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In Dummit and Foote, the authors state that "Free $\mathbb{Z}$-modules have no nonzero elements of finite order."

Why is this true? If $M$ is a free $\mathbb{Z}$-module, then for any $x\in M$, there exists a subset $A\subseteq M$, unique nonzero elements $a_1,...,a_n\in A$ and unique $z_1,.,,,.z_n\in\mathbb{Z}$ such that $x=z_1a_1+\cdots +z_na_n$. Why does this imply that $x$ has infinite order assuming $x$ is nonzero?

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  • $\begingroup$ You should really spend some time looking for elements of finite order in, say, $\Bbb Z^2=\Bbb Z\oplus\Bbb Z$ $\endgroup$ – Lubin Apr 30 '18 at 3:57
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I am assuming that the set $A$ you refer to is the basis of the module $M$. Since $x$ is nonzero, you may assume without loss of generality that $z_1\ne0$. Let $n\in\Bbb N$ satisfy $nx=0$, then $$ 0\cdot a_1 +\cdots + 0\cdot z_n = 0 = nx = nz_1\cdot a_1 +\cdots + nz_n\cdot a_n. $$ This implies $nz_1=0$ because of the uniqueness of this expression and because $z_1\ne0$ we can conclude that $n=0$. Hence, there is no positive $n$ that annihilates $x$, which means that $x$ has infinite order.

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