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Let $M=\mathbb Z^{\mathbb N}$ be the product of a countable number of copies of the group $\mathbb{Z}$ and let $N=\mathbb Z^{(\mathbb N)}$ be the direct sum of a countable number of copies of $\mathbb{Z}$. Why is it true that $M$ is not isomorphic to $N \oplus M/N$?

Thoughts I've had so far:

  • I found out that $M$ is not a free or even projective $\mathbb{Z}$-module (note: projective modules are free over PIDs, as a commenter pointed out), but that fact doesn't preclude the possibility that the sequence $0 \to K \to M \to K \oplus M/K \to 0$ can still split some of the time for some $\mathbb{Z}$-submodule $K \subseteq M$. So the result does not follow directly from $M$ not being projective.
  • I don't know much about $M/N$ besides that it consists of infinite sequences of integers, with sequences that only differ by a finite number of entries being identified. If $M/N$ is free or projective then the result follows from the fact that $M$ is not projective, but I don't have any intuition as to the freeness or projectiveness of $M/N$.
  • I tried a proof considering $M$ as a ring with multiplication defined component-wise, and then consider $M$,$N$,$M/N$ as $M$-modules. But this didn't get me very far.
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I should note that the question whether $M \cong N \oplus M/N$ and the question whether the corresponding sequence splits are not necessarily the same. (Although the splitting implies the isomorphism.) (Maybe that is the point you are making in your first bullet.) –  Myself Mar 17 '13 at 22:39
    
Arg! I've made the mistake of thinking precisely that multiple times and keep forgetting it. If I made that point it was unintentional - thanks for the reminder :) –  Jon Paprocki Mar 17 '13 at 22:53
    
So what is the precise question you mean to ask? My first idea would be that the sequence does not split, but that they are in fact isomorphic (namely both free modules of uncountable rank). –  Myself Mar 17 '13 at 22:59
    
I'm just interested in showing that $M$ is not isomorphic to $N \oplus M/N$. Can you elaborate on your idea? $M$ isn't free so I'm puzzled what the two free modules of uncountable rank you are referring to are. –  Jon Paprocki Mar 17 '13 at 23:25
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Concerning "is not a free or even projective $\mathbb{Z}$-module": Over PIDs, projective modules are free. –  Martin Brandenburg Mar 18 '13 at 14:50

1 Answer 1

up vote 8 down vote accepted

If $A$ is an abelian group, then $\cap_{n \geq 0} 2^n A$ is a subgroup of $A$, whose elements may be called $2^{\infty}$-divisible. Note that $0$ is the only $2^{\infty}$-divisible element of $\mathbb{Z}$. Therefore, the same is true for $\mathbb{Z}^{\mathbb{N}}$. But the element represented by $(2^0,2^1,2^2,\dotsc)$ is $2^{\infty}$-divisible in $\mathbb{Z}^{\mathbb{N}}/\mathbb{Z}^{\oplus \mathbb{N}}$. Hence, there is no monomorphism $\mathbb{Z}^{\mathbb{N}}/\mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$.

Alternatively, it is a well-known result by Baer that $\mathbb{Z}^{\oplus \mathbb{N}} \to \hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$, $e_n \mapsto \mathrm{pr_n}$ is an isomorphism. In particular (and actually this is the main step in the proof) a homomorphism $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}$ vanishes when it does vanish on the direct sum, i.e. $\hom(\mathbb{Z}^{\mathbb{N}}/\mathbb{Z}^{\oplus \mathbb{N}},\mathbb{Z})=0$.

For the sake of completeness, here is the argument: If $f : \mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}$ is a homomorphism which vanishes on the direct sum, and $x \in \mathbb{Z}^\mathbb{N}$, then for every $n \in \mathbb{N}$ we choose $u_n,v_n \in \mathbb{Z}$ with $x_n = 2^n \cdot u_n + 3^n \cdot v_n$. Then one observes that $f(2^n u_n)_n \in \mathbb{Z}$ is $2^{\infty}$-divisible, thus vanishes. Likewise $f(3^n v_n)_n$ vanishes, so that $f(x)=0$.

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