# An indecomposable $\mathbf{Z}$-module whose injective hull is not indecomposable

I'd like to find an indecomposable $\mathbf{Z}$-module whose injective hull is not indecomposable, and I'm running out of ideas:

The only indecomposable $\mathbf{Z}$-modules I know are $\mathbf{Z}$, $\mathbf{Q}$ (which is the injective hull of $\mathbf{Z}$), $\mathbf{Z}/p^n\mathbf{Z}$, $\mathbf{Z}(p^\infty)$ (which is the injective hull of $\mathbf{Z}/p^n\mathbf{Z}$).

All of these have indecomposable injective hulls, so I don't really know what to do. Does someone have an idea?

-

Short answer: Amongst many examples, the $p$-adic integers are a good example.
You have found all indecomposable injective modules. Over a commutative noetherian ring, the indecomposable injectives are in one-to-one correspondence with the prime ideals, $\mathfrak{p} \mapsto E(R/\mathfrak{p})$. For $R=\mathbb{Z}$ the prime ideals correspond to prime positive integers $p$ and 0, and as you mentioned the injective hull $E(\mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}(p^\infty)$ are the Prüfer groups and the the injective hull of $\mathbb{Z}/0\mathbb{Z}$ is $E(\mathbb{Z})\cong \mathbb{Q}$.
A theorem of Kulikov shows that the only indecomposable non-torsion-free abelian groups are the subgroups of the Prüfer groups, namely $\mathbb{Z}/p^n\mathbb{Z}$ and $\mathbb{Z}(p^\infty)$ itself (every non-torsion-free group contains one of these as a direct summand).
Hence you are looking for a torsion-free abelian group that is not “rank one” but yet is indecomposable. A good example of this is the additive group $p$-adic integers. It is indecomposable, but its injective hull is the direct sum of uncountably many copies of $\mathbb{Q}$. This shows in particular, that the number of direct summands of a subgroup can be vastly different.
If you have trouble verifying the $p$-adics $J$ are indecomposable, then write $J=A\oplus B$ and consider $\mathbb{Z}/p\mathbb{Z} \cong J/pJ \cong A/pA \oplus B/pB$, so that we must have $A=pA$ or $B=pB$ is $p$-divisible. However if $A =pA$ then $a=pa_1 =p^2a_2 = p^n a_n$, but the only element of the $p$-adics that is divisible by $p$ infinitely often is $a=0$. Hence $A=0$, and $B=J$ is indecomposable.