= Solution
We first give the common reduction for the three conditions, then prove the cycle from (i) to (ii), from (ii) to (iii), and from (iii) back to (i). Assume here that $\operatorname{char}K\ne p$. The <Galois module of roots of unity> $\mu_p$ fits into the Kummer sequence
$$
1\longrightarrow\mu_p\longrightarrow K_s^\times\xrightarrow{\ p\ }K_s^\times\longrightarrow1,
$$
which is surjective on the right because $p$th-root polynomials are separable. Restrict it to the <absolute Galois group> of any algebraic extension $L/K$. The long exact sequence gives the <Kummer cohomology divisibility criterion>
$$
0\longrightarrow H^n(L,K_s^\times)/pH^n(L,K_s^\times)
\longrightarrow H^{n+1}(L,\mu_p)
\longrightarrow H^{n+1}(L,K_s^\times)[p]\longrightarrow0.
$$
<Continuous cohomology of a profinite group> in positive degree with discrete coefficients is torsion: a continuous normalized cocycle becomes zero on a sufficiently small open subgroup, and restriction followed by corestriction multiplies its class by the finite index. For an abelian group, <p-primary torsion> consists of elements killed by powers of $p$, while <p-divisibility> means that multiplication by $p$ is surjective. For a torsion group $A$, $A\{p\}=0$ if and only if $A[p]=0$, because a nonzero element of $p$-power order has a multiple of exact order $p$. Consequently the two conditions on the multiplicative cohomology groups are jointly equivalent to
$$
\boxed{H^{n+1}(L,\mu_p)=0}.
$$
The coefficient module $K_s^\times$ is used with its restricted Galois action as in the question. Purely inseparable algebraic extensions give equivalent <absolute Galois groups>; the $\mu_p$ reduction therefore also covers those extensions.
The <cohomological dimension at a prime> condition $\operatorname{cd}_p(K)\le n$ means that $H^q(G_K,M)=0$ for every discrete $p$-primary torsion <Galois module> $M$ and every $q>n$. Cohomological dimension does not increase on a closed subgroup. Since <absolute Galois groups> of algebraic extensions identify with closed subgroups, up to the purely inseparable identification just noted, condition (i) gives $H^{n+1}(L,\mu_p)=0$ for every algebraic $L/K$. The Kummer reduction proves \b[(i) implies (ii)].
Back to article page
