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I am looking at the following:

Show that a torsion-free divisible group $G$ is a vector space over $\mathbb{Q}$.

I have no problem verifying the axioms of vector spaces after noting that the term divisible group $G$ implies that a solution to $ny=x$ exist where $n$ is an integer and is unique by the fact that the group $G$ is torsion-free.

What does vector spaces over $\mathbb{Q}$ mean? And why does the question say that it is a vector spaces over $\mathbb{Q}$ instead of $\mathbb{Z}$?

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4 Answers 4

up vote 9 down vote accepted

A vector space is defined over a field. If you want to define something over a ring, it's known as a module.

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And why does the question say that it is a vector spaces over $\mathbb Q$ instead of $\mathbb Z$?

Notice that an Abelian group is a module over $\mathbb Z$ in a canonical way. The problem you quote says that this $\mathbb Z$-module structure extends canonically to a $\mathbb Q$-module or $\mathbb Q$-vector space structure provided that your group is also torsion-free and divisible .

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What does vector spaces over $\mathbb{Q}$ means?

Not sure, but are you asking what one actually means by a vector space $V$ over a field $F$? The field is the set of acceptable scalars for your vector space. So if the vector space is over $\mathbb{Q}$, you couldn't have $\pi$ as a scalar, for example, whereas you could have $3/4$.

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ok. i think i get your point. On the other hand the question ask why a torsion-free divisible group is a vector spaces over $Q$ and not vector spaces over $Z$. After looking at the answer above and checking the definition of a field only i realize that $Z$ is not a field. While the smallest field containing the integer is $Q$, right? sorry. next time i should do a bit more research when posing question like this. –  Seoral Jan 14 '11 at 5:38
@Seoral Yes, $\mathbb{Q}$ is the smallest field into which $\mathbb{Z}$ may be embedded. Look into field of fractions if you're curious to see this more generally. –  yunone Jan 14 '11 at 5:45
Ok, thanks for the verification. –  Seoral Jan 14 '11 at 6:15

Because it is talking about vector spaces over $\mathbb{Q}$ (the rational numbers). This is know as a $\mathbb{Q}$-vector space or more generally an $F$-vector space (where $F$ is an arbitrary field). See this for more details.

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