The existence of non-trivial homomorphisms $\prod_{n=1}^{\infty}\mathbb{Z}/\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$

Question

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 $p$?

Answer 1

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.

Answer 2

Here's an elementary proof that doesn't require ultrafilters, but uses axiom of choice.

The group \begin{equation*} \prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z} \end{equation*}

clearly surjects onto \begin{equation*} \prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}). \end{equation*}

The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.

Answer 3

Let $F$ be any non-principal ultrafilter on $\mathbb{N}$. Let $I$ be the subset of the ring $R:=\prod_1^{\infty}\mathbb{Z}$ consisting of all functions that vanish on some element of $F$. Then $I$ is an ideal containing $J:=\coprod_1^{\infty} \mathbb{Z}$, whence we get a surjective homomorphism $R/J\to R/I$. But $R/I$ is isomorphic to the ultrapower of the integers determined by $F$. By Łoś's theorem, $R/I$ is an elementary extension of $\mathbb{Z}$. It follows that the quotient $(R/I)/p(R/I)$ is isomorphic to $\mathbb{Z}/p \mathbb{Z}$. For terminology and details, see The Use of Ultraproducts in Commutative Algebra, by Hans Schoutens.