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Abstract Algebra

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Dual of a Z-Module (Abelian Group): Trivial or Non-Trivial?

ABSALG-OCJVQU

Modules over the ring of integers $\mathbb{Z}$ are the same thing as abelian groups (written additively). The $\mathbb{Z}$-module homomorphisms between $\mathbb{Z}$-modules are exactly the abelian group homomorphisms between them.

The dual $\mathbb{Z}$-module $A^\ast:={\rm Hom}_\mathbb{Z}(A,\mathbb{Z})$ of a $\mathbb{Z}$-module $A$ consists, therefore, of the group homomorphisms from $A$ to $\mathbb{Z}$. $A^\ast$ is indeed an abelian group under pointwise addition of homomorphisms and​ is often denoted by $A^\vee$.

In the following question, we consider several abelian subgroups $A$ of $\mathbb{Q}$. We ask whether or not $A^\ast$ is the zero module $0$, and, if it is not zero, whether or not $A^\ast$ equals $\mathbb{Z}$.

Let $A_1$ be the abelian subgroup of $\mathbb{Q}$ with basis $\mathcal{P}_{N,+}=\{p^n\mid n\ge N\}$ for a fixed prime $p\ge 2$.

Then $A_1^\ast$ is

and is

.

Let $A_2$ be the abelian subgroup of $\mathbb{Q}$ with basis $\mathcal{P}_-=\{p^{-n}\mid n\ge 0\}$ for a fixed prime $p\ge 2$.

Then $A_2^\ast$ is

and is

.

Let $A_3$ be the abelian subgroup of $\mathbb{Q}$ with basis $\mathcal{P}_N=\{p^n\mid -N\le n\le N\}$ for a fixed prime $p\ge 2$ and integer $N\ge 1$.

Then $A_3^\ast$ is

and is

.