Countable infinite direct product of $\mathbb{Z}$ modulo countable direct sum

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.

Solution 1:

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 $f(x)=0$.