Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

1 Answer 1

(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.

share|improve this answer

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.