Solution: Day 2, problem 3

We define a map F: Hom(P, Z) to P by assigning to a homomorphism f: P --> Z the sequence (r₁, r₂, ... ) with r_n = f( 0, ... ,0,1,0, ... ) (1 in the nth place). We must show that F is injective and that its image is exactly S. Suppose that F(f) = 0. Then f vanishes on any finite sequence (i.e. on the subgroup S of P). Let x = (x₁, x₂, ... ) be an arbitrary element of P and define a decomposition x = a + b by writing x_n = a_n + b_n with 2ⁿ | a_n, 3ⁿ | b_n. Then for each positive integer n the sequence a differs by a finite sequence from a sequence divisible by 2ⁿ, so f(a) must be divisible by 2ⁿ for every n and hence must vanish. Similarly f(b) = 0, so f(x) = f(a) + f(b) also vanishes. This proves that F is injective.

Clearly S is a subset of Im(F), since a sequence (r₁, r₂, ... ) with almost all r_n equal to 0 is the image under F of the homomorphism (a₁, a₂, ... ) |--> Sigma r_na_n from P to Z.
To show the reverse inclusion,let f be in Hom(P, Z) and r_n = f(0, ... ,0,1,0, ... ) (n >= 1). We must show that almost all r_n vanish. Assume not, and define a = (a₁,a₂, ... ) where each a_n is of the form +or- 2^s_n with the sign chosen so that r_na_n >= 0 and the exponent so that 2^s_n > 2^{s_{n - 1} + 1}|r_n|. Then the integer f(a) is congruent modulo 2^{r_{n + 1}} to 2^s₁|r₁| + ... + 2^s_n |r_n|, and the binary expansion (in standard form, with all coefficients 0 or 1) of this is the concatenation of the binary expansions of r₁, ... , r_n, with 0 interspersed. Taking the 2 - adic limit of this statement as n goes to infinity gives a contradiction, since f(a) is an ordinary integer and hence has a (standard) binary expansion consisting of all 0's or all 1's from some point onwards.

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Don Zagier 1996