Freedom

Every morphism f:\Bbb{Z}^\Bbb{N}\to A of abelian groups vanishing on \Bbb{Z}^{(\Bbb{N})} is identically zero.

Use this fact to prove that \Bbb{Z}^\Bbb{N} is not a free abelian group.

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Freedom

2 thoughts on “Freedom

  1. As for the first part: suppose f is a morphism as described in the statement. Since 2^n and 3^n are coprime for all n, for any sequence (a_n)\in\Bbb{Z}^\Bbb{N} we can write (a_n) = (2^n b_n) + (3^n c_n). Now, we have that f((2^n b_n)) = f((b_0, 2b_1, 4b_2, \dots, 2^k b_{k+1},0,0,\dots)) + f((0,0,\dots,0,2^{k+1}b_{k+2},2^{k+2}b_{k+3},\dots)). The first summand vanishes, since f is zero over \Bbb{Z}^{(\Bbb{N})}. Now, the second summand is equal to 2^{k+1}f((0,0,\dots,b_{k+2}, 2b_{k+3},\dots), so 2^{k+1} divides f((2^n b_n)) for all k, and so f((2^n b_n))=0. A similar argument shows that f((3^n c_n))=0 as well, and so f is identically zero, as we wanted to prove.

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