Proposition 2

— Let K be a field, $L$ a p-radical extension of height $\leq 1$ of K and S a subset of L. For S to be p-free it is necessary and sufficient that $K(T) \neq K(S)$ for every subset $T \neq S$ of S.

Suppose first that S is p-free and let $T \neq S$ be a subset of S. Denote by A the set of families with finite support $a = (\alpha_s)_{s \in S}$ of integers between 0 and $p-1$; let $A'$ be the subset of A consisting of the families $a = (\alpha_s)_{s \in S}$ such that $a_s = 0$ for all s in $S - T$. We also put $u_a = \prod_{s \in S} s^{\alpha_s}$ for $a \in A$. Then (V, p. 98, Prop. 1) the family $(u_a)_{a \in A'}$ is a basis of $K(S)$ over K and the subfamily $(u_a)_{a \in A'}$ is a basis of $K(T)$ over K. Since $A' \neq A$, we have $K(T) \neq K(S)$.

Suppose now that S is not p-free. Then there exists an integer $n \geq 1$ and a sequence of elements $x_1, ..., x_n$ of S such that $(x_1, ..., x_{n-1})$ is p-free but $(x_1, ..., x_n)$ is not. We have $[K(x_1, ..., x_{n-1}):K] = p^{n-1}$ and $[K(x_1, ..., x_n):K] < p^n$. Since $[K(x_1, ..., x_n):K]$ is a multiple of $[K(x_1, ..., x_{n-1}):K]$, we thus have

whence $x_n \in K(x_1, ..., x_{n-1})$. This shows that $K(S - \{x_n\}) = K(S)$.