Define \(M\) as the group ring \(R = \ZZ[\ZZ_2]\) with the action \((\cdot) \times -1\). Construct a free resolution of \(M\) and compute \(\tor_R^*(M, M)\).
Solution
Let \(G=C_2=\{1,\sigma\}\), let \[ R=\mathbb Z[G]=\mathbb Z[\sigma]/(\sigma^2-1), \] and let \(M\cong\mathbb Z\) be the sign module, so \(\sigma m=-m\).
1. The \(R\)-linear surjection\[ \epsilon_-:R\to M, \qquad \epsilon_-(a+b\sigma)=a-b, \]has kernel \((1+\sigma)R\).
The map is \(R\)-linear because multiplication by \(\sigma\) interchanges \(a\) and \(b\), while \(\sigma\) acts by \(-1\) on \(M\). Moreover, \[ \epsilon_-(a+b\sigma)=0\iff a=b\iff a+b\sigma=a(1+\sigma). \]
2. Multiplication by \(1+\sigma\) has kernel \((1-\sigma)R\), and multiplication by \(1-\sigma\) has kernel \((1+\sigma)R\).
For \(r=a+b\sigma\), \[ r(1+\sigma)=(a+b)(1+\sigma), \] so this vanishes exactly when \(b=-a\), i.e. \(r=a(1-\sigma)\). Likewise \[ r(1-\sigma)=(a-b)(1-\sigma), \] which vanishes exactly when \(a=b\), i.e. \(r=a(1+\sigma)\).
3. Therefore a free resolution of \(M\) is\[ \cdots\xrightarrow{\,1+\sigma\,}R \xrightarrow{\,1-\sigma\,}R \xrightarrow{\,1+\sigma\,}R \xrightarrow{\epsilon_-}M\to0, \]so, with \(F_0=R\), the differentials are\[ d_{2j+1}=1+\sigma, \qquad d_{2j}=1-\sigma\quad(j\ge1). \]
Exactness at \(F_0\) follows from 1, and exactness at every higher term follows from the two alternating kernel computations in 2.
4. Tensoring this resolution with \(M\) over \(R\) gives the chain complex\[ \cdots\xrightarrow{-2}\mathbb Z \xrightarrow{0}\mathbb Z \xrightarrow{-2}\mathbb Z \xrightarrow{0}\mathbb Z\to0, \]where the rightmost copy is degree \(0\).
Under \(R\otimes_RM\cong M\cong\mathbb Z\), multiplication by \(1+\sigma\) becomes multiplication by \(1+(-1)=0\), while multiplication by \(1-\sigma\) becomes multiplication by \(1-(-1)=2\) (up to the harmless sign convention \(-2\) if one writes \(\sigma-1\) instead).
5. Hence\[ \operatorname{Tor}_0^R(M,M)\cong\mathbb Z. \]
At degree \(0\) the outgoing differential is zero and the incoming differential \(d_1\otimes M\) is also zero, so \[ H_0(F_\bullet\otimes_RM)=\mathbb Z/0\cong\mathbb Z. \] Equivalently, \(M\otimes_RM\cong R/(1+\sigma)\cong\mathbb Z\).
6. For \(j\ge0\),\[ \operatorname{Tor}_{2j+1}^R(M,M)\cong\mathbb Z/2. \]
At odd degree the outgoing differential is \(0\), so the kernel is all of \(\mathbb Z\), while the incoming even differential is multiplication by \(2\). Thus the homology is \(\mathbb Z/2\mathbb Z\).
7. For \(j\ge1\),\[ \operatorname{Tor}_{2j}^R(M,M)=0. \]
At positive even degree the outgoing differential is multiplication by \(2\), whose kernel in \(\mathbb Z\) is zero; hence the homology vanishes.
8. Therefore\[ \boxed{ \operatorname{Tor}_n^R(M,M)\cong \begin{cases} \mathbb Z,&n=0,\\ \mathbb Z/2,&n\text{ odd},\\ 0,&n>0\text{ even}. \end{cases}} \]
Combine <1>5–<1>7.