Let $A$ be an artinian ring and $f : X \rightarrow \bigoplus\_{j=1}^{n}I\_{j}$ be a morphism of $A$-modules, where each $I\_{j}$ is injective and indecomposable. If $f$ is a monomorphism, then can we conclude that there is an injective envelope $g : X \rightarrow \bigoplus\_{t=1}^{m}I\_{j\_{t}}$? (Here $\{ j\_1, \ldots, j\_m \} \subset \{1, \ldots, n \}$)



Can we extract an injective envelope from a monomorphism?
Yes: given a monomorphism $f\colon X\to I$ with $I$ injective, as in the question, you can find a decomposition $I=I\_0\oplus I\_1$ such that $f=\begin{pmatrix}f\_0&0\end{pmatrix}$ and $f\_0\colon X\to I\_0$ is left minimal, so it is an injective envelope.


Getting this decomposition does not require any assumption on the ring $R$, and only much weaker assumptions on $I$, for example that it is pure injective. A good reference is [*On minimal approximations of modules*](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.554.9503&rep=rep1&type=pdf) by Krause and Saorín.


