we clearly have $K = K^{p^{-n}} = K^{p^{-\infty}}$ for all n. If K is imperfect, we have $K \neq K^P$, whence

$^1$ A p-radical extension is also called a purely inseparable extension, or a radical extension.

for $n \geq 1$; the subfields $K^{p^{-n}}$ of E are then pairwise distinct and $K^{p^{-\infty}}$ is an algebraic extension of infinite degree of K.

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 $E_0 = K$.

The Corollary is trivial when $p = 1$; suppose then that $p \neq 1$. The stated conditions are necessary by V, p. 5, Th. 3 ; they are sufficient by Prop. 3.

Since E is a p-radical extension of finite degree of K, there are elements $a_1, \ldots, a_m$ of E, p-radical over K, such that $E = K(a_1, \ldots, a_m)$. Let i be in the range 1 to m ; since $a_i$ is a fortiori p-radical over $K(a_1, \ldots, a_{i-1})$, the degree

is a power of p (V, p. 24, Prop. 1). We have $[E : K] = n_1 \ldots n_m$ by V, p. 18, Th. 2, whence the result.