# Why is the Herbrand quotient of the dual $\hat{A}$ equal to the inverse of the Herbrand quotient of $A$ in this situation?

I'm reading Serre's Local Fields, and I'm trying to understand the proof of Prop. 9 in $\S$5 of Chap. 8 (p.136). First, the setup:

• $p$ is a prime number
• $G$ is a cyclic group of order $p$
• $A$ is a $G$-module
• $h(A)$ is the Herbrand quotient of $A$
• $\varphi(A)$ is the Herbrand quotient of $A$ if it were acted on by $G$ trivially

Prop. 9 states that

When $\varphi(A)$ is defined, the quantities $\varphi(A^G)$, $\varphi(A_G)$, and $h(A)$ are all defined, and $$h(A)^{p-1}=\varphi(A^G)^p/\varphi(A)=\varphi(A_G)^p/\varphi(A).$$

In the proof, one of the cases that we end up reducing to is when $A$ is a $G$-module that has the following properties:

• $A$ is $p$-divisible (for every $y\in A$, there is an $x\in A$ such that $px=y$)
• every element of $A$ is $p$-power torsion (for every $y\in A$, there is an $n\in\mathbb{N}$ such that $p^ny=0$)
• the $p$-torsion subgroup $\{y\in A\mid py=0\}$ of $A$ is finite

Serre says at this point that:

... it is quicker to use Pontryagin duality: it transforms $A$ into a compact group $\hat{A}$, which is a free module of finite type over the ring $\mathbb{Z}_p$ of $p$-adic integers, on which $G$ acts. It can be immediately verified that $$h(A)=h(\hat{A})^{-1},\qquad\varphi(A)=\varphi(\hat{A})^{-1},\qquad\varphi(A^G)=\varphi(\hat{A}_G)^{-1}.$$

Not having any more than a Wikipedia-level knowledge of Pontryagin duality, this passage is unfortunately rather opaque to me. From this section of the Wikipedia page, I deduce that we have implicitly put the discrete topology on $A$. Okay, that's reasonable. But what is the justification for the displayed formulas? Noting the multiplicativity of the Herbrand quotient on exact sequences, I assume the first equation is true because there is an exact sequence involving $A$, $\hat{A}$, and $G$-module $B$ with $h(B)=1$, and similarly with the second equation, but I can't seem to figure out what $B$ ought to be, nor why the third equation is true.

If you are feeling generous, I would also appreciate a hint/explanation as to why $\hat{A}$ is a finitely generated free $\mathbb{Z}_p$-module.

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In that context, duality usually is taken to mean the continuous $\hom(-,\mathbb Q_p/\mathbb Z_p)$, with the target endowed with the discrete topology. – Mariano Suárez-Alvarez Sep 20 '11 at 5:17
@Mariano: So we have put the discrete topology on both the source and target, so all homomorphisms are continuous, correct? Are these abelian group homomorphisms, or $G$-module homomorphisms (in which case, what is the $G$-module structure on $\mathbb{Q}_p/\mathbb{Z}_p$)? I feel like am being particularly dumb with this question. I cannot see how to fit $A$ and $\text{Hom}_{\mathbb{Z}}(A,\mathbb{Q}_p/\mathbb{Z}_p)$ into an exact sequence, much less one whose third term has Herbrand quotient equal to 1 (I do know that any finite group has this property). – Zev Chonoles Sep 22 '11 at 5:24
I also realize that the three properties of $A$ make it very similar to $\mathbb{Q}_p$, but the right way to use this observation escapes me. – Zev Chonoles Sep 22 '11 at 5:33

I don't know the $p$-version of this, but I can tell you how things work in the "non-$p$"-case. It should not be hard to tweak this into what you want.
Write $\hat H^\bullet(G,\mathord-)$ for Tate cohomology of a finite group $G$. For each abelian group $A$ let $A^\wedge=\hom(A,\mathbb R/\mathbb Z)$ be the abelian group of all homomorphisms of abelian groups; if $A$ happens to be a $G$-module, then $A^\wedge$ is also a $G$-module in the usual way. Then there is a canonical isomorphism $$\hat H^{p-1}(G,A^\wedge)\cong\left(\hat H^{-p}(G,A)\right)^{\wedge}.$$ In other words, $(\mathord-)^\wedge$ flips the cohomological degree with a twist, and dualizes. This can be proved from general nonsense using the fact that $(\mathord-)^\wedge$ is a duality---it is done in detail in Cartan-Eileberg.
Now, for example, suppose $A$ is a $G$-module for which $h(A)$ is defined. We have (using periodicity: I am assuming from now one that here $G$ is cyclic, or, to be silly, that $G$ has period $2$) $$\hat H^0(G,A^\wedge)=\hat H^{-1}(G,A)^\wedge=\hat H^1(G,A)^\wedge$$ and $$\hat H^1(G,A^\wedge)=\hat H^{-2}(G,A)^\wedge=\hat H^0(G,A)^\wedge.$$ Since the groups $\hat H^p(G,A)$ are finite, they are (non-canonically) isomorphic to their duals, so $h_0(A^\wedge)=h_1(A)$ and $h_1(A^\wedge)=h_0(A)$. This gives the first formula you want.
In your situation, you have only groups which play nicely with $p$, so the special duality in my old comment to your question gives a duality in cohomology, and you can play the same game. C-E proves the duality theorem in sufficient generality, if I recall correctly, that you can use it immediately. – Mariano Suárez-Alvarez Sep 22 '11 at 6:32