Cohomology ring of a product

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)

Solution 1:

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.

Solution 2:

Have you seen this paper? He starts by giving an example showing that the cohomology ring isn't necessarily determined by the integral cohomology rings of the factors, but analyzes conditions under which you can compute the cohomology ring of the product from knowledge of the cohomology of the factors.