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I'm trying to solve this for a problem and I need to know if what I have done is right:

Let $B$ be a torsion-free abelian group. Then we consider the set $$A=\{(b,n):b\in B,n\in \mathbb Z,n\neq 0\}$$ and define $$(b,n)\sim (a,m) \text{ iff } bm=an.$$ This yields an equivalence relation, now you can define addition of classes by $(b,n)+(a,m)=(am+bn,nm)$. Then $(A,+)$ is a torsion-free abelian group and $B$ can be embedded into $(A,+)$, but also $(A,+)$ is divisible, so $(A,+)$ can be embedded in a direct sum of copies of $\mathbb Q$.

Thank you for your time.

darij grinberg
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Родион Раскольников
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    You should also know that $(A,+)$ torsion-free and divisible implies it is uniquely divisible (for any $a\in A$ and $n\in{\bf Z}$ nonzero there is precisely one $\bar{a}$ such that $n\bar{a}=a$; check that this allows a well-defined action of $\bf Q$) and hence it is a $\bf Q$-vector space. – anon Jun 02 '13 at 01:11
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    Not only can your $A$ be embedded in a direct sum of copies of $\mathbb Q$, it's actually isomorphic to a direct sum of copies of $\mathbb Q$. As anon said, $A$ has the structure of a $\mathbb Q$-vector space. Any vector space (over any field) has a basis, which gives you an isomorphism to a direct sum (indexed by the basis) of copies of the field. – Andreas Blass Jun 02 '13 at 01:16
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    How can I embed $\prod_{i\in\mathbb{Z}}\mathbb{Q}$ in a direct sum of some $\mathbb{Q}$'s? – Ash GX Jun 02 '13 at 08:55
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    @AshGX The product is a $\mathbb{Q}$-vector space, so it has a basis (if you assume the axiom of choice). – egreg Jun 02 '13 at 09:17
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    The construction you're doing is exactly the same as building $A\otimes_{\mathbb{Z}}Q$. – egreg Jun 02 '13 at 09:18
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    @РодионРаскольников: Another method can be found in http://math.stackexchange.com/a/425493/36434 – Seirios Jun 24 '13 at 05:29
  • Just to add to Seirios's comment: this question has already been asked and answered here. – Matt E Jul 26 '13 at 04:15

1 Answers1

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Here is an answer using tensor products, inspired in egreg's comment above.

By corollary 4.27 in this expository article by Keith Conrad, the map $B\to B\otimes_\mathbb{Z} \mathbb{Q}$, $b\mapsto b\otimes 1$ is injective, since $B$ is torsion-free. Now, the abelian group $B\otimes_\mathbb{Z} \mathbb{Q}$ has a $\mathbb{Q}$-vector space structure (it is the $\mathbb{Q}$-extension of scalars of $B$), hence it is isomorphic to a direct sum of copies of $\mathbb{Q}$.

darij grinberg
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Bruno Stonek
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