Proposition 3

— Let K be a field, L a p-radical extension of height $\leq 1$ of K and S, T two subsets of L. The following conditions are equivalent: a) The subset S is p-free over K and T is p-free over K(S). b) $S \cap T = \varnothing$ and S U T is p-free over K. If T is p-free over K(S), we have $T \cap K(S) = \varnothing$ and a fortiori $S \cap T = \varnothing$. We may therefore assume S and T disjoint. Let A be the subset of $N^{(S \cup T)}$ consisting of all families $a = (a_x)_{x \in S \cup T}$ with $a_x < p$ for all $x \in S \cup T$, and define subsets $A'$ of $N^{(S)}$ and $A''$ of $N^{(T)}$ in similar fashion. We can in a natural way, identify $N^{(S \cup T)}$ with $N^{(S)} \times N^{(T)}$, and then A is identified with $A' \times A''$. For $a \in A$ put $u_a = \prod_{x \in S \cup T} x^{\alpha_x}$, and similarly define $u'_\beta$ and $u''_\gamma$ for $\beta \in A'$ and $\gamma \in A''$. We have $u_a = u'_\beta u''_\gamma$ for $a = (\beta, \gamma)$ in $A = A' \times A''$. Moreover $(u'_\beta)_{\beta \in A'}$ generates the vector K-space K(S) (V, p. 94, Prop. 1). For S U T to be p-free over K it is necessary and sufficient that the family $(u'_\beta u''_\gamma)_{\beta \in A', \gamma \in A''}$ should be free over K (V, p. 98, Prop. 1); it comes to the same (II, p. 222) to suppose that the family $(u'_\beta)_{\beta \in A'}$ is free over K and the family $(u''_\gamma)_{\gamma \in A''}$ free over K(S). Now the equivalence of a) and b) follows from Prop. 1 (V, p. 98).