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I have been told that the $\mathbb{Z}$ module $\mathbb{Z}/2\mathbb{Z}=\{0,1\}$ isn't free.

For a module to be free, there must exist a subset such that every element is expressible as a finite linear combination of elements from that subset, and that subset must be linearly independent. So for the case of $\mathbb{Z}/2\mathbb{Z}$, why isn't the set $\{1\}$ a basis, since $0=0\cdot1$ and $1=1\cdot1$?

Thanks for any replies

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up vote 6 down vote accepted

The set $\{1\}$ isn't independent, because $2(1)=1+1=0$. (The $2$ here lives in $\Bbb{Z}$, so the "multiplication" is the module action; all $1$s and $0$s live in $\Bbb{Z}/2\Bbb{Z}$.)

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Ah ok I see, thanks! – Lammey Aug 11 '14 at 6:31

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