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Let $G = \langle x\rangle$ be the infinite cyclic group generated by an element $x$. The group ring $R = Q(G)$ consists of finite sums of the form $r_{−m}x^{−m} + r_{−m+1}x^{−m+1} + · · · + r_{−1}x^{−1} + r_0 + r_1x + · · · + r_nx^n$ where $m, n ≥ 0$ and $r_{−m}, r_{−m+1},\ldots, r_n$ are rational numbers. The ring of polynomials $Q[x]$ is naturally embedded in R. I need to:

(a) Prove that any element $f ∈ R$ is an associate to a polynomial $\overline f ∈ Q[x]$.

(b) Prove that R is an integral domain

I was able to achieve a by setting $\overline f = x^mf$, but I am a little stuck on b.

We are given the definition, a ring R with identity is an integral domain if the identity element of R is non-zero and R has no zero divisors. So obviously the identity is 1, which is nonzero. Now to show that there are no zero divisors, I need to show that for $f,g\in R, (fg = 0 \rightarrow f= 0$ or $g = 0$). So let $f = \sum_{i=-m}^nr_ix^i$ and $g = \sum_{j=-k}^ps_jx^k$. Then multiplying them together gives $fg = \sum_{i=min\{m,k\}}^{max\{n,p\}}c_ix^i$ (I think. Honestly, this could be wrong.) Now I have no idea where to go from here

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  • $\begingroup$ For integral domainness just think about leading coefficients. $\endgroup$ – Lord Shark the Unknown Dec 16 '17 at 19:13
  • $\begingroup$ @LordSharktheUnknown so here the coefficients $r_is_j$ would need to sum up to 0, but I don't know how to guarantee that since we're not given they have to be positive. Could you elaborate? $\endgroup$ – Vinny Chase Dec 16 '17 at 19:18
  • $\begingroup$ Shark said the leading coefficients. There is no 'summing to zero' condition on them. | Alternatively, you know every element is associate to a polynomial... do you know the polynomial ring $\Bbb Q[x]$ is a domain? $\endgroup$ – anon Dec 16 '17 at 19:40
  • $\begingroup$ @anon Oh I read it wrong. Yes thank you I got it! $\endgroup$ – Vinny Chase Dec 16 '17 at 20:20

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