Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{Z}$?

share|cite|improve this question
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.

share|cite|improve this answer

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 .

share|cite|improve this answer

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.

share|cite|improve this answer

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 $3/4$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.