I know cup product structure on $H^*(\mathbb{R}P^n;\mathbb{Z}_2)= \mathbb{Z}_2[\alpha]/(\alpha^{n+1})$. How to get $H^*(\mathbb{R}P^n;\mathbb{Z})$ from this? I have two cochain complexes for two coefficient rings.
Now my question is what will be the induced map between these two cochain complexes and what will be $H^*(\mathbb{R}P^n;\mathbb{Z})$?
Best Answer
I think, $H^*(\mathbb RP^{2n+1})=\mathbb Z[\alpha,\beta]/(2\alpha,\alpha^{n+1},\alpha\beta,\beta^2)$, where $\alpha$ has degree $2$ and $\beta$ has degree $2n+1$.
You can see it when you write down spectral sequence for $S^1$-fibration $\mathbb RP^{2n+1}\to\mathbb CP^n$ and look on multiplicative structure (as you doing for calculating ring $H^*(\mathbb CP^{n})$).
And for odd dimension, $H^*(\mathbb RP^{2n})=\mathbb Z[\alpha]/(2\alpha,\alpha^{n+1})$, where $\alpha$ has degree $2$. It makes cleare when you consider inclusion $\mathbb RP^{2n}\to\mathbb RP^{2n+1}$.
EDIT: this spectral sequence just shows that when you project $\mathbb RP^{2n+1}\to\mathbb CP^n$, the pullback maps $\mathbb Z\to\mathbb Z_2$ in cohomology are surjections, and ring structure is induced.