Cohomology ring of $\mathbb R P^\infty$ with $\mathbb Z_{2k}$ coefficients Let $k$ be a positive integer. I am trying to show that as rings, $H^*(\mathbb RP^\infty ; \mathbb Z_{2k}) \cong \mathbb Z_{2k}[a,b]/(2a  , 2b  , a^2 - kb)$. This is exercise 3.2.5 in Hatcher. The hint is to "Use the coefficient map $\mathbb Z_{2k} \rightarrow\mathbb Z_2$ and the proof of theorem 3.12". I tried to adapt the proof of theorem 3.12 (the computation of $H^*(\mathbb RP^\infty, \mathbb Z_{2k})$) but was unable to do so: the proof basically shows that cup products of a generator of $H^1$ generate all of the higher cohomology groups. In the $\mathbb Z_{2k}$ case the cohomology ring doesn't have such a simple description, so I couldn't find a way to make an inductive proof work. Does anyone have a proof? 
 A: Let's use the proof of the theorem as Hatcher suggested. It is easy to see that in the diagram below the map $H^1(P^n) \to H^1(P^1)$  is a multiplication by $k$ ( injectivity and $H^1(P^1) = Z_{2k}$). The same applies to the two downwards maps on the right
Proceeding as in the Hatcher's proof, we conclude that $H^1(P^n,P^n-P^1) \to H^1(R^n,R^n - R^1)$ has to be a multiplication by $k$ in order to make the diagram commute:

Therefore in the first cohomology the left downwards map is a multiplication by $k$.

So $a\cup a = kb$
A: I realize this question is old but, since nobody ever answered, I decided to post a solution.
Consider the short exact coefficient sequence
$$0\to \mathbb{Z}_2\overset{\times k}{\to} \mathbb{Z}_{2k}\to \mathbb{Z}_{2}\to 0\;.$$
This induces a long exact Bockstien sequence
$$... \to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_2)\to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_2)\overset{\beta}{\to} H^{*+1}(\mathbb{R} P^{\infty};\mathbb{Z}_2)\to...$$ 
Recall that 
$$H^*(\mathbb{R}P^{\infty};\mathbb{Z}_{2})\simeq \mathbb{Z}_2[t]$$
This gives the exact segment
$$\mathbb{Z}_2(t^n)\to H^n(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to \mathbb{Z}_2(t^n) \to \mathbb{Z}_2(t^{n+1})\;.$$
Since $\beta$ satisfies the Leibniz rule and we are working over $\mathbb{Z}_2$, we have $\beta(t^{n})=0$ is $n$ is even and $t^{n+1}$ if $n$ is odd. This gives two possibilities. In the odd case, we have an iso
$$0\to \mathbb{Z}_2(t^{2n+1})\to H^{2n+1}(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to 0\;$$
and in the even case, we have the iso
$$0\to H^{2n}(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to \mathbb{Z}_2(t^{2n}) \to 0\;.$$
By the first isomorphism we conclude that $H^1$ is generated by a single element $a=kt$ satisfying $2a=0$. From the second isomorphism, we conclude that $H^2$ is generated by a single element $b$, satisfying $2b=0$. Since $b$ mod $k$ must be $t^2$, we have $b=kt^2$ and $kb=k^2t^2=a^2$. The full ring structure can be deduced from these two cases.
