# Cohomology of the Baer-Specker group

Let $$A = \prod_{i \in \mathbb{N}} \mathbb{Z}$$ be the Baer-Specker group; that is, a countably infinite product of the integers. We will consider this as a discrete abelian group.

Are the higher cohomology groups $$H^{*}(A, \mathbb{Z})$$ known?

In the first degree, we have $$H^{1}(A, \mathbb{Z}) \simeq \operatorname{Hom}(A, \mathbb{Z})$$, which is known to be a countably infinite direct sum of $$\mathbb{Z}$$. Is $$H^{*}(A, \mathbb{Z})$$ isomorphic to an exterior algebra over $$H^{1}(A, \mathbb{Z})$$, like we would expect from a finite product?

No, $$H^*(A,\mathbb{Z})$$ is not isomorphic to an exterior algebra over $$H^1(A,\mathbb{Z})$$. If it was, then it would be countable in each degree, being an exterior algebra over a countable abelian group. But $$H^i(A,\mathbb{Z})$$ has cardinality at least (and I think exactly) $$2^\mathfrak{c}$$ for each $$i\geq 2$$, where $$\mathfrak{c}$$ denotes the cardinality of the continuum. To see this, first note that $$H_*(A,\mathbb{Z})$$ contains a summand isomorphic to $$A$$ in each positive degree, by applying the Künneth theorem to the decomposition $$A \cong \mathbb{Z}^i \times A$$ for each $$i$$. Thus, for $$i\geq 2$$, the universal coefficient theorem implies that $$H^i(A,\mathbb{Z})$$ contains a direct summand isomorphic to $$\mathrm{Ext}(A,\mathbb{Z})$$. By [Nunke, "Slender groups", Bull. AMS, 1961], the group $$\mathrm{Ext}(A,\mathbb{Z})$$ is isomorphic to a direct sum of $$2^\mathfrak{c}$$ copies of $$\mathbb{Q}$$ and $$2^\mathfrak{c}$$ copies of $$\mathbb{Q}/\mathbb{Z}$$. (See also https://mathoverflow.net/a/441395.)