# The cohomology $H^*(BU(n); R)$

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?

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).

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