Compute:
\(\tor(\ZZ \oplus \ZZ_2 \oplus \ZZ_8, \ZZ \oplus \ZZ_4 \oplus \ZZ_4)\)
\(\ext(\ZZ \oplus \ZZ_2 \oplus \ZZ_3, \ZZ \oplus \ZZ_4 \oplus \ZZ_5)\)
Solution
1. Over \(\mathbb Z\),\[ \operatorname{Tor}_1^{\mathbb Z}(\mathbb Z/m,\mathbb Z/n)\cong\mathbb Z/\gcd(m,n), \]and \(\operatorname{Tor}_1^{\mathbb Z}(\mathbb Z,-)=0\).
Resolve \(\mathbb Z/m\) by \(0\to\mathbb Z\xrightarrow{m}\mathbb Z\to\mathbb Z/m\to0\). After tensoring with \(\mathbb Z/n\), the Tor group is the kernel of multiplication by \(m\) on \(\mathbb Z/n\), which is cyclic of order \(\gcd(m,n)\).
2. Hence\[ \boxed{\operatorname{Tor}_1^{\mathbb Z}(\mathbb Z\oplus\mathbb Z_2\oplus\mathbb Z_8, \mathbb Z\oplus\mathbb Z_4\oplus\mathbb Z_4) \cong (\mathbb Z/2)^2\oplus(\mathbb Z/4)^2.} \]
Tor is additive in each variable. The two \(\mathbb Z_4\) summands paired with \(\mathbb Z_2\) each contribute \(\mathbb Z_2\), and paired with \(\mathbb Z_8\) each contribute \(\mathbb Z_4\). Every term involving a free \(\mathbb Z\) summand vanishes.
3. Over \(\mathbb Z\),\[ \operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n,B)\cong B/nB, \]and \(\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z,B)=0\).
Apply \(\operatorname{Hom}_{\mathbb Z}(-,B)\) to the same length-one free resolution of \(\mathbb Z/n\).
4. For \(B=\mathbb Z\oplus\mathbb Z_4\oplus\mathbb Z_5\),\[ B/2B\cong(\mathbb Z/2)^2, \qquad B/3B\cong\mathbb Z/3. \]
For \(B/2B\), the free summand contributes \(\mathbb Z/2\), the \(\mathbb Z_4\) summand contributes \(\mathbb Z/2\), and multiplication by \(2\) is an automorphism of \(\mathbb Z_5\). For \(B/3B\), only the free summand survives, because multiplication by \(3\) is an automorphism on both \(\mathbb Z_4\) and \(\mathbb Z_5\).
5. Therefore\[ \boxed{\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z\oplus\mathbb Z_2\oplus\mathbb Z_3, \mathbb Z\oplus\mathbb Z_4\oplus\mathbb Z_5) \cong (\mathbb Z/2)^2\oplus\mathbb Z/3.} \]
Ext is additive in the first variable over finite direct sums. Apply <1>3–<1>4.