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I'm trying to determine whether $\mathbb Q$ under addition has a basis. My naive idea, based on what I have covered so far, would be to try and show that if we assume $\mathbb Q$ has a basis then it would be finitely generated which I believe would then lead to a contradiction. But I don't think this is a valid approach and I'm not sure what other approaches I could take.

I'd appreciate some ideas/approaches about determining whether $\mathbb Q$ has a basis or not :)

Hai
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  • What do you mean by a "basis" in this context? That's not normally a group theoretic concept. If you are thinking of $\mathbb Q$ as a vector space over itself, then of course it has a basis (any non zero element). – lulu Nov 11 '20 at 13:21
  • As a Z-module ? – Anthony Nov 11 '20 at 13:21
  • A vector space is defined over a scalar field. – Wuestenfux Nov 11 '20 at 13:21
  • It seems like this may be what you're trying to do. – Cameron Buie Nov 11 '20 at 13:23
  • I am talking about a basis in the same sense that every finitely generated abelian group has a basis. I believe what @CameronBuie is referring to. – Hai Nov 11 '20 at 13:24
  • But as you remark, $\mathbb Q$ is not finitely generated. – lulu Nov 11 '20 at 13:26
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    As a Z-module its not free (has no basis) as demonstrated by the link. – Wuestenfux Nov 11 '20 at 13:27
  • Not even two nonzero elements of $\mathbb Q$ can be linearly independent over $\mathbb Z$: take $p/q, r/s\in\mathbb Q\setminus{0}$, then $rq(p/q)-ps(r/s)=0$, with $rq, ps\in\mathbb Z\setminus{0}$. Thus, whatever set you come up with, it won't be a base. –  Nov 11 '20 at 13:28
  • @lulu Yes, I just mean a basis in the sense that there is a set ${x_i}_{i \in I}$ generating the group $Q$ where the elements are linearly independent with coefficients from $\mathbb Z$. I am not sure if this relates to modules since I have not covered them yet. – Hai Nov 11 '20 at 13:31
  • @Hai: it's not true that every finitely generated abelian group has a basis. The ones that do are exactly the free (equivalently, torsion-free) ones. $\mathbb{Q}$ does not have a basis either: it's not generated by one element but any two elements are linearly dependent. – Qiaochu Yuan Nov 11 '20 at 19:11

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You haven't defined what a "basis" is.

Here's, however, a statement showing that $\mathbf{Q}$ is far to having a basis as abelian group for any reasonable notion of basis as group:

For every generating subset $S$ of $\mathbf{Q}$, for every finite subset $F$ of $S$, the subset $S-F$ generates $\mathbf{Q}$.

(I assume that a basis would at least have the property that it's a generating subset in which no element is redundant; the above shows it can't exist.)

The claim is true: indeed, the quotient $H=\langle S\rangle/\langle S-F\rangle$ is a finitely generated quotient of $\mathbf{Q}$. Hence, if nonzero, it admits for some prime $p$ a cyclic group $C_p$ of order $p$ as quotient. But since $p\mathbf{Q}=\mathbf{Q}$, it cannot admit $C_p$ as quotient, so $H=0$, that is, $\langle S-F\rangle=\langle S\rangle$.

YCor
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