$R$ a non-unitary ring such that $R^+ \cong \mathbb{Q}/\mathbb{Z}$

Question

We are given $R$ a non-unitary ring such that $R^+ \cong \mathbb{Q}/\mathbb{Z}$. Prove: $ab=0$ for all $a,b \in R$.

Here is what I have attempted so far:

We take two arbitrary elements of $\mathbb{Q}/\mathbb{Z}$. We can write them as $\frac{k}{r},\frac{k'}{r'}$, with the relations $k=a-rq, k'=b-r'q'$ for some $a,b,r,r',q,q',k,k' \in \mathbb{Z}$.

Multiplying them, we obtain $$\frac{kk'}{rr'} = \frac{(a-rq)(b-r'q')}{rr'} = \frac{ab}{rr'} - \frac{aq'}{r} - \frac{bq}{r'} + qq' \equiv \frac{ab}{rr'} - \frac{aq'}{r} - \frac{bq}{r'} \in \mathbb{Q}/\mathbb{Z}$$ Since $qq' \in \mathbb{Z}$.

From here, I cannot manage to show that $\frac{kk'}{rr'} \equiv 0$ since the term $\frac{ab}{rr'}$ contains the term $\frac{kk'}{rr'}$. Filling this relation in just ends up giving $0 \equiv -2qq'$, which makes sense since $qq'\in \mathbb{Z}$.

But then I thought more about what $\mathbb{Q}/\mathbb{Z}$ looks like, and I am now convinced that it is equal to $(-1,1) \subset \mathbb{Q}$. Taking $\frac{1}{3},\frac{1}{2} \in \mathbb{Q}/\mathbb{Z}$ we have $\frac{1}{2}\frac{1}{3} = \frac{1}{6} \not \equiv 0$. This leads me to believe that the claim to be proven is in fact not true.

I also have the feeling that I should be using the non-unitary-ness of $R$. I read elsewhere that $\mathbb{Q}/\mathbb{Z}$ is an abelian group that is not isomorphic to the additive group of any ring, so here it must hold because of the non-unitary-ness of $R$.

I am very curious to hear any insights anyone out there might have about this claim, thanks in advance!

EDIT: 20/07/2021 I have marked Mindlack's answer as "accepted" since it is the first one that I understood, and intuitively made the most sense to me. I would have marked tomasz's answer as well if I could, as they provided a general way of looking at this problem and a neat proof. Also check the comments for more!

Answer 1

This translates to proving that there is no nonzero bilinear map $\beta: \mathbb{Q}/\mathbb{Z} \times \mathbb{Q}/\mathbb{Z} \rightarrow \mathbb{Q}/\mathbb{Z}$. But, if $m,n \geq 1$, $\beta(1/m,1/n)=\beta(n/mn,1/n)=n\beta(1/mn,1/n)=\beta(1/mn,n/n)=0$.

Answer 2

The problem (both in your solution and the thing that makes you doubt) is that you confound the ring multiplication with the usual multiplication of rational numbers.

The assumption that the ring is non-unitary does not really mean much. In this case, the adjective weakens the definition: usually, every ring is a non-unitary ring, but not every non-unitary ring is a ring (most authors assume that rings are unitary by definition).

$\mathbf Q/\mathbf Z$ does not equal $(-1,1)\cap \mathbf Q$. Every element has a representative there (in fact, it has one in $[0,1)\cap \mathbf Q$, but it does not allow you to define any sort of ring structure.

I think the more general statement is the following: a (non-unitary) ring (in fact, a distributive algebra is enough, you don't need associativity) whose additive group is both divisible and torsion has to be a zero ring (meaning, its multiplication has to be identically zero). I think it's easier, since (by virtue of abstraction) it's harder to confuse the ring elements with rational numbers.

The proof is simple, and a bit similar to what you have written: take any $a,b$. You want to show that $ab=0$. Let $n\in \mathbf N$ be such that $na=0$ and let $b'$ be such that $b=nb'$. Then by distributivity $ab=a(nb')=n(ab')=(na)b'=0$.