$\{a x^{z}: a\in \Bbb{Q}, z \in \Bbb{Z}\} \approx \Bbb{Q}^{\times} \otimes_{\Bbb{Z}} \Bbb{Z}^+ \implies$? what about $\Bbb{Q}$-linear sums?

Question

Consider all functions $f: \Bbb{Q} \to \Bbb{Q}$ of the form $f(x) = a x^z$ where $a \in \Bbb{Q}, z \in \Bbb{Z}$, call it $G$. It forms an abelian group under usual multiplication. I think it's isomorphic to $\Bbb{Q}^{\times} \otimes_{\Bbb{Z}} \Bbb{Z}^+$. If so, then what is the ring of $\Bbb{Q}$-linear sums of $G$ isomorphic to, $\Bbb{Q}[G]$?

Answer 1

The thing is, $\mathbb{Q}^*\otimes_{\mathbb{Z}} \mathbb{Z} = \mathbb{Q}^*$ (and in general for any abelian group $A$, $A\otimes_{\mathbb{Z}} \mathbb{Z} = A$). So that can't be your answer. In fact, you see that any function $f$ correspond to a couple $(a,z)\in \mathbb{Q}^*\times \mathbb{Z}$, not to a combination of such objects.

So this suggests that what you are looking for is just $\mathbb{Q}^*\times \mathbb{Z}$. And indeed, you can check that $\mathbb{Q}^*\times \mathbb{Z}\to G$ given by $(a,z)\mapsto (x\mapsto ax^z)$ is an isomorphism (surjectivity is by definition, and injectivity is easy).

As for $\mathbb{Q}$-linear combinations, well the elements of $G$ are basically Laurent monomials, so it is easily seen to be the Laurent polynomial ring $\mathbb{Q}[X,X^{-1}]$, which is isomorphic to the group ring $\mathbb{Q}[\mathbb{Z}]$.

PS : Your notation for groups is extremely weird, I don't know if it is standard in some area of mathematics, but I think most people will find it confusiong.