Tensor Product, Exterior Power and Splitting Let $M$ be a $\mathbb{Z}$-module and consider the submodule $K=\langle m\otimes m\mid m\in M\rangle$ of $M\otimes M$. Under what conditions does the SES $$0\to K\to M\otimes M\to M\wedge M\to 0$$ split? In other words when can we write $M\otimes M\cong K \times (M\wedge M)$. 
Thanks.
 A: This is true if $M$ is finitely generated. By the structure theorem, there exists a presentation $M = \mathbb{Z} \langle e_i \rangle$ where the only relations are of the form $n_ie_i = 0$, for some $n_i \in \mathbb{Z}$. Then $M \otimes M = \mathbb{Z}\langle e_i \otimes e_j \mid (n_i, n_j) \ne 1 \rangle$, and to define a group homomorphism on $M \otimes M$ it suffices to check the torsion conditions on the images of the generators. Define $s : M \otimes M \to K$ by
$$s(e_i \otimes e_j) = \begin{cases}
e_i \otimes e_j + e_j \otimes e_i \qquad & i < j \\
e_i \otimes e_i \qquad & i = j \\
0 \qquad & i > j
\end{cases}$$
extended $\mathbb{Z}$-linearly. Notice that $e_i \otimes e_j + e_j \otimes e_i = (e_i + e_j) \otimes (e_i + e_j) - (e_i \otimes e_i) - (e_j \otimes e_j)$, so this does land in $K$. As the torsion conditions on the generators are satisfied, $s$ is a well-defined group homomorphism, and $s \circ i = \text{id}_K$, where $i : K \hookrightarrow M \otimes M$ is the inclusion. By the splitting lemma, $0 \to K \to M \otimes M \to \wedge^2 M \to 0$ is split exact.
The same reasoning also applies to any $\mathbb{Z}$-module $M$ whose matrix of relations can be "diagonalized", after choosing a well-order on the generating set: e.g. any direct sum of finitely generated $\mathbb{Z}$-modules, such as any $\mathbb{F}_2$-vector space.
Another class of examples where the sequence splits is if $2$ acts as a unit on $M$, in which case symmetrization is a splitting map $M \otimes M \to K$:
$$\text{sym}(e_i \otimes e_j) = \frac{1}{2}(e_i \otimes e_j + e_j \otimes e_i)$$
which gives the result for e.g. any $\mathbb{Q}$-vector space, or more generally any injective $\mathbb{Z}$-module with no $2$-torsion. The result holds for the Prufer $2$-group $E_2 := E_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z}) = \mathbb{Z}[1/2]/\mathbb{Z}$ as well, since $E_2 \otimes E_2 = 0$ (as $E_2$ is $2$-divisible, but every element is killed by a power of $2$), which shows that the result also holds for any injective $\mathbb{Z}$-module (by the structure theorem of injectives over $\mathbb{Z}$, as no terms with $E_2$ will appear in the tensor square or exterior square).
