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 $\{ j_1, \ldots, j_m \} \subset \{1, \ldots, n \}$)
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1$\begingroup$ For $n=1$, this would mean that any monomorphism into an injective module is an injective envelope... $\endgroup$– abxCommented May 9, 2020 at 14:41
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$\begingroup$ @abx you are right... I changed the question now in a clever way. Thank you. $\endgroup$– RobertCommented May 9, 2020 at 14:46
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1$\begingroup$ @abx it is not true that the injective envelope is indecomposable $\endgroup$– RobertCommented May 9, 2020 at 14:58
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$\begingroup$ You are right, sorry. I delete my comment. $\endgroup$– abxCommented May 9, 2020 at 15:45
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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 by Krause and Saorín.
answered May 9, 2020 at 19:06