Compute \(\tor(\QQ, A)\)
- Compute \(\tor(\QQ/\ZZ, A)\)
Solution
1. For every abelian group \(A\),\[ \boxed{\operatorname{Tor}_1^{\mathbb Z}(\mathbb Q,A)=0.} \]
The \(\mathbb Z\)-module \(\mathbb Q\) is a localization of \(\mathbb Z\), hence is flat. Tor in positive degree with a flat module vanishes.
2. Apply \(-\otimes_{\mathbb Z}A\) to\[ 0\longrightarrow\mathbb Z\longrightarrow\mathbb Q\longrightarrow\mathbb Q/\mathbb Z\longrightarrow0. \]The resulting long exact Tor sequence contains\[ 0\longrightarrow \operatorname{Tor}_1^{\mathbb Z}(\mathbb Q/\mathbb Z,A) \longrightarrow A \longrightarrow \mathbb Q\otimes A. \]
The term \(\operatorname{Tor}_1(\mathbb Q,A)\) on the left is zero by 1, and \(\mathbb Z\otimes A\cong A\).
3. The kernel of the localization map \(A\to\mathbb Q\otimes A\) is exactly the torsion subgroup \(A_{\mathrm{tors}}\).
Under \(\mathbb Q\otimes A\cong S^{-1}A\) with \(S=\mathbb Z\setminus\{0\}\), an element \(a\in A\) maps to zero iff some nonzero integer \(s\) annihilates \(a\), which is precisely the definition of torsion.
4. Hence\[ \boxed{\operatorname{Tor}_1^{\mathbb Z}(\mathbb Q/\mathbb Z,A)\cong A_{\mathrm{tors}}.} \]
Combine 2 and 3.