I am trying to calculate $H^*(\mathbb{R}P^3 \times \mathbb{C}P^5,\mathbb{Z})$ as a cohomology ring.
I know that
$$H^*(\mathbb{R}P^3,\mathbb{Z}) = \frac{\mathbb{Z}[\alpha,\beta]}{(2 \alpha, \alpha^2,\beta^2,\alpha \beta)}$$
with $\alpha$ the generator of $H^2(\mathbb{R}P^3)$ and $\beta$ generating $H^3(\mathbb{R}P^3)$
and
$$H^*(\mathbb{C}P^5) = \frac{\mathbb{Z}[\gamma]}{(\gamma^6)}$$
with $\gamma$ the generator of $H^2(\mathbb{C}P^5)$
Initially I thought the cross-product would just give an isomorphism, but the presence of the $\mathbb{Z}/2\mathbb{Z}$ term in the cohomology of $\mathbb{R}P^3$ kills that idea.
Now I could calculate the cohomology groups since both space have a nice CW structure, it would be possible to use
$$H^n(X \times Y) = \sum_{i+j=n}H^i(X) \otimes H^j(Y) \oplus \sum_{p+q=n+1} \operatorname{Tor}(H^p(X),H^q(Y))$$
but how can we deduce the ring/cup product structure?
(I am guessing the answer is to somehow use the cross product, but I just can't see it)
Best Answer
Hatcher has as Theorem $3.16$ that the cross product is an isomorphism if $X$ and $Y$ are CW complexes and $H^k(Y)$ is finitely generated free for all $k$. So, you should still be able to apply it.