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As field of reals $\mathbb{R}$ can be made a vector space over field of complex numbers $\mathbb{C}$ but not in the usual way. In the same way can we make the ring of integers $\mathbb{Z}$ as a vector space the field of rationals $\mathbb{Q}$? It is clear if it forms a vector space, then $\dim_{\mathbb{Q}}\mathbb{Z}$ will be finite. Now i am stuck. Please help me. Thanks in advance.

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Assuming that you want the usual addition operation on $\mathbb{Z}$ to be its addition operation as a $\mathbb{Q}$-vector space, then no, it is not possible for it to be a $\mathbb{Q}$-vector space. For example, we would have a scalar $\frac{1}{2}\in\mathbb{Q}$ and an element $\mathbf{1}\in\mathbb{Z}$ of our supposed vector space (using bold-face to distinguish elements of $\mathbb{Z}$ from elements of $\mathbb{Q}$), so we must be able to form their product $\frac{1}{2}\cdot \mathbf{1}\in\mathbb{Z}$. By the distributivity of scalar multiplication, $$\mathbf{1}=1\cdot\mathbf{1}=(\tfrac{1}{2}+\tfrac{1}{2})\cdot \mathbf{1}=(\tfrac{1}{2}\cdot \mathbf{1})+(\tfrac{1}{2}\cdot \mathbf{1})$$ However there is no integer $\mathbf{n}\in\mathbb{Z}$ such that $\mathbf{1}=\mathbf{n}+\mathbf{n}$, so this is a contradiction.

If instead we allow an arbitrary addition operation on $\mathbb{Z}$, then yes, we can make it a $\mathbb{Q}$-vector space, by "transport of structure" (Wikipedia link) via a bijection, e.g., a bijection $\varphi:\mathbb{Z}\to\mathbb{Q}$.

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  • $\begingroup$ Yes i got the point ....thanking you...nicely explained... $\endgroup$
    – neelkanth
    Jul 11, 2015 at 4:50
  • $\begingroup$ +1 for that "transport of structure" thing. $\endgroup$
    – Silent
    Apr 11, 2018 at 4:54

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