Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 21 / 4

We use the standard characteristic-$p$ absolute Galois-group theorem, ${\mathrm{cd}}_p(K)\le 1$, and the cohomological identification $\mathrm{Br}(K)=H^2(K,K_s^{\times })$. Both are clearly stated inputs permitted for this question.

In characteristic $p$, the map $a\mapsto a^p$ on $K_s^{\times }$ is injective, although it need not be surjective: $p$th roots are purely inseparable. Thus it gives an exact sequence of discrete Galois modulesThe quotient $Q$ is killed by $p$. Since ${\mathrm{cd}}_p(K)\le 1$, $H^2(K,Q)=0$. The long exact cohomology sequence containsTherefore multiplication by $p$ on $\mathrm{Br}(K)$ is surjective. This is Brauer p-divisibility in characteristic p. The quotient-module sequence is the relevant one in characteristic $p$; the separable Kummer sequence with a nontrivial ${\mu }_p$ is not available there.

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 $\mathrm{char}K\ne p$. The Galois module of roots of unity ${\mu }_p$ fits into the Kummer sequencewhich 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 criterionContinuous 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 toThe 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 ${\mathrm{cd}}_p(K)\le n$ 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 (i) implies (ii).

Every finite extension of degree prime to $p$ is an algebraic extension. Thus the universal vanishing established for all algebraic extensions in (ii) holds in particular for the extensions appearing in (iii). Equivalently, the Kummer reduction gives $H^{n+1}(L,{\mu }_p)=0$ for each of these finite extensions. Hence (ii) implies (iii). No transfer or separability assumption is required for this implication.

Let $G=G_K$ and choose a profinite Sylow subgroup $P$ at the prime $p$. Its image in every finite quotient of $G$ is a Sylow $p$-subgroup. Put $F=K_s^P$. Every finite intermediate field $K\subseteq L\subseteq F$ has degree prime to $p$, because its corresponding open subgroup contains $P$. Condition (iii), together with the preceding Kummer reduction, gives $H^{n+1}(L,{\mu }_p)=0$ for each such field. The cohomology continuity at closed subgroups property saysfor a discrete $G$-module $M$. Applying it to $M={\mu }_p$ gives $H^{n+1}(P,{\mu }_p)=0$.

The action of the pro-p group $P$ on ${\mu }_p$ is trivial. Indeed it maps continuously to $\mathrm{Aut}({\mu }_p)\cong {\mathbb{F}}_p^{\times }$, whose order $p-1$ is prime to $p$; every finite image of $P$ is a $p$-group, so that image is trivial. Choosing a primitive root identifies this module with the trivial module ${\mathbb{F}}_p$. Thus $H^{n+1}(P,{\mathbb{F}}_p)=0$.

We spell out why this controls arbitrary $p$-primary coefficients. A finite discrete $p$-primary $P$-module has a composition series with trivial ${\mathbb{F}}_p$ factors: its action factors through a finite $p$-group, whose only simple module in characteristic $p$ is the trivial one. Induction through the long exact cohomology sequence therefore gives vanishing of $H^{n+1}$ for every such finite module. Any discrete $p$-primary $P$-module is the filtered union of finite $P$-stable submodules. The orbit of an element is finite by continuity, and its orbit generates a finite abelian $p$-group. Continuous cohomology commutes with these filtered unions, sofor every discrete $p$-primary module $M$. This is the mechanism behind trivial coefficients detect the cohomological dimension of a pro-p group.

Now let $M$ be any discrete $p$-primary $G$-module and $a\in H^{n+1}(G,M)$. Its restriction to $P$ is zero. Continuity at the closed subgroup $P$ makes its restriction zero in some open $U\supseteq P$. The index $m=[G:U]$ is prime to $p$, and the restriction-corestriction identity in group cohomology for the corestriction map in group cohomology givesBut $a$ has $p$-power order, so multiplication by $m$ is invertible on the cyclic group it generates. Hence $a=0$. This proves $H^{n+1}(G,M)=0$ for all discrete $p$-primary $M$. Dimension shifting through an acyclic coinduced module gives vanishing in every higher degree, which is exactly ${\mathrm{cd}}_p(K)\le n$.

Therefore (iii) implies (i), completing the equivalence of all three conditions. The key reason prime-to-$p$ extensions suffice is that they approximate a pro-$p$ Sylow subgroup, where the cyclotomic module becomes trivial; prime-to-$p$ transfer then detects every $p$-primary cohomology class.