# 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.

1. If such module $M$ were free, then $\hom(M,Z)$ would be uncountable. But by the statement, every map $M\to\mathbb Z$ is determined in a countable set, so such homset is at most countable.
2. 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.