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Solution: Day 2, problem 3

We define a map ϕ:Hom(P,Z)\phi : Hom(P, \mathbb{Z}) to PP by assigning to a homomorphism
f:P→Zf: P \rarr \mathbb{Z} the sequence (λ1,λ2,...)( \lambda _{1}, \lambda _{2}, ... ) with λn=f(0,...,0,1,0,...)\lambda _{n} = f( 0, ... ,0,1,0, ... ) (1 in the nnth place). We must show that ϕ\phi is injective and that its image is exactly SS. Suppose that ϕ(f)=0\phi (f) = 0. Then ff vanishes on any finite sequence (i.e. on the subgroup SS of PP). Let x=(x1,x2,...)x = (x_{1}, x_{2}, ... ) be an arbitrary element of PP and define a decomposition x=a+bx = a + b by writing xn=an+bnx_{n} = a_{n} + b_{n} with 2n∣an,3n∣bn2^{n} | a_{n}, 3^{n} | b_{n}. Then for each positive integer nn the sequence aa differs by a finite sequence from a sequence divisible by 2n2^{n}, so f(a)f(a) must be divisible by 2n2^{n} for every nn and hence must vanish. Similarly f(b)=0f(b) = 0, so f(x)=f(a)+f(b)f(x) = f(a) + f(b) also vanishes. This proves that ϕ\phi is injective.

Clearly S⊆S \subseteq Im(φ), since a sequence (λ1,λ2,...)( \lambda _{1}, \lambda _{2}, ... ) with almost all λn\lambda _{n} equal to 0 is the image under ϕ\phi of the homomorphism (a1,a2,...)↦∑λnan(a_{1}, a_{2}, ... ) \mapsto \sum \lambda _{n}a_{n} from PP to Z\mathbb{Z}.
To show the reverse inclusion, let ff be in Hom(P,Z)Hom(P, \mathbb{Z}) and λn=f(0,...,0,1,0,...)(n≥1)\lambda _{n} = f(0, ... ,0,1,0, ... ) (n ≥ 1). We must show that almost allλn\lambda _{n} vanish. Assume not, and define a=(a1,a2,...)a = (a_{1},a_{2}, ... ) where each ana_{n} is of the form ± 2sn with the sign chosen so that λnan≥0\lambda _{n}a_{n} ≥ 0 and the exponent so that 2sn>21+sn−1∣λn∣2^{s_{n}} > 2^{1+s_{n-1}} | \lambda _{n}|. Then f(a)∈Zf(a) \isin \mathbb{Z} is congruent modulo 2sn+12^{s_{n+1}} to 2s1∣λ1∣+...+2sn∣rn∣2^{s_{1}}| \lambda _{1}| + ... + 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 λ1,...,λn\lambda _{1}, ... , \lambda _{n}, with 0 interspersed. Taking the 2 - adic limit of this statement as n→∞n \rarr ∞ gives a contradiction, since f(a)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.