13

Is there a surjective morphism $\mathbb{Z}^I\to \mathbb{Z}^{(J)}$ for some $I,J$?

i) I'm asking about group morphisms

ii) $\mathbb{Z}^{(J)}$ denotes the direct sum of $J$ copies of $\mathbb Z$

iii) Of course, I want $J$, and thus $I$ to be infinite, otherwise it's trivial.

I want to know if there is such a surjection for any $J$, so it suffices to find one $J$ with no such surjection and it'll be done. However it seems that the answer won't really depend on $J$.

Thoughts : I think there is no surjective morphism, and so was trying to find a contradiction. Of course such a morphism is split, so I was trying to analyze summands of $\mathbb{Z}^I$ but got essentially nowhere. I also tried studying maps $\mathbb{Z}^I \to \mathbb{Z}$, to study the projections. I know that if such a map vanishes on almost zero sequences, then it vanishes, so I was trying to prove that it would do that here, but with no success.

I don't really know how to proceed further.

EDIT: the link that was given in the comments can help. Indeed, if there were such a surjection, then by passing to hom's into $\mathbb{Z}$, we would have an injection $\mathbb{Z}^J \to \hom (\mathbb{Z}^I, \mathbb{Z})$. Now I don't know how general the quoted Baer's result is, but if it generalizes to any exponent, then this would provide an injection $\mathbb{Z}^J\to \mathbb{Z}^{(I)}$, which is clearly contradictory, by looking for instance at "almost" $2^\infty$-divisible elements of $\mathbb{Z}^J$ (if this isn't clear I can sketch the proof here, but it's not that complicated)

Alternatively, if Baer's result isn't that general, it might be possible to use some trickery to reduce to $I= \mathbb{N}$, at least when assuming that $J=\mathbb{N}$.

Maxime Ramzi
  • 43,598
  • 3
  • 29
  • 104
  • 1
    Have you tried looking at the problem from a categorical perspective? – Shaun Apr 01 '19 at 16:03
  • 1
    You defined $\mathbb{Z}^{(J)}$, but not $\mathbb{Z}^{I}$. Are they the same thing, or do your brackets have a meaning? – user1729 Apr 01 '19 at 16:07
  • @Shaun it depends on what you mean, but I have tried to think categorically about it. I do think it's something specifically group-theoretic though – Maxime Ramzi Apr 01 '19 at 16:09
  • @user1729 $\mathbb{Z}^I$ is just the power of $\mathbb{Z}$, this notation is more standard than the other, that's why I didn't define it. If they were the same, the question would be sort of empty – Maxime Ramzi Apr 01 '19 at 16:10
  • 1
    Okay (but call it "the direct product" or something, rather than just "the power" :-p) – user1729 Apr 01 '19 at 16:13
  • Why should it be split? (Is this obvious? I haven't thought hard about it.) If it is indeed split, then the answer is "no" for $I, J=\mathbb{N}$. – user1729 Apr 01 '19 at 16:26
  • 2
    @user1729 : it is split because $\mathbb{Z}^{(I)}$ is free – Maxime Ramzi Apr 01 '19 at 16:27
  • @user1729 : thank you for the link, even if it does not actually answer directly (because the question says that there is no splitting of $\mathbb{Z}^{(\mathbb{N})} \to \mathbb{Z}^\mathbb{N}$, which isn't exactly the same thing, but the answer gives more information which could help – Maxime Ramzi Apr 01 '19 at 16:29
  • The question in the link is: "Why is it true that $\mathbb Z^{\mathbb N}$ is not isomorphic to $\mathbb Z^{(\mathbb N)} \oplus \mathbb Z^{\mathbb N}/\mathbb Z^{(\mathbb N)}$?" Which corresponds to a surjection $\mathbb Z^{\mathbb N}\twoheadrightarrow\mathbb Z^{(\mathbb N)}$, no? Surject onto the left factor. – user1729 Apr 01 '19 at 16:38
  • @user1729 : no, because it is a specific injection $\mathbb{Z}^{(\mathbb{N})}\to \mathbb{Z}^\mathbb{N}$ that is used there, not any injection – Maxime Ramzi Apr 01 '19 at 16:39
  • Ah, right, I understand. – user1729 Apr 01 '19 at 16:44

1 Answers1

13

By the Łoś–Eda Theorem, $\operatorname{Hom}(\mathbb{Z}^I,\mathbb{Z})$ is a free abelian group for any set $I$ (namely, it is free on the homomorphisms given by the countably complete ultrafilters on $I$). If a surjective homomorphism $\mathbb{Z}^I\to\mathbb{Z}^{(J)}$ existed, it would induce an injective homomorphism $\operatorname{Hom}(\mathbb{Z}^{(J)},\mathbb{Z})\to\operatorname{Hom}(\mathbb{Z}^I,\mathbb{Z})$, and so $\operatorname{Hom}(\mathbb{Z}^{(J)},\mathbb{Z})$ would be free since $\operatorname{Hom}(\mathbb{Z}^I,\mathbb{Z})$ is free. But $\operatorname{Hom}(\mathbb{Z}^{(J)},\mathbb{Z})\cong\mathbb{Z}^J$ is not free if $J$ is infinite, so this is a contradiction.

Eric Wofsey
  • 330,363
  • I have a problem : this paper seems to state that $\hom (\mathbb{Z}^I, \mathbb{Z}) = \mathbb{Z}^{(I)}$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well) – Maxime Ramzi Apr 01 '19 at 21:19
  • What paper are you referring to? – Eric Wofsey Apr 01 '19 at 21:20
  • Of course I forgot to paste the link, here it is : https://www.lms.ac.uk/sites/lms.ac.uk/files/1955%20On%20direct%20sums%20of%20free%20cycles.pdf – Maxime Ramzi Apr 01 '19 at 21:25
  • Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals. – Maxime Ramzi Apr 01 '19 at 21:40
  • But I thought I had a proof of the result stated there so I'll have to find where it goes wrong – Maxime Ramzi Apr 01 '19 at 21:41
  • Well I found the mistake in my proof so it's all well and good. Thank you for your answer ! – Maxime Ramzi Apr 01 '19 at 22:08