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This question was asked in my module theory assignment and I need help in proving it.

Every $\mathbb{Q}$-vectorspace $V \neq 0$ is not free over the subring $\mathbb{Z} \subseteq \mathbb{Q}$.

I assumed that V is free over $\mathbb{Z}$ in hope of getting a contradiction. So, there exists a set {$ a_1,..., a_n$} all in $\mathbb{Z}$ such that $v= a_1 x_1 +... + a_n x_n$, {$v_1,..., v_n$} $\in V$.

I am not able to find a contradiction as I am unable to proceed foreward. All I can say is that V is now not a vector space as $\mathbb{Z} $ is not a field.

Can you help me completing the proof?

3 Answers3

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Dietrich Burde essentially already proved the result. Here are some more details. First we show the claim for finite vector spaces. So let $V$ be a finite-dimensional $\mathbb{Q}$ vector space, $n=\dim_\mathbb{Q} V$. Suppose $V$ was free as a $\mathbb{Z}$-module, then $\mathbb{Q}\subset V=\mathbb{Q}^n$ is a submodule of a free $\mathbb{Z}$-module hence it is itself free. But $\mathbb{Q}$ is not free as a $\mathbb{Z}$-module, since any two rationals are linearly dependent and clearly $\mathbb{Q}$ is not generated by one rational as a $\mathbb{Z}$-module.

To deal with the infinite dimensional case it should suffice to write $V$ as the union of its finite-dimensional subspaces.

nilsw
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Every $\mathbb{Q}$-vector space $V$ is divisible as an abelian group: for every $x\in V$ and every nonzero integer $n$, there exists $y\in V$ such that $ny=x$.

Claim. No nonzero free abelian group is divisible.

Indeed, if $G$ is a divisible abelian group, then the only homomorphism $G\to\mathbb{Z}$ is the trivial one, so $\mathbb{Z}$ cannot be a homomorphic image of $G$. But $\mathbb{Z}$ is a homomorphic image of every nonzero free abelian group.

egreg
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Consider a $\mathbb Q$-vector space $V\neq0$ and suppose, toward a contradiction that it is free over $\mathbb Z$, say with basis $B$. Consider any basis vector $b\in B$. Since $V$ is a $\mathbb Q$-vector space, we can consider the element $\frac12b$ of $V$, and we can express it as a $\mathbb Z$-linear combination $\sum_in_ib_i$ of some elements $b_i$ of $B$ (possibly including $b$ itself as one of the $b_i$'s). So we have a linear dependence relation $b-\sum_i2n_ib_i=0$ among the basis vectors $b$ and $b_i$, in which the basis vector $b$ has coefficient $1$ (if $b$ is not among the $b_i$'s) or $1-2n_i$ (if $b=b_i$). In any case, the coefficient of $b$ is odd and, in particular, not $0$. That contradicts the fact that a basis $B$ has to be linearly independent.

Andreas Blass
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