Why $\zeta^m \alpha \in K[\zeta]$? In the following lemma from "The Algebraic structures of group rings" : by D.S. Passman,



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*What does $K[G]$ contained isomorphically between $K[\zeta_1, \ldots \zeta_n] $ and $K(\zeta_1, \ldots \zeta_n)$ means? Does it mean $K[G] \cong K(\zeta_1, \ldots \zeta_n)$?

*Now let me take the case where we have only one variable say $\zeta$ i.e. $G=\langle x \rangle$ , an infinite cyclic group. Then $K[\zeta] \hookrightarrow K\langle x \rangle= K[G]$ and according to the red line towards the end, if $\alpha = \sum_{-\infty}^{\infty} a_ix^i \in K<x>$ then there should exist a large enough $m$ such that $\zeta^m \alpha \in K[\zeta]$. Why does this $m$ exist?


I think as supp{$\alpha$} is finite, we can find one $m$ such that $x^m \alpha \in K\langle x \rangle =\ K[G]$, but why $\zeta^m \alpha \in K[\zeta]$? We have embedded $K[\zeta] $ in our group ring but it is not onto. 
May be if we first write $K[[]\zeta]]\ (\text{Laurent poly ring})\ \cong K \langle x \rangle$ , then writing  $\zeta^m \alpha \in K[\zeta]$ makes sense. May be this what assumed here in obvious way?
 A: (1) Let $\mu:K[\zeta_1,\ldots,\zeta_n]\to K[G]$ be the natural homomorphism given by $\zeta_i\mapsto x_i$. Then $\mu$ is an embedding and we will identify $K[\zeta_1,\ldots,\zeta_n]$ and $K[x_1,\ldots,x_n]$, i.e. $\zeta_i=x_i$. Further Passman proves, that $K[G]$ is an integral domain, hence we can consider field of fractions of any subring of $K[G]$, and it turns out that $K[G]$ contained in the field of fractions of $K[\zeta_1,\ldots,\zeta_n]$, i.e. in $K(\zeta_1,\ldots,\zeta_n)$.
(2) Let $\alpha\in K[G]$, then 
$$
\alpha=\sum_{i=1}^l \alpha_i\prod_{j=1}^n x_j^{e_{i,j}}=
\sum_{i=1}^l \alpha_i\prod_{j=1}^n \zeta_j^{e_{i,j}}
$$ for some $l\in\mathbb{Z}_{>0}$, $\alpha_i\in K$ and $e_{i,j}\in\mathbb{Z}$. If all $e_{i,j}\geq 0$, then $m:=0$, otherwise let $m=-\min\{e_{i,j}\}$. Denote $\zeta=\zeta_1\ldots\zeta_n$. Then $\zeta^m\alpha\in K[\zeta_1,\ldots,\zeta_n]$. 
Now we show that $K[G]$ is an integral domain. Let $\alpha,\beta\in K[G]$ and $\alpha\beta=0$. Then there exists $p,q\in\mathbb{Z}_{\geq 0}$ such that $\zeta^p\alpha,\zeta^q\beta\in K[\zeta_1,\ldots,\zeta_n]$, therefore $0=\zeta^p\alpha\zeta^q\beta$. But $K[\zeta_1,\ldots,\zeta_n]$ is an integral domain, hence $\zeta^p\alpha=0$ or $\zeta^q\beta=0$. Let $\zeta^p\alpha=0$. Obviously, $\zeta^p\in G$, hence $\zeta^p$ invertible and $\zeta^p\alpha=0$ follows $\alpha=0$.
Lastly let $\alpha\in K[G]$. Then $\zeta^m\alpha\in K[\zeta_1,\ldots,\zeta_n]$ for some $m\in\mathbb{Z}_{\geq 0}$
and $\alpha\in\frac{1}{\zeta^m}K[\zeta_1,\ldots,\zeta_n]\subseteq K(\zeta_1,\ldots,\zeta_n)$.
