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Is $M=\left\{\frac{a}{2^n}| a \in \mathbb{Z}, n \in \mathbb{N}\right\}$ a free $\mathbb{Z}$-module? I am struggling with free modules and am not sure how I would check this? I know that the basis cannot be finite from another problem I have worked before. However, I am not sure about the infinite basis.

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Since you already know that a basis, if one existed, would have to be infinite, you can complete the proof by noting that your group doesn't contain even two (let alone infinitely many) linearly independent elements.

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Suppose $A$ is a free $\Bbb{Z}$-module, and $b$ is an element of a basis. Then there is no element $a\in A$ such that $2a=b$, since otherwise $b$ would have two different representations in terms of basis elements. Thus in your $M$ there is nothing that could possibly be a basis element, so there is no basis.

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  • $\begingroup$ But you did not assume $a$ or $2$ was a basis element so why is your second sentence valid? $\endgroup$ – Leo Spencer Jul 3 '13 at 18:23
  • $\begingroup$ $2$ isn't an element of the module at all, it's an element of $\Bbb{Z}$. And $a$ need not be a basis element, but it is a linear combination of basis elements. $\endgroup$ – Chris Eagle Jul 3 '13 at 18:24
  • $\begingroup$ of course on both points. $\endgroup$ – Leo Spencer Jul 3 '13 at 18:25

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