Show that no commutative ring with identity has additive group isomorphic to \(\mathbb Q/\mathbb Z\).
Solution
1. Every element of the additive group \(\QQ/\ZZ\) has finite order.
Let \[ \frac ab+\ZZ \in \QQ/\ZZ, \] where \(a\in\ZZ\) and \(b\in\NN\) with \(b>0\). Then \[ b \left( \frac ab+\ZZ \right) = a+\ZZ = 0. \] Thus every element of \(\QQ/\ZZ\) is torsion.
2. Suppose, toward a contradiction, that \(R\) is a ring with identity\(1_R\) whose additive group is isomorphic to \(\QQ/\ZZ\). Then there is aninteger \(n\geq1\) such that\[ n1_R=0. \]
Under an additive-group isomorphism \[ (R,+)\cong\QQ/\ZZ, \] the element \(1_R\) corresponds to an element of \(\QQ/\ZZ\). By step 1, that element has finite additive order. Hence some positive integer \(n\) satisfies the displayed equality.
3. Every element of \(R\) is annihilated additively by \(n\):\[ nr=0 \]for every \(r\in R\).
For any \(r\in R\), distributivity and the identity property give \[ \begin{aligned} nr &= (n1_R)r\\ &= 0r\\ &= 0. \end{aligned} \]
4. The additive group \(\QQ/\ZZ\) is not annihilated by any positiveinteger.
Fix \(n\geq1\). Consider \[ \frac1{n+1}+\ZZ \in \QQ/\ZZ. \] Its additive order is exactly \(n+1\): multiplying by \(n+1\) gives \(0\), while for \(1\leq k<n+1\), \[ \frac{k}{n+1}\notin\ZZ. \] In particular, \[ n \left( \frac1{n+1}+\ZZ \right) \neq 0. \] Thus multiplication by \(n\) does not annihilate all of \(\QQ/\ZZ\).
5. No ring with identity can have additive group isomorphic to\(\QQ/\ZZ\).
Step 3 says that, under the supposition in step 2, multiplication by \(n\) annihilates the entire additive group of \(R\). An additive-group isomorphism would then imply that multiplication by \(n\) annihilates \(\QQ/\ZZ\), contradicting step 4.
6. Q.E.D.
Step 5 proves the required nonexistence statement. In particular, no commutative ring with identity has the stated additive group.