Cohomology of the Baer-Specker group

Question

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?

Answer 1

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.)