4

This is an exercise in Hungerford. But can somebody explain why is the following not a counter-example?

Let $G$ be the direct sum of $|\mathbb{R}|$ copies of $\mathbb{Q}$. Let $K$ be the direct sum of $|\mathbb{N}|$ copies of $\mathbb{Q}$. Then $G \oplus \mathbb{Q} \cong G \oplus K$ but $\mathbb{Q}$ is not isomorphic to $K$.

Indeed, suppose $f : \mathbb{Q} \rightarrow K$ is a $\mathbb{Z}$-module isomorphism. We may show that $f$ is a $\mathbb{Q}$-module isomorphism obtaining thus a contradiction. For any $v \in Q$ non-zero and a non-zero natural $b$ there is a unique $w$ such that $bw = v$, that is $w = (1/b)v$. We have $b[(1/b)v] = v$ so $b f((1/b) v) = f(v)$. The same uniqueness argument applies in $K$ since it is torsion-free, so $f((1/b) v) = (1/b) f(v)$. Hence $f$ is a $\mathbb{Q}$-module isomorphism.

Pedro
  • 6,518
  • Can you produce an actual isomorphism $;G\oplus\Bbb Q\to G\oplus K;$ ? I'm afraid that what may be clear setwise is not that straightforward groupwise... – DonAntonio Nov 02 '13 at 20:06
  • 3
    @DonAntonio: Same number of $\mathbb{Q}$s. – Martin Brandenburg Nov 02 '13 at 20:08
  • Can you cite the specific exercise so others can look it up? – anon Nov 02 '13 at 20:09
  • It's exercise 12, part b, in page 199 – Pedro Nov 02 '13 at 20:10
  • I don't see that clear, @Martin: an element in $;G\oplus\Bbb Q;$ is of the form $;(a,q);,;a\in G,,,q\in\Bbb Q;$ , whereas an element in $;G\oplus K;$ is of the form $;(a,k);,;a\in G,,,k\in K;$ . The second coordinate may be a little problematic to deal with, although I think both direct product could be isomorphic to a general uncountable direct product. Still, I can't see it that clear, yet. And "same number of $;\Bbb Q$'s" is a little unaccurate here. Perhaps the problem is there, with "same cardinality" instead. – DonAntonio Nov 02 '13 at 20:15
  • @DonAntonia: Do you agree that $\Bbb R\times \Bbb N\cong \Bbb R$ as sets? Just extend that to coordinates. – anon Nov 02 '13 at 20:16
  • Proposition 2.9 in Hungerford, page 185, shows that (or asks us to show that) two modules of the same rank are isomorphic. So I think it does follow. – Pedro Nov 02 '13 at 20:17
  • @anon: One needs $\mathbb{R} \sqcup \mathbb{N} \cong \mathbb{R} \sqcup $. (and my answer is $\mathbb{N} \sqcup \cong \mathbb{N}$) – Martin Brandenburg Nov 02 '13 at 20:22
  • Oops yes. Too late to edit unfortunately. | OP, the exercise is incorrect, as you've discovered. It should probably say something like finite rank or finitely generated in order to be true. – anon Nov 02 '13 at 20:22
  • what do you mean by rank in this case? a divisible group is never free – Pedro Nov 02 '13 at 20:37
  • Possibly related: http://math.stackexchange.com/questions/27744 – Watson Aug 22 '16 at 13:38

1 Answers1

2

Yes. A more simple counterexample is $\mathbb{Q}^{\oplus \mathbb{N}} \oplus \mathbb{Q} \cong \mathbb{Q}^{\oplus \mathbb{N}} \oplus 0$. Are you sure that the exercise in Hungerford is exactly as stated?

Martin Brandenburg
  • 163,620
  • Here's a copy of the page: http://i.imgur.com/1xIp21v.png – Pedro Nov 02 '13 at 20:14
  • Thanks. I assume that Exercise 11 is the classification of divisible abelian groups. There is also a reference to II.2.11. What is this? – Martin Brandenburg Nov 02 '13 at 20:27
  • You are correct in your assumption. The other exercise is http://i.imgur.com/TH6wLQD.png – Pedro Nov 02 '13 at 20:29
  • Ok. I thought that I could reconstruct what Hungerford actually had in mind with this exercise, since I don't think that he would ask for a proof of a claim which is obviously false (after all, it's the same as in II.2.11 (c) but with $\mathbb{Q}$ instead of $\mathbb{Z}$!). But I'm not sure what Hungerford intended. Maybe the claim is true if $G,H,K$ are direct sums of finitely many "standard" divisible abelian groups ($\mathbb{Q}$ and $\mathbb{Z}/p^{\infty}$)? – Martin Brandenburg Nov 03 '13 at 09:33
  • In that case yes, it would be true, by the same argument that is used for part (a). – Pedro Nov 03 '13 at 13:19