Let $\prod\_{n=1}^{\infty}\mathbb{Z}$ be the Baer-Specker group and $\bigoplus\_{n=1}^{\infty}\mathbb{Z}$ be the natural free abelian subgroup. It is known that if $G$ is a countable abelian group with no infinitely divisible elements (e.g. $\mathbb{Z}$), then every homomorphism $\prod\_{n=1}^{\infty}\mathbb{Z}/\bigoplus\_{n=1}^{\infty}\mathbb{Z}\to G$ is trivial.


I've heard by word of mouth of a result on fundamental groups which would imply the existence of non-trivial homomorphisms $\prod\_{n=1}^{\infty}\mathbb{Z}/\bigoplus\_{n=1}^{\infty}\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}$ for any prime $p\geq 2$. What is an explicit construction of such a homomorphism for given $p$?



The existence of non-trivial homomorphisms $\prod_{n=1}^{\infty}\mathbb{Z}/\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$
One can skip the ultraproduct part: fix any nonprincipal ultrafilter $F$, as in the SJR's construction. For each $(a\_n)\in R$, split $\mathbb{N}$ into $N\_0,N\_1,\ldots,N\_{p-1}$ so that $n\in N\_k$ iff $a\_n\cong k \mathop{\rm mod} p$. Then send $(a\_n)$ to $k$ such that $N\_k\in F$. But if by "explicit construction" you mean without the axiom of choice - I believe you can't get one.


You may find more related information in Fuchs *'Infinite Abelian Groups I'*, for example Section 42, Exercise 7:
$$
\prod\_{n=1}^{\infty}\mathbb{Z}/\bigoplus\_{n=1}^{\infty}\mathbb{Z}\cong Q\oplus \prod\_p A\_p
$$
where $Q$ is a rational vector space of cardinality continuum and $A\_p$ is the $p$-completion of a free abelian group of rank continuum. The product runs over all primes. 


**Edit:** This might be a good place to advertise an open problem, printed in 1986 in Kourovka Notebook as problem 10.54a and attributed to Rüdiger Göbel:
For a cardinal number $\mu$, let 
$$
\mathbb{Z}^{<\mu}=\{f\in\mathbb{Z}^\mu\mid |\mathop{\rm supp}f|<\mu\}
$$
Let
$$
G\_\mu=\mathbb{Z}^\mu/\mathbb{Z}^{<\mu}
$$
Does there exist a nonzero direct summand $D$ of $G\_{\omega\_1}$ such that $D\ncong G\_{\omega\_1}$?


The answer is still unknown.


