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I am convinced about this, directly from the definition, but I found it strange that I could not find a reference. Every field of characteristic zero contains $\mathbb{Q}$ and hence given and natural number $n$ and any element field element, say $g$, we can write $f=g*\frac{1}{n}$ and then, $nf=g$. Am I doing something wrong here?

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nope. your right on. – jspecter Jul 3 '11 at 4:18
Wrong "your." Ahhhhh. (I would say the "F-word" but this is a family site.) – jspecter Jul 3 '11 at 4:30
@Patrick Da Silva: I am not a mathematics major and I needed the result for a specific field of characteristic zero, but then noticed it should work for any field of characteristic zero. As I said, I was convinced of the proof (elementary as it is), but my advisor wanted a reference and I could not find anything. – B M Jul 3 '11 at 5:04
Dear Brittany, it is quite praiseworthy to have generalized your result from a specific example to a field of characteristic zero. Asking for confirmation here is a sign of modesty and a quite healthy initiative, definitely not "weird". +1 and welcome to math.stackexchange. – Georges Elencwajg Jul 3 '11 at 9:29
Oh, actually I thought it was weird because I was more on the "exercise" point of view. If you generalized this from an example then I understand the possible lack of confidence. Not weird at all. Even though people might say "yes, this is true" or "yeah I know what you mean", they might not find that or think of noticing that themselves, which is why sites like mathstackexchange exist, so that our eyes can see. +1! – Patrick Da Silva Jul 3 '11 at 15:06

(To remove this from "unanswered".)

No, you are right. Every $\mathbb{Q}$-algebra, including every ring containing $\mathbb{Q}$ as a (unital) subring (with the same unity), is a divisible group precisely because it contains an element that behaves like $\tfrac{1}{n}$.

More generally an $R$-algebra is an $R$-module, which gives information on the additive structure of the algebra. A $\mathbb{Q}$-module is a vector space over $\mathbb{Q}$, and so must be a direct sum of copies of $\mathbb{Q}$, and so is a divisible, torsion-free abelian group.

A similar argument shows that every ring of characteristic p has an additive group that is a direct sum of copies of $\mathbb{Z}/p\mathbb{Z}$ and so is divisible by every integer coprime to p.

Oddly a (unital, associative) algebra with a divisible additive group must in fact a torsion-free divisible additive group and be a $\mathbb{Q}$-algebra. Some proofs are given as answers to this question, but your argument will work just as well here in reverse.

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