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I'm currently reading the book "Galois theory of $p$-extensions" by Helmut Koch.

There, we calculate the cohomological dimension of the galois group $G(K/k)$ where $K$ is the maximal (normal) $p$-extension of $k$.

(Here $p$ is a prime and $k$ is a local field or global field of finite type, i.e finite extension of $\mathbb{Q}$ of $\mathbb{Q}_p$.)

As $G(K/k)$ is a pro-p group, we study $H^2(G(K/k), \mathbb{F}_p)$ where $\mathbb{F}_p$ is the finite field with $p$ elements with trivial group action.

Let $k'$ be the field generated by $k$ and the $p$'th roots of unity, and let $K'$ be the maximal $p$-extension of $k'$. Then there is a canonical group homomorphism from $G(K'/k')$ to $G(K/k)$ (restriction map). This induces a homomorphism from $H^2(G(K/k),\mathbb{F}_p)$ to $H^2(G(K'/k'),\mathbb{F}_p)$.

The question is, is this map injective?

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I believe this map is always injective. Here is a quick argument: first note that $K'$ is normal over $k$ (because it is invariant under any automorphism of the algebraic closure of $k$ which preserves $k'$, and $k'$ is normal over $k$). This means that we can view the map $G(K'/k') \to G(K/k)$ as a composition of two maps $$ G(K'/k') \stackrel{f}{\to} G(K'/k) \stackrel{g}{\to} G(K/k) .$$ It will hence suffice to show that both $f^*:H^2(G(K'/k),\mathbb{F}_p) \to H^2(G(K'/k'),\mathbb{F}_p)$ and $g^*:H^2(G(K/k),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p)$ are injective. Now $f: G(K'/k') \to G(K'/k)$ is an inclusion of a subgroup of finite index $[k':k]$, and hence there exists a corestriction map $f_!:H^2(G(K'/k'),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p)$ such that $f_! \circ f^*$ is multiplication by $[k':k]$. Since $[k':k]$ is coprime to $p$ and $H^2(G(K'/k),\mathbb{F}_p)$ is a $p$-torsion group it follows that $f^*$ is injective. To show that $g^*$ is injective, we note that $g: G(K'/k) \to G(K/k)$ is a surjective group homomorphism with kernel $G(K'/K)$, and so the Hochschild-Serre spectral sequence gives an exact sequence $$ H^1(G(K'/K),\mathbb{F}_p)^{G(K/k)} \to H^2(G(K/k),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p) .$$ To show that $g^*$ is injective it will hence suffice to show that $H^1(G(K'/K),\mathbb{F}_p)^{G(K/k)} = 0$, i.e., that $K$ does not admit any non-trivial cyclic $p$-extensions inside $K'$ which are normal over $k$. But this is just because $K$ does not admit any non-trivial cyclic $p$-extensions which are normal over $k$: indeed, it is the maximal $p$-extension of $k$.

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  • $\begingroup$ Thank you for answering the question. I guess your intention is to show that H^1(G(K'/K),F_p)=0 by showing that K is the maximal elementary abelian p-subextension of K'/K. Can you explain a little bit more about how to use the maximality of K to show that K'/K does not admit nontrivial cyclic p-extension of K? (For example there can be non-trivial cyclic p-extension say L of K contained in K' such that L/k is not normal.) $\endgroup$
    – gualterio
    Commented May 23, 2018 at 8:23
  • $\begingroup$ If $K \subseteq L \subseteq K'$ is a cyclic $p$-extension of $K$ in $K'$ which is normal over $k$, then $L$ is in particular a normal $p$-extension of $k$. Since $K$ is the maximal normal $p$-extension of $k$, it would have to contain $L$, and so $L$ must be trivial. In cohomological terms, cyclic $p$-extensions of $K$ in $K'$ which are normal over $k$ are classified by the group $H^1(G(K'/K),\mathbb{F}_p)^{G(K/k)}$, and so this group must be trivial. $\endgroup$
    – Yonatan Harpaz
    Commented May 23, 2018 at 8:49
  • $\begingroup$ Thanks again. So the last sentence is the one that I needed the most. Can you please suggest me any references on the theorem?, because I haven't seen them before and maybe in someday I might need those kinds of theorems. $\endgroup$
    – gualterio
    Commented May 23, 2018 at 9:40
  • $\begingroup$ Well, $H^1(G(K',K), \mathbb{F}_p)$ is the group of homomorphisms $G(K'/K) \to \mathbb{F}_p$. To each such homomorphism $f: G(K'/K) \to \mathbb{F}_p$ we can associate the fixed field $K \subseteq K_f \subseteq K'$ of its kernel $Ker(f) \subseteq G(K'/K)$. Then $K_f$ is either $K$ (if $f=0$) or a cyclic p-extension of $K$ (if $f \neq 0$).--> $\endgroup$
    – Yonatan Harpaz
    Commented May 23, 2018 at 11:27
  • $\begingroup$ --> There is a natural action of $G(K/k)$ on the set of cyclic $p$-extensions of $K$ inside $K'$ (because there is a natural action of $G(K'/k)$ on this set, but any automorphism of $K'$ which fixes $K$ must map any normal extension of $K$ to itself). There is also a natural action of $G(K/k)$ on $H^1(G(K',K),\mathbb{F}_p)$, and the association $f \mapsto K_f$ is $G(K/k)$-equivariant. This means that if $f$ is a non-zero $G(K/k)$-invariant element of $H^1(G(K',K),\mathbb{F}_p)$ then $K_f$ is a non-trivial cyclic $p$-extension of $K$ which invariant under $G(K/k)$.--> $\endgroup$
    – Yonatan Harpaz
    Commented May 23, 2018 at 11:34

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