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Why I can use Kunneth formula to say that $H^{*}(\mathbb{C}P^{\infty} \times \mathbb{C}P^{\infty})= \mathbb{C}[x_{1}] \otimes \mathbb{C}[x_{2}]$?

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Are you sure you mean $\mathbb C[x_1]$ and not $\mathbb Z[x_1]$? Are you sure you mean "Why..." and not "How..."? – Rasmus Jan 19 '13 at 23:13
Yes, I take coefficients of cohomology in $\mathbb{C}$. Also how satisfies my question. – ArthurStuart Jan 19 '13 at 23:21
up vote 6 down vote accepted

The statement of the Kunneth formula from Hatcher is

The cross product $H^*(X; R) \otimes_R H^*(Y; R)$ is an isomorphism of rings if X and Y are CW complexes and $H^k(Y; R)$ is a finitely-generated free $R$-module for all $k$.

$\mathbb{C}P^\infty$ is (homeomorphic to...) a CW complex; you can find a description of the entire structure in Hatcher's Vector Bundles or Milnor's Characteristic Classes (and probably many other books). For any $k$, $H^k(\mathbb{C}P^\infty; \mathbb{C})$ is isomorphic either to $0$ or $\mathbb{C}$. So the product $\mathbb{C}P^\infty \times \mathbb{C}P^\infty$ satisfies the hypotheses.

Working over $\mathbb{C}$, we immediately have $H^*(\mathbb{C}P^\infty \times \mathbb{C}P^\infty; \mathbb{C}) \simeq H^*(\mathbb{C}P^\infty) \otimes_\mathbb{C} H^*(\mathbb{C}P^\infty) \simeq \mathbb{C}[x] \otimes \mathbb{C}[y]$, where $x$ and $y$ are the generators of each copy of $H^2(\mathbb{C}P^\infty; \mathbb{C})$.

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very nice answer! – Bombyx mori Jan 20 '13 at 1:28
But from Hilbert basis theorem we have that $\mathbb{C}[x]$ is finitely-generate as $\mathbb{C}$-module. So can I apply kunnet formula directly for $H^{*}(\mathbb{C}P^{\infty}) \simeq \mathbb{C}[x]$... – ArthurStuart Jan 20 '13 at 16:01
@ArthurStuart Sure. I stuck to the exact statement of the Kunneth Formula above, so I verified the hypotheses to the letter. – Adam Saltz Jan 20 '13 at 17:29
@ArthurStuart Moreover, a submodule of a finitely-generated module need not be finitely-generated, so I guess in principle you could run into trouble. Certainly everything works fine with coefficients over $\mathbb{C}$ or $\mathbb{Z}$. If you had some reason to compute cohomology with coefficients in a non-Noetherian ring, you'd have to check that each $H^k$ is finitely-generated separately. – Adam Saltz Jan 20 '13 at 17:31

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