# Determining the generators of cohomology (as a ring)

I am working on a problem to show that the cohomology graded rings of $\mathbb{C}P^3$ and $S^2$ x $S^4$ are not isomorphic (unreduced with integer coefficients)

I have already calculated the graded groups of both and shown that they are isomorphic and both have a $\mathbb{Z}$ in dimensions 0,2,4,6 and zeros elsewhere.

Now from what I have seen in class I know that the cohomology of $\mathbb{C}P^3$ is isomorphic to $\mathbb{Z} [x_2] / (x^4)$ where $x_2$ is the generator of $H^2(\mathbb{C}P^3)$ This tells me that the fourth dimension is generated by $x_2^2$ and the sixth dimension is generated by $x_2^3$.

However this is not the case for $S^2$ x $S^4$. Say that the second dimension is generated by $u$, the fourth by $v$ and the sixth by $w$. I gather from a hint we were given in class that $uv = w$ but I am unsure how to prove this. It also appears to me that I need to verify that $u^2 \neq v$, but I am unsure how to do this as well. The notion of multiplication in these rings is still a bit fuzzy to me.

Any help you could offer on this would be appreciated. I could benefit from some general advice on how to determine the generators in an arbitrary cohomology ring.

Try Künneth. When $\textrm{Tor}_1$ vanishes, as it will here since each graded piece of $H^*(S^n; \mathbb{Z})$ is free abelian for any $n$, Künneth gives you an isomorphism of algebras.
• Got it thanks. I had used the Kunneth theorem to calculate the group but I guess I didn't look carefully enough the first time, because it also tells us how the multiplication works. So for example when we end up with a $\mathbb{Z} \otimes \mathbb{Z}$ in some dimension, then the generator in that dimension is the cup product of the generators of the two copies of $\mathbb{Z}$ that are being tensored. – Elliot Feb 14 '14 at 22:16