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I am reading this post, Prop. 2.1. It seem that none of the argument is dependent on the we are working with coefficient in $\Bbb Z$.

Hence, let $R$ be a unital commutative ring.

(I) Do the results for $H^*(BU, \Bbb Z)$ hold for $H^*(BU;R)$?

(II) Are the arguments the same?

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Yes. Alternatively, you can deduce the computation of $H^*(BU;R)$ from the computation of $H^*(BU;\mathbb{Z})$, since the canonical ring-homomorphism $H^*(BU;\mathbb{Z})\otimes R\to H^*(BU;R)$ is an isomorphism (this is true more generally for any space whose homology with coefficients in $\mathbb{Z}$ is free and finitely generated in each degree, by the universal coefficient theorem).

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  • $\begingroup$ Hi Eric, also, is it for Prop 2.1, that the generators are at degree $2k$? $\endgroup$ – W. Zhan Apr 26 '19 at 21:03
  • $\begingroup$ Yes, it's a polynomial ring on generators in each even degree. $\endgroup$ – Eric Wofsey Apr 26 '19 at 21:13

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