Show the ring \(H^*(\mathbb{RP}^\infty; \mathbb{Z}_{2k})\) is isomorphic to \(\mathbb{Z}_{2k}[\alpha, \beta] / (2\alpha, 2\beta, \alpha^2 - k\beta)\) where \(|\alpha| = 1\) and \(|\beta| = 2\).
Solution
Put \(R=\mathbb Z_{2k}\). With the standard CW structure on \(\mathbb{RP}^\infty\), there is one cell in each dimension and the cellular boundary is multiplication by \(2\) in even dimensions and \(0\) in odd dimensions. Hence the cellular cochain differential with coefficients in \(R\) is \[ \delta^q= \begin{cases} 2,&q\text{ odd},\\ 0,&q\text{ even}. \end{cases} \] Therefore \[ H^q(\mathbb{RP}^\infty;R)\cong \begin{cases} R,&q=0,\\ \ker(2:R\to R)=\{0,k\}\cong\mathbb Z_2,&q\text{ odd},\\ R/2R\cong\mathbb Z_2,&q>0\text{ even}. \end{cases} \]
Let \(\alpha\in H^1\) be represented by the cellular \(1\)-cocycle taking the value \(k\) on the unique \(1\)-cell, and let \(\beta\in H^2\) be the class of the cellular \(2\)-cochain taking value \(1\) on the unique \(2\)-cell. Then \[ 2\alpha=2\beta=0. \] The cellular diagonal used in the proof of Hatcher’s Theorem 3.12 shows that the cup product of the basic degree-one cellular cochain with itself represents the basic degree-two cochain. Since our degree-one cocycle has coefficient \(k\), its square is represented by \(k^2\) times the basic degree-two cochain. Thus \[ \alpha^2=k^2\beta. \] But \(\beta\) has additive order \(2\), and \(k^2\equiv k\pmod 2\), so \[ \boxed{\alpha^2=k\beta.} \]
Reduction of coefficients \(R\to\mathbb Z_2\) sends \(\beta\) to the nonzero class in \(H^2(\mathbb{RP}^\infty;\mathbb Z_2)\), so \(\beta^j\) is nonzero for every \(j\ge1\). Likewise \(\alpha\beta^j\) generates the odd-dimensional group in degree \(2j+1\). Hence \(\alpha,\beta\) generate the whole ring and there are no further relations. Therefore \[ \boxed{ H^*(\mathbb{RP}^\infty;\mathbb Z_{2k}) \cong \mathbb Z_{2k}[\alpha,\beta]/(2\alpha,2\beta,\alpha^2-k\beta), \quad |\alpha|=1,\ |\beta|=2. } \]