Is every field of characteristic zero a divisible group (under addition)?

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?

(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.