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Let G be the (non-principal) ultraproduct of all finite cyclic groups of orders n!, n=1,2,3,... . Is there a homomorphism from G onto the infinite cyclic group?

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Is this ultraproduct unique (i.e., it doesn't depend on the ultrafilter)? It seems to me as though it should be, but I can't prove it. Then again, I don't have much experience with these things. –  Peter Shor Aug 31 '10 at 15:38
    
@Peter Shor: the ultrapower of the reals (i.e. the hyperreals) is independent of the non-principal ultrafilter you choose...so long as you believe GCH. Maybe something similar is needed here. –  Matthew Towers Sep 9 '11 at 22:01
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4 Answers 4

up vote 20 down vote accepted

I think the answer is no. The ultraproduct $U$ is naturally a quotient of ${\mathbb Z}^{\infty}$, the direct product of countably many copies of ${\mathbb Z}$. In the obvious quotient map, the image of the direct sum is zero. Now, it is enough to show that:

Claim: Any homomorphism $ \phi: {\mathbb Z}^{\infty} \to {\mathbb Z}$ that vanishes on the direct sum is identically zero.

Proof: (I learned this from a book by T.Y.Lam):

For a prime number $p$, let $A_p$ be the set of elements in ${\mathbb Z}^{\infty}$ of the form $(a_0, pa_1, p^2a_2,...)$, i.e. the elements whose $i$-th coordinate is divisible by $p^i$. Any element $x \in A_p$ can be decomposed as

$x= y+z= (a_0, pa_1, \dots, p^{n-1}a_{n-1}, 0,0, \dots ) + p^n (0,0,..,0, a_n, pa_{n+1},..)$

Now, $y$ is in the direct sum, hence $\phi(y)=0$. Also $\phi(z) \in p^n {\mathbb Z}$, which implies that $\phi (x) \in \cap_{n=1}^{\infty} p^n {\mathbb Z} = \{ 0 \}$

Now, choose two distinct primes $p$ and $q$. Since $\gcd(p^n,q^n)=1$, it is easy to see that $A_p+A_q= {\mathbb Z}^{\infty}$. This implies that $\phi \equiv 0$.

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Why can you assume that $x$ is of the form $(a_0,2a_1,4a_2,\dots)$? –  Stefan Geschke Aug 31 '10 at 10:53
    
I have just edited the question for clarity. I had seen this argument before, but I had forgotten it. Note that it actually implies that the dual of the product is the sum. –  Tom Goodwillie Aug 31 '10 at 11:36
    
Sorry, I was in a run, so the proof was not clear. I just rewrote it. –  Keivan Karai Aug 31 '10 at 11:41
    
Sorry Tom, I think we started to edit almost at the same time! –  Keivan Karai Aug 31 '10 at 11:42
    
I guess I was in a hurry, too, when I wrote "question" for "answer". –  Tom Goodwillie Aug 31 '10 at 12:27
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One can at least show that there exists a non-trivial homomorphism to the rationals.

Call a sequence $a=(a_n)_{n \in {\mathbb N}}$ with $a_n \in {\mathbb Z}/n! {\mathbb Z}$ bounded if there exists a constant $C$ such that $|\tilde a_n| \leq C$ for all $n$, where $\tilde a_n$ is the representative of $a_n$ with $-n!/2 < \tilde a_n \leq n!/2$. Let $L$ be the set of bounded elements in the ultraproduct.

For $a \in L$, $\phi(a):= \lim_{n \to \omega} \tilde a_n$ defines a homomorphism $\phi \colon L \to {\mathbb Z}$. Considered as a homomorphism to ${\mathbb Q}$ it has an extension to the whole ultraproduct, since ${\mathbb Q}$ is an injective abelian group. Hence, there exists a non-trivial homomorphism from the ultraproduct to the rationals.

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Do you mean the ultrafilter limit with respect to the ultrafilter used to form the ultraproduct when you write $\lim_{n\to\omega}\tilde a_n$? –  Stefan Geschke Aug 31 '10 at 13:26
    
Yes. However, to get the conclusion, one can also take an arbitrary non-torsion element and build a similar extension. (Which looks even easier.) –  Andreas Thom Aug 31 '10 at 14:10
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I think in certain cases, say if the set of prime numbers belongs to the ultrafilter, $G$ becomes a vector space over the field of rational numbers.

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The orders of the cyclic groups in the question are $n!$ not $n$. Hence, the ultra-product always contains a lot of torsion. –  Andreas Thom Aug 31 '10 at 9:53
    
Oh! Thank you for the correction. –  Narutaka OZAWA Aug 31 '10 at 11:08
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I agree with the first answer.

The ultraproduct in question is pure-injective (by a theorem of Jan Mycielski, Some compactifications of general agebras, Coll. Math., 13 (1964), 1-9). If an abelian group G contains a non-zero homomorphic image of a pure-injective group, then G is not slender. However, the infinite cyclic group is slender (by a theorem of Specker from 1950 (E. Specker, Additive Gruppen von folgen ganzer Zahlen, Portugaliae Math. 9 (1950), 131-140)), which is proved above as the Claim.

One might add: by a theorem of Sasiada (in L. Fuchs, Infinite Abelian Groups, vol.2, 1973), no reduced torsion-free abelian group of power less than continuum contains a non-trivial homomorphic image of the ultraproduct in question.

The results quoted above are also to be found in P.C. Eklof, A.H. Mekler, Almost Free Modules, rev. ed., 2002.

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