# What tells rational cohomology about integral cohomology?

Say we have a finite CW complex with cells only in even degrees. For example a $\mathbb {CP}^n$ or a complex flag variety. If we know the rational cohomology ring, does it also determine the integral cohomology ring?

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Form two CW-complexes $X_f$ and $X_g$ by choosing attaching maps $f, g: S^3 \to S^2$ with Hopf invariant $H(f) = 1$ and $H(g) = 2$. Then, $H^*(X_f, \mathbb Q)$ and $H^*(X_g, \mathbb Q)$ are isomorphic as graded rings, but $H^*(X_f, \mathbb Z) = \mathbb Z[x_2]/(x_2^3)$ is not isomorphic to $H^*(X_g, \mathbb Z) = \mathbb Z[x_2, y_4]/(x_2^2 - 2y_4, y_4^2, x_2y_4)$.
Yes, they both have 1 cell each in dimensions 0, 2 and 4. (Using the appropriate CW structure on $S^2$). –  Justin Young Oct 27 '11 at 19:04
Ah, I see, I misunderstood how your construction worked. I now understand you're simply describing an attaching map of a 4-cell onto $S^2$. –  MartianInvader Oct 27 '11 at 19:13