If $A$ and $B$ are summands of injective module $Q$ then $A+B$ is a sumand of $Q$.

Question

Let $Q$ be an injective module over $\mathbb Z$. Suppose $A$ and $B$ are sumands of $Q$. Prove $A+B$ is a sumand of $Q$.

I already saw the solution, but I can't remember how. I think the trick was to write $A+B$ as a direct sum of sumands of $Q$ but I don't remember how.

Answer 1

This is not generally true over any ring: it is if the ring is hereditary, which means that submodules of projective modules are projective or, equivalently, that quotients of injective modules are injective. Over non commutative rings the notion is not symmetric: a ring can be right hereditary but not left hereditary.

Let $Q$ be an injective right module over the right hereditary ring $R$. If $A$ and $B$ are summands of an injective module $Q$, they are injective, so also $A\oplus B$ (external direct sum) is injective and, by hereditariness, the obvious epimorphism $A\oplus B\to A+B$ yields that $A+B$ is injective.

Now, $\mathbb{Z}$ is hereditary.

Answer 2

Hint:

The fact that the injective $\Bbb Z$-modules are precisely the divisible $\Bbb Z$-modules comes in very handy at times like this.

This, along with the elementary fact that injective submodules split out of the modules that contain them, will bring you to your solution.