# A question about Group Rings

Let $R:=\mathbb{Z}_p[C_{p^\infty}]$ be the group ring of a Prufer group over the field of integers module a prime $p$.

We have $C_{p^\infty}=\langle u_1, u_2, ..., u_n, ... |\,\,\,\, u_1^p=1,\,\,u_{i+1}^p=u_i,\,\,\, i=1,2,...\rangle$.

$(u_{k}-1)^m=0$ if and only if $m\geq p^{k}$.

I solved one implication ($m\geq p^{k}$ implies $(u_{k}-1)^m=0$) but I cannot get the other.

If $(x-1)^m=0$, then taking the minimum $n\in \mathbb{N}$ such that $m\leq p^n$ we find that $(x-1)^{p^n}=(x^{p^n}-1)=0$ and hence that $n\geq k$. Hence we can assume that $p^{k-1}<m<p^k$.

If $m$ is not coprime with $p$ then $m=p^ch$ and $h<p^{k-c}$. Then (working by induction on the order of $x$) from $0=(x-1)^{p^ch}=(x^{p^c}-1)^h$ we deduce that $h\geq p^{k-c}$ that is absurd. Hence we can also assume that $m$ is coprime with $p$.

• For the first problem: note that if $x^m = 0$ in a ring, then $x^{p^n} = 0$ for some $n$ (just pick $n$ large enough that $p^n\geq m$. – Tobias Kildetoft Mar 28 '14 at 11:54
• I wrote "assume for simplicity" because I was thinking at some generalization of it, but maybe it will be better without it. About the first problem, I don't get your suggestion. The $k$ is fixed in my statement: $p^k$ is the order of $u_k$. I want to prove that $(u_k-1)^m=0$ implies that $m$ must be greater than $p^k$. Have I misunderstood? – W4cc0 Mar 28 '14 at 12:01
• So, you have that it is $0$ for some $m$, and hence that it is $0$ for some $p^n$. Now go back to conclude that $n\geq k$. – Tobias Kildetoft Mar 28 '14 at 12:02
• I think I get it now: we have to take the minimum $n$ such that $m\leq p^n$ and then we have the result by contradiction because $(u_k-1)^{p^n}=(u_k^{p^n}-1)$ which is different from $0$ when n is strictly lower than $k$. Thanks! (If I understood that well) – W4cc0 Mar 28 '14 at 12:07
• Yes, precisely. – Tobias Kildetoft Mar 28 '14 at 12:10

Let $m \ge 0$. Write $$m = \sum_{i=0}^{\ell} a_i p^i, \quad 0 \le a_i < p, \quad a_{\ell} > 0.$$ This decomposition is just $m$ in base $p$, thus is unique. It follows that $$(u_k-1)^m = (u_k-1)^{\sum_{i=0}^{\ell} a_i p^i} = \prod_{i=0}^{\ell} (u_k^{p^i} - 1)^{a_i} = \prod_{i=0}^{\ell} (u_{k-i} - 1)^{a_i}.$$ Since all these factors are zero for $i \ge k$, we see that $(u_k-1)^m = 0$ if $\ell \ge k$, i.e. $m \ge p^k$.
If $m < p^k$, note the interesting fact that since each term in the expansion of $(u_{k-i}-1)^{a_i}$ has the coefficient $\binom{a_i}j (-1)^j$ for $0 \le j \le a_i$ (which is non-zero in $\mathbb Z/p\mathbb Z$), writing $\prod_{i=0}^{\ell} (u_{k-i}-1)^{a_i}$ over the basis you gave for the group ring gives a non-zero term of the form $b u_k^{b_0} \cdots u_{k-\ell}^{b_{\ell}}$ with $0 < b < p$ and $0 \le b_i \le a_i$. In other words, $$(u_k-1)^m = \prod_{i=0}^{\ell} (u_{k-i}-1)^{a_i} = \prod_{i=0}^{\ell} \left( \sum_{b_i=0}^{a_i} \binom{a_i}{b_i} (-1)^{b_i} u_{k-i}^{b_i} \right) = \underset{0 \le b_i \le a_i}{\sum_{0 \le i \le \ell}} \underset{\neq 0}{\underbrace{\left( \prod_{i=0}^{\ell} \binom{a_i}{b_i} (-1)^{b_i} \right) }}\left( \prod_{i=0}^{\ell} u_{k-i}^{b_i} \right).$$ The basis elements $\left( \prod_{i=0}^{\ell} u_{k-i}^{b_i} \right)$ being linearly independent, you see that $(u_k-1)^m \neq 0$.