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The following question came up in my research. Since lots of clever people post here, I thought I'd ask it.

Recall that the group ring of a group $G$ is the abelian group $\mathbb{Z}[G]$ consisting of linear combinations of formal symbols $[g]$, where $g$ ranges over elements of $G$ (the abelian group $\mathbb{Z}[G]$ also has an obvious ring structure, but that's not important for this question).

Consider the group ring $\mathbb{Z}[\mathbb{Q}]$ of the rational numbers $\mathbb{Q}$ (considered as an additive group). There is a natural projection $\pi : \mathbb{Z}[\mathbb{Q}] \rightarrow \mathbb{Z}[\mathbb{Q}/\mathbb{Z}]$. It has a large kernel; for instance this kernel contains $[n]-[0]$ for integers $n$ and things like $[3/2]-[1/2]$. There is also a natural involution $i : \mathbb{Z}[\mathbb{Q} \setminus \{0\}] \rightarrow \mathbb{Z}[\mathbb{Q} \setminus \{0\}]$ defined by $i([q]) = [1/q]$. Here by $\mathbb{Z}[\mathbb{Q} \setminus \{0\}]$ I just mean formal sums of $[q]$ where $q$ is a nonzero rational number. We have a natural inclusion $\mathbb{Z}[\mathbb{Q} \setminus \{0\}] \subset \mathbb{Z}[\mathbb{Q}]$.

Question. What is $\text{ker}(\pi) \cap \text{ker}(\pi \circ i)$? It clearly contains things like $[1]-[-1]$, but I don't know if it contains any more "exotic" elements.

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What's $i([0])$? –  anon Mar 28 '12 at 3:22
    
Good point! I'll edit the question to make this make more sense. –  Lewis Mar 28 '12 at 3:30
    
Is this intersection meant to take place in $\mathbb{Z}[\mathbb{Q} \setminus \{ 0 \} ]$? –  Qiaochu Yuan Mar 28 '12 at 3:41
    
@QiaochuYuan : Yes. –  Lewis Mar 28 '12 at 3:45

1 Answer 1

I believe that that intersection of kernels contains, for any integer $k\notin\{0,-1\}$, the element $[1] - [k] + [\frac k{k+1}] - [\frac{-1}{k+1}]$.

I also found (just by messing around) the element $[\frac52] - [\frac57] + [\frac{-2}7] - [\frac23] + [\frac53] - [\frac{-5}2]$.

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Thanks! This is more complicated than I thought. I'll wait a couple of days to see if anyone gives generators for the kernel and then I'll accept your answer. –  Lewis Mar 28 '12 at 14:20

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