Show that \(H^*(K(\mathbb{Z}_m, 1); \mathbb{Z}_k)\) is isomorphic as a ring to \(H^*(K(\mathbb{Z}_m, 1); \mathbb{Z}_m) \otimes \mathbb{Z}_k\) if \(k\) divides \(m\). In particular, if \(m/k\) is even, this is \(\Lambda_{\mathbb{Z}_k}[x] \otimes \mathbb{Z}_k[y]\).
Solution
Let \[ X=K(\mathbb Z_m,1), \qquad k\mid m. \] The standard infinite lens-space CW model for \(X\) has one cell in every dimension. With coefficients in a ring \(R\) on which \(m\) acts as zero, its cellular cochain differential is zero, since the integral cellular boundary maps alternate between \(0\) and multiplication by \(m\). Hence \[ H^q(X;\mathbb Z_m)\cong\mathbb Z_m, \qquad H^q(X;\mathbb Z_k)\cong\mathbb Z_k \] for every \(q\ge0\).
The coefficient reduction \[ \mathbb Z_m\longrightarrow\mathbb Z_k \] induces a ring homomorphism \[ H^*(X;\mathbb Z_m)\otimes_{\mathbb Z_m}\mathbb Z_k \longrightarrow H^*(X;\mathbb Z_k). \] In each degree the source and target are both free rank-one \(\mathbb Z_k\)-modules, and the cellular generator reduces to the cellular generator. Thus the map is an isomorphism degreewise, hence a ring isomorphism: \[ \boxed{H^*(X;\mathbb Z_k) \cong H^*(X;\mathbb Z_m)\otimes_{\mathbb Z_m}\mathbb Z_k.} \]
Hatcher’s calculation of the \(\mathbb Z_m\) ring gives generators \[ x\in H^1(X;\mathbb Z_m), \qquad y=\beta(x)\in H^2(X;\mathbb Z_m), \] with \(y\) polynomial and, when \(m\) is even, \[ x^2=\frac m2\,y, \] the unique element of order \(2\) in \(H^2(X;\mathbb Z_m)\). If \(m/k\) is even, write \(m=2rk\). After reduction to \(\mathbb Z_k\), \[ \frac m2=rk=0\in\mathbb Z_k, \] so \(x^2=0\). Therefore in this case \[ \boxed{H^*(K(\mathbb Z_m,1);\mathbb Z_k) \cong\Lambda_{\mathbb Z_k}[x]\otimes\mathbb Z_k[y], \quad |x|=1,\ |y|=2.} \]