# 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)

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.
• @Qwirk but $H^*(\mathbb{C}P^5;\mathbb{Z})$ is. – wckronholm Jun 15 '11 at 3:48
• So can we simplify $$\frac{\mathbb{Z}[\gamma]}{(\gamma^6)} \otimes \frac{\mathbb{Z}[\alpha,\beta]}{(2 \alpha, \alpha^2,\beta^2,\alpha \beta)} = \frac{\mathbb{Z}[\alpha,\beta,\gamma]}{(2 \alpha, \alpha^2,\beta^2,\alpha \beta,\gamma^6)}$$ or do I need to check the cohomology groups and resulting product? – Juan S Jun 15 '11 at 4:05
• The theorem tells you the resulting cohomology ring is the tensor product of the two cohomology rings you mentioned, which is the ring you describe in the comment above. I should add that the isomorphism is not just as rings, but as graded rings. It would be good to note the degree of each of $\alpha$, $\beta$, and $\gamma$. – wckronholm Jun 15 '11 at 4:11