Identify all units of the group ring $\mathbb{Q}(G)$ where $G$ is an infinite, cyclic group. This question is related to this one, in that I am asking about the same problem, but not necessarily about the same aspect of the problem.
I need to identify all units of the group ring $\mathbb{Q}(G)$ where $G$ is an infinite cyclic group.
Now, as I understand it, if $R$ is a ring and $G$ is a group, then if we consider the set of all formal sums 
$$r_{1}g_{1}+r_{2}g_{2}+\cdots + r_{k}g_{k},$$ 
$r_{i} \in R$, $g_{i} \in G$, where we allow the empty sum to play the part of the zero element $0$, 
then if we consider two formal sums to be equivalent if they have the same reduced form, the group ring $R(G)$ refers to the set of equivalence of classes of such sums with respect to this equivalence relation.
In this case, then, since $\mathbb{Q} = R$ and $G$ is some infinite cyclic group, say $\langle x \rangle$ (although, if it is an infinite cyclic group, couldn't we say that it is isomorphic to $\mathbb{Z}$?), so our sums look like 
$$q_{1}x_{1} + q_{2}x_{2} + \cdots + q_{k}x_{k}$$
for some rationals $q_{i}$ and elements of $\langle x \rangle$, $x_{i}$. 
Now, the units of this group ring are the nonzero, invertible elements, and that various relationships exist among units, principal ideals, and associate elements. I am not sure how to apply any of this information to this situation, though, as I am relatively inexperienced with working with group rings.
Moreover, I did find the answered question I linked to above, but this answer uses some terminology that I am unfamiliar with: for example, I do not know what it means to be a "localization of \mathbb{Q}[x]", and I only know a little bit about Laurent polynomials from Complex Analysis, which I'm assuming is where he is getting the negative powers from in his answer.
Now, among my questions is: 1. How do you know that $\mathbb{Q}(G)$ is isomorphic to $\mathbb{Q}[x, x^{-1}]$? That it is seems weird to me, since $G$ here is supposed to be cyclic, and he seems to be saying that a group ring on a cyclic group is isomorphic to a group ring on a group with two generators, but perhaps my confusion just stems from my inexperience with group rings? If someone could please explain this to me, I would be forever grateful. Also, 2. what is the actual isomorphism used or how do you show that the two group rings are isomorphic? 3. How does this tell us what the units are?
I'm extremely confused and I thank you very much in advance for your time, help, and patience!
 A: To clarify $\mathbb{Q}G$ is a group ring and $\mathbb{Q}[x,x^{-1}]$ is a ring of Laurent polynomials $p(x) = \sum_{i\in \mathbb{Z}} q_i x^i$. Since $G$ is a cyclic infinite group we write each $a\in G$ as $a = c^i, \,\, i\in \mathbb{Z}$, in the multiplicative form where $c$ is a generator of $G$. The map $c^i \mapsto x^i$ induces a ring iso between $\mathbb{Q}G$ and $\mathbb{Q}[x,x^{-1}]$. You should be able to check the details from here.
For the general notion of a localization of a ring there is much to say, see e.g. https://en.wikipedia.org/wiki/Localization_of_a_ring
A: For a field $K$ and a group $G$, consider the formal finite sums of the form $$\sum_{g \in G} a(g) [g], \qquad a(g) \in K $$
Then this set becomes a ring $K[G]$ with the obvious addition : 
$$\sum_{g \in G} a(g) [g] +\sum_{g \in G} b(g) [g]= \sum_{g \in G} (a(g)+b(g))[g]$$
And the  multiplication generated by $[g][g'] = [gg']$ the group law of the ring, so that :
$$(\sum_{g \in G} a(g) [g] )(\sum_{g' \in G} b(g') [g'])= \sum_{g \in G} [g]\sum_{h \in G} a(h)b(h^{-1}g)$$
(this multiplication is commutative iff $G$ is a commutative group)
Assume that $G = (\mathbb{Z},+)$ then we have a ring isomorphism $$\varphi : K[G] \to K[x,x^{-1}], \qquad \varphi(\sum_{g \in \mathbb{Z}} a(g) [g]) = \sum_{g \in \mathbb{Z}} a(g) x^g$$
And its units are the elements of the form $a(g) x^g, a(g) \in K^\times$, because the only polynomials being inversible in $K[x,x^{-1}]$ are the powers of $x$ times a constant (think in term of zeros/poles in the algebraic closure)
