Identify the following groups up to isomorphism
\(\ZZ_m \tensor \ZZ_n\)
\(\ZZ_{60}^4 \tensor (\ZZ_{24}^3 \oplus \ZZ_8^4 \oplus \ZZ_{120})\)
\(\ZZ_n \tensor \QQ\)
\((\ZZ \oplus \ZZ_n) \tensor (\QQ/\ZZ)\)
Solution
Goal: Compute the following tensor products of abelian groups (\(\mathbb{Z}\)-modules) up to group isomorphism:
\(\mathbb{Z}_m \otimes_\mathbb{Z} \mathbb{Z}_n\),
\(\mathbb{Z}_{60}^4 \otimes_\mathbb{Z} (\mathbb{Z}_{24}^3 \oplus \mathbb{Z}_8^4 \oplus \mathbb{Z}_{120})\),
\(\mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Q}\),
\((\mathbb{Z} \oplus \mathbb{Z}_n) \otimes_\mathbb{Z} (\mathbb{Q}/\mathbb{Z})\).
1. Compute Part 1: \(\mathbb{Z}_m \otimes_\mathbb{Z} \mathbb{Z}_n \cong \mathbb{Z}_{\gcd(m, n)}\).
1.1. By the standard presentation of cyclic modules, \(\mathbb{Z}_m \cong \mathbb{Z} / m\mathbb{Z}\).
1.2. For any abelian group \(A\), \((\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} A \cong A / mA\).
1.3. Applying this to \(A = \mathbb{Z}/n\mathbb{Z}\) gives: \[\mathbb{Z}_m \otimes_\mathbb{Z} \mathbb{Z}_n \cong (\mathbb{Z}/n\mathbb{Z}) / m(\mathbb{Z}/n\mathbb{Z}) \cong \mathbb{Z} / (n\mathbb{Z} + m\mathbb{Z}) = \mathbb{Z} / \gcd(m, n)\mathbb{Z} \cong \mathbb{Z}_{\gcd(m, n)}.\]
1.4. Proof: Standard right-exactness of tensor products with cyclic modules.Q.E.D.
2. Compute Part 2: \(\mathbb{Z}_{60}^4 \otimes_\mathbb{Z} (\mathbb{Z}_{24}^3 \oplus \mathbb{Z}_8^4 \oplus \mathbb{Z}_{120})\).
2.1. Tensor product distributes over direct sums: \[(A^{\oplus 4}) \otimes (B_1^{\oplus 3} \oplus B_2^{\oplus 4} \oplus B_3) \cong (A \otimes B_1)^{\oplus 12} \oplus (A \otimes B_2)^{\oplus 16} \oplus (A \otimes B_3)^{\oplus 4}.\]
2.2. Compute each component using 1:
\(\mathbb{Z}_{60} \otimes \mathbb{Z}_{24} \cong \mathbb{Z}_{\gcd(60, 24)} = \mathbb{Z}_{12}\).
\(\mathbb{Z}_{60} \otimes \mathbb{Z}_8 \cong \mathbb{Z}_{\gcd(60, 8)} = \mathbb{Z}_4\).
\(\mathbb{Z}_{60} \otimes \mathbb{Z}_{120} \cong \mathbb{Z}_{\gcd(60, 120)} = \mathbb{Z}_{60}\). <2>3. Summing all components: \[\mathbb{Z}_{60}^4 \otimes (\mathbb{Z}_{24}^3 \oplus \mathbb{Z}_8^4 \oplus \mathbb{Z}_{120}) \cong \mathbb{Z}_{12}^{12} \oplus \mathbb{Z}_4^{16} \oplus \mathbb{Z}_{60}^4.\]
2.4. Tensor product distributes over direct sums, so the tensor product of \(\mathbb{Z}_{60}^4\) with the direct sum is the direct sum of the pairwise tensor products; each pairwise product is computed by 1, and the multiplicities come from the exponents.
3. Compute Part 3: \(\mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Q} \cong 0\) (for \(n \ge 1\)).
3.1. In \(\mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Q}\), for any \(a \in \mathbb{Z}_n\) and \(q \in \mathbb{Q}\): \[a \otimes q = a \otimes \left( n \cdot \frac{q}{n} \right) = (n \cdot a) \otimes \frac{q}{n} = 0 \otimes \frac{q}{n} = 0.\]
3.2. Since the tensor product is generated by simple tensors, \(\mathbb{Z}_n \otimes_\mathbb{Z} \mathbb{Q} = 0\).
3.3. Every simple tensor \(a \otimes q\) vanishes because \(a\) is \(n\)-torsion and \(q\) is \(n\)-divisible; since simple tensors generate the tensor product, the whole group is \(0\).
4. Compute Part 4: \((\mathbb{Z} \oplus \mathbb{Z}_n) \otimes_\mathbb{Z} (\mathbb{Q}/\mathbb{Z}) \cong \mathbb{Q}/\mathbb{Z}\).
4.1. By distributivity: \[(\mathbb{Z} \oplus \mathbb{Z}_n) \otimes (\mathbb{Q}/\mathbb{Z}) \cong (\mathbb{Z} \otimes (\mathbb{Q}/\mathbb{Z})) \oplus (\mathbb{Z}_n \otimes (\mathbb{Q}/\mathbb{Z})).\]
4.2. For the first term: \(\mathbb{Z} \otimes_\mathbb{Z} A \cong A\) for any abelian group \(A\), so \(\mathbb{Z} \otimes (\mathbb{Q}/\mathbb{Z}) \cong \mathbb{Q}/\mathbb{Z}\).
4.3. For the second term: \(\mathbb{Z}_n \otimes (\mathbb{Q}/\mathbb{Z}) \cong (\mathbb{Q}/\mathbb{Z}) / n(\mathbb{Q}/\mathbb{Z})\).Since \(\mathbb{Q}/\mathbb{Z}\) is divisible, for any \(x \in \mathbb{Q}/\mathbb{Z}\) there exists \(y \in \mathbb{Q}/\mathbb{Z}\) such that \(n y = x\), so \(n(\mathbb{Q}/\mathbb{Z}) = \mathbb{Q}/\mathbb{Z}\).Thus \((\mathbb{Q}/\mathbb{Z}) / n(\mathbb{Q}/\mathbb{Z}) = 0\).
4.4. Therefore: \[(\mathbb{Z} \oplus \mathbb{Z}_n) \otimes (\mathbb{Q}/\mathbb{Z}) \cong (\mathbb{Q}/\mathbb{Z}) \oplus 0 \cong \mathbb{Q}/\mathbb{Z}.\]
4.5. Distributivity splits the tensor product into two summands; the first is \(\mathbb{Q}/\mathbb{Z}\) because \(\mathbb{Z} \otimes A \cong A\), and the second vanishes because \(\mathbb{Q}/\mathbb{Z}\) is divisible, so \(n(\mathbb{Q}/\mathbb{Z}) = \mathbb{Q}/\mathbb{Z}\).
5. Q.E.D.
5.1. 1 through 4 determine all four tensor product groups.