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?
Is every field of characteristic zero a divisible group (under addition)?
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5nope. your right on. – 2011-07-03
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0@jspecter: Thanks, the confirmation helps a lot. – 2011-07-03
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0I 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? =) – 2011-07-03
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1Wrong "your." Ahhhhh. (I would say the "F-word" but this is a family site.) – 2011-07-03
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1@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. – 2011-07-03
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13Dear 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. – 2011-07-03
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0What Georges said! – 2011-07-03
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2Oh, 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! – 2011-07-03
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0@Georges, @Alex and @Patrick: Thanks for the vote of confidence. – 2011-07-04
1 Answers
(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.