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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 $p$?

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  • $\begingroup$ Does Baer-Specker group just mean product topology on the direct product? $\endgroup$ – Amritanshu Prasad Apr 27 '14 at 15:38
  • $\begingroup$ And presumably “countable” means countably infinite? $\endgroup$ – Lubin Apr 27 '14 at 15:38
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    $\begingroup$ This should not depend on topologies on the groups since any such homomorphism could not possibly be continuous but yes the Specker group is the infinite direct product. Countable could mean finite. $\endgroup$ – Jeremy Brazas Apr 27 '14 at 16:24
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    $\begingroup$ It's equivalent to SJR's answer, but if want a more "explicit" answer, given a nonprincipal ultrafilter, you map a sequence $(u_n)$ to the element $\lim_F(u_n \mod p)$, where the limit is in the cyclic group of order $p$. $\endgroup$ – YCor Apr 27 '14 at 16:49
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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.

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    $\begingroup$ I'm not sure what you mean: if $F$ is not an ultrafilter there is some $(a_n)$ such that no $N_k$ belongs to $F$. $\endgroup$ – YCor Apr 27 '14 at 18:30
  • $\begingroup$ @Yves - thanks, I missed the "ultra" part. $\endgroup$ – Adam Przeździecki Apr 27 '14 at 20:05
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    $\begingroup$ OK. In a sense, you retrieve this way the ultraproduct, since the homomorphism factors through the ultrapower of $\mathbf{Z}$ with respect to $F$. So basically it's a restatement of the ultrapower argument. (The ultrapower has a unique nonzero homomorphism onto $\mathbf{Z}/p\mathbf{Z}$ up to scalar multiplication in $\mathbf{Z}/p\mathbf{Z}^*$, [and unique if prescribed to have the value 1 on the constant sequence 1], so SJR's answer provides the same homomorphism.) $\endgroup$ – YCor Apr 27 '14 at 21:25
  • $\begingroup$ Perhaps a naive question: if you also send $(b_n)$ to $\ell$ where $N_{\ell}\in F$, how to you know that $N_{(k+\ell) \text{mod}p }=\{n|a_n+b_n=(k+\ell) \text{mod}p\}\in F$? $\endgroup$ – Jeremy Brazas Apr 27 '14 at 23:17
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    $\begingroup$ @JeremyBrazas Write $N_k(a)$, $N_\ell(b)$ and $N_{k+\ell}(a+b)$ for the sets. Clearly $N_{k+\ell}(a+b)$ contains $N_k(a) \cap N_\ell(b)$ and this intersection is in $F$. Since $F$ is a filter $N_{k+\ell}(a+b) \in F$. $\endgroup$ – Mark Wildon Apr 27 '14 at 23:56
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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 $e_i$.

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    $\begingroup$ Note that this is a bit weaker than the construction with ultrafilters, because that construction actually can be lifted to a homomorphism $\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}\to\widehat{\mathbb{Z}}$. $\endgroup$ – Eric Wofsey Apr 28 '14 at 12:04
  • $\begingroup$ Nice. Alternatively, we can think of the last displayed item as a ring $R$, every element of which satisfies the equation $x^p=x$. So if $I$ is any maximal ideal of $R$, then the elements of the field $R/I$ all satisfy the same equation, whence $R/I$ is the $p$-element field. $\endgroup$ – Sidney Raffer Apr 28 '14 at 12:20
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    $\begingroup$ @SJR: The maximal ideals in that ring are naturally in bijection with ultrafilters on $\mathbb{N}$; your special case of Kevin's construction is the same as the ultrafilter construction. $\endgroup$ – Eric Wofsey Apr 28 '14 at 12:48
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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.

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