Is the quotient group $ℝ/ℚ$ the additive group of a ring

hbghlyj

1. Show that if there is $u ∈ R$ with $2u = 1_R$ then $R$ has no element of order 2 in its additive group. Hence deduce that the quotient group $ℝ/ℤ$ is not the additive group of a ring.
2. Is the quotient group $ℝ/ℚ$ the additive group of a ring?

The following is my proof, unsure of part 2.

1. If $2x=0$, then $x = x1 = x(2u) = 2xu = 0u = 0$, so $(R, +)$ has no element of order 2.
   $(\frac12+ℤ)+(\frac12+ℤ)=ℤ⇒\frac12+ℤ$ has order 2 in $ℝ/ℤ$.
   Let $R$ be a ring with additive group $ℝ/ℤ$ and $1_R = e + ℤ$, then $2(\frac e2+ ℤ) = e + ℤ$, contradiction.
   The additive group ℚ/ℤ
2. $ℝ$ and $ℝ/ℚ$ are $ℚ$-vector spaces of same cardinality so $ℝ ≅ ℝ/ℚ$ is a ring.

Czhang271828

Question: Does the additive group $A:=\{x+\mathbb Q\mid x\in \mathbb R\}$ admit a ring structure with an identity?

Answer: Yes. Since $|\mathbb R|=2^{\aleph_0}=2^{\aleph_0}-1=|A|$, $A$ and $\mathbb R$ are isomorphic as $\mathbb Q$-linear spaces, given by the bijection
$$
\varphi:\mathbb R\overset \sim \to A, rs\mapsto ra\quad (\forall r\in \mathbb Q).
$$
Define the multiplication on $A$ by
$$
a\cdot a'=\varphi (\varphi ^{-1}(a)\cdot \varphi^{-1}(a')).
$$
Then we can verify

* $(a a') a''=\varphi (\varphi ^{-1}(a) \varphi^{-1} (a')) a''=\varphi ^{-1}(a) \varphi^{-1} (a')\varphi ^{-1}(a'')$, thus $(aa')a''=a(a'a'')$;
* $\varphi (1)a=\varphi(1\varphi^{-1}(a))=a$, thus $\varphi(1)a=a=a\varphi(1)$;
* the commutativity also holds,
* $a(a'+a'')=aa'+aa''$, by $\mathbb Q$-linearity of $\varphi$ and $\varphi^{-1}$.