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

Question

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?

I must admit it's kind of weird that with the proof in hand you don't understand that you got it, but yes, you got it. In a field of characteristic $0$, it is usually understood that $n$ (the "number" we wish to have in those fields) is defined recursively by $1 + (n-1)$ so that we have a copy of $\mathbb Q$ that comes from it as you stated. $n$ has an inverse, and writing $g = n (n^{-1})g = n f$ with $f$ defined as above , you have the exact definition of the divisibility of the group. Have you written $0$ characteristic because of an exercise or because you noticed what happens then? =) — Patrick Da Silva, Jul 03 '11 at 04:27
Wrong "your." Ahhhhh. (I would say the "F-word" but this is a family site.) — jspecter, Jul 03 '11 at 04:30
@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 03 '11 at 05:04
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 03 '11 at 09:29
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 03 '11 at 15:06
@Georges, @Alex and @Patrick: Thanks for the vote of confidence. — B M, Jul 04 '11 at 17:09

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

(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.

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

1 Answers1