Timeline for Should cohomology of $\mathbb{C} P^\infty$ be a polynomial ring or a power series ring?

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Jul 25, 2020 at 21:03	comment	added	D.-C. Cisinski		$R\Gamma(\mathbb{C}P^\infty,\mathbb{Z})\cong R\lim_nR\Gamma(\mathbb{C}P^n\mathbb{Z})$ by descent. We have $R\Gamma(\mathbb{C}P^\infty,\mathbb{Z})\cong\prod_{n\geq 0}\mathbb{Z}[2n]$ from the theory of Chern classes, and then $\bigoplus_{n\geq 0}\mathbb{Z}[2n]\cong\prod_{n\geq 0}\mathbb{Z}[2n]$ for formal reasons in the derived category of abelian groups. The power series formulation is more natural from a sheaf theoretic perspective, but, in the derived category, $R\Gamma(\mathbb{C}P^\infty,\mathbb{Z})$ is both a polynomial ring and ring of power series.
Jul 25, 2020 at 18:00	comment	added	Lennart Meier		If you want to have all degrees in one degree, you can build either the cohomology theory sending $X$ to $\bigoplus_{i\in \mathbb{Z}}H^i(X)$ in every degree or to $\prod_{i\in \mathbb{Z}}H^i(X)$. The first is not represented by a spectrum as it does not satisfy the wedge axiom, while the second one is.
Jul 24, 2020 at 23:56	comment	added	Liviu Nicolaescu		To avoid confusion I write $H^$ when I think of cohomology as a direct sum and $H^{*}$ when I think of it as a direct product.
Jul 24, 2020 at 17:47	history	edited	PowerToThePeople	CC BY-SA 4.0	added 1117 characters in body
Jul 24, 2020 at 17:26	review	First posts
Jul 24, 2020 at 17:27
Jul 24, 2020 at 17:25	answer	added	Denis Nardin		timeline score: 13
Jul 24, 2020 at 17:20	history	asked	PowerToThePeople	CC BY-SA 4.0