Q:

Cohomology of the Eilenberg-Maclane space $K(\mathbb{R},1)$

Let $\mathbb{R}$ be the reals as an abelian group. A connected topological space $X$ is called an Eilenberg–MacLane space of homotopy type $K(\mathbb{R},1)$, if it has fundamental group isomorphic to $\mathbb{R}$ and all other homotopy groups trivial. Such a space exists, is a CW-complex, and is unique up to a weak homotopy equivalence.
1)Where can I find references to the cohomology of $K(\mathbb{R},1)$? 
2)Could we calculate it from basic algebraic topology tools?
My interest on this space relies in the fact that $K(\mathbb{R},1)=B\mathbb{R}^\delta$, i.e., it is also the classifying space for the group $\mathbb{R}^\delta$, which is the real numbers with the discrete topology. 
Also, it has relations with:
Friedlander-Milnor's Conjecture: Let $G$ be a Lie group, and let denote $G^\delta$ the same group with the discrete topology. Then the map $H^*(BG,\mathbb{Z}_p)\to H^*(BG^\delta,\mathbb{Z}_p)$ is an isomorphism for any $p$.
The map $\mathbb{R}^{\delta}\to \mathbb{R}$ is continuous, then we have also a continuous map at classifying space level $B\mathbb{R}^{\delta}\to B\mathbb{R}$. It is known that the map $H^*(B\mathbb{R},\mathbb{Z})\to H^{*}(B\mathbb{R}^{\delta},\mathbb{Z})$ is injective, but the proof comes from the solution of the Friedlander-Milnor's Conjecture for nilpotent Lie groups. I would like to see how this woks in this easiest example of $\mathbb{R}$.

A:

Let's start with of $K(\mathbb{Q},1)$.  We can explicitly construct a $K(\mathbb{Q},1)$ as the mapping telescope of a sequence of maps $K(\mathbb{Z},1)\to K(\mathbb{Z},1)\to\dots$ which parallel writing $\mathbb{Q}$ as the union of a sequence of cylic subgroups.  From this construction we can compute that $H_1(K(\mathbb{Q},1);\mathbb{Z})\cong\mathbb{Q}$ and $H_n(K(\mathbb{Q},1);\mathbb{Z})=0$ for $n>1$, since $K(\mathbb{Z},1)$ is just $S^1$.  See Group homology of the rationals for more details.
Now, $\mathbb{R}$ is just a direct sum of uncountably many copies of $\mathbb{Q}$.  So, we can write it as a filtered colimit of copies of $\mathbb{Q}^n$ for finite $n$, and thus write $K(\mathbb{R},1)$ as a filtered homotopy colimit of $K(\mathbb{Q}^n,1)$'s.  By the Künneth formula, we find that $H_i(K(\mathbb{Q}^n,1);\mathbb{Z})$ is a vector space over $\mathbb{Q}$ for all $i>0$.  It follows that the same is true of $K(\mathbb{R},1)$, since any filtered colimit of abelian groups which are $\mathbb{Q}$-vector spaces is a $\mathbb{Q}$-vector space.  It then follows immediately that $H^i(K(\mathbb{R},1);\mathbb{Z}_p)$ is trivial for $i>0$, since if $V$ is a $\mathbb{Q}$-vector space then $\operatorname{Hom}(V,\mathbb{Z}_p)$ and $\operatorname{Ext}(V,\mathbb{Z}_p)$ are trivial since they are both $p$-divisible and $p$-torsion.
With torsion-free coefficients, describing the cohomology is more complicated.  Keeping track of the computations with $H_i(K(\mathbb{Q}^n,1);\mathbb{Z})$ above more carefully, we can see that for $i>0$, $H_i(K(\mathbb{R},1);\mathbb{Z})$ has as a basis over $\mathbb{Q}$ corresponding to subsets of size $i$ from a fixed basis for $\mathbb{R}$ over $\mathbb{Q}$.  You can then use this to show that $H^*(K(\mathbb{R},1);\mathbb{Q})$ is a completed exterior algebra on generators corresponding to a a basis for $\mathbb{R}$ over $\mathbb{Q}$ (i.e., the inverse limit of the exterior algebras on finitely many of the generators at a time).  Or, without choosing a basis, $H^n(K(\mathbb{R},1);\mathbb{Q})$ can be described as the space of alternating $n$-linear forms on $\mathbb{R}$ as a vector space over $\mathbb{Q}$, with the cup product corresponding to the natural product on such forms.
With coefficients in $\mathbb{Z}$ things are more complicated and I don't know a simple description of $H^*(K(\mathbb{R},1);\mathbb{Z})$.  Note though that the fact that $H^*(B\mathbb{R},\mathbb{Z})\to H^{*}(B\mathbb{R}^{\delta},\mathbb{Z})$ is injective is totally trivial, since $B\mathbb{R}$ is contractible so its cohomology is trivial.

