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Consider the abelian group $\mathbb{Z}^\mathbb{Z}$ under component-wise addition, i.e. the group of formal power series under addition. I am interested in finding a subgroup $H$ of this group that satisfies

  1. Any two elements of $H$ differ in infinitely many places; that is, the difference between them is never a polynomial;
  2. Any element of $\mathbb{Z}^\mathbb{Z}$ can be written as an element of $H$ plus a polynomial.

(1) implies that the representation in (2) is unique.

I had attempted to find $H$ using the fact that every vector space has a basis on the vector space of power series over $\mathbb{Q}$, but under closer inspection this doesn't work. (I can achieve (2) as long as the basis includes $1, x, x^2,$ etc., by throwing them out; however, there's no way to satisfy (1).)

My other attempt was to show there is no such $H$ by pigeonhole. We know $|H| = |\mathbb{R}|$. Thus by pigeonhole, find $|\mathbb{R}|$ elements of $H$ which have the same last digit. Among those elements, find $|\mathbb{R}|$ which have the same last two digits. This can continue on for any finite number of digits. I think we may assume that these finite number of shared digits are all nonzero, but I don't know what to do next.

I am fairly certain the idea of such a subset $H$ is a common group theory notion but I only know basic group theory myself. This also seems like it would be a well-known problem. Can someone help me here?

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up vote 16 down vote accepted

If such an $H$ exists, it must be isomorphic (as a group) to the quotient $\mathbb{Z}[[x]]/\mathbb{Z}[x]$. In the quotient, the element $$ \alpha:=\sum_{n\geq 0}n!x^n+\mathbb{Z}[x]\in\mathbb{Z}[[x]]/\mathbb{Z}[x] $$ has the property that $\alpha\neq 0$, but for any integer $n\geq 1$, $\alpha$ is a multiple of $n$. No element of $\mathbb{Z}[[x]]$ has this property, so the subgroup $H$ cannot exist.

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+1 A great idea! – Hagen von Eitzen Jul 23 '13 at 16:59

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