We say that a module $M$ is indecomposable if for $M=M_{1} +M_{2}$ (not direct sum) we have that $M_{1}=M$ or $M_{2}=M$. Let $\sigma:P \to M$ a projective cover of $M$, this means that $\sigma$ is an epimorphism and $\ker(\sigma)$ is superfluous on $P$ which means that if there is some $P_{1} \leq P$ such $P_{1} + \ker(\sigma)=P$ then $P_{1}=P$ and $M$ semi-perfect, i.e, every quotient of $M$ has a projective cover, I want to prove that if P is indecomposable then $M$ is indecomposable.

Lets take $M$ such $M=M_{1} +M_{2}$ and as $P$ is projective and $M$ semiperfect I got by some result I found that every submodule of $P$ is supplemented which means that for every $P_{1} \leq P$ there is some $P' \leq P$ such $P_{1} +P' =P$, so considering $\ker(\sigma)$ and its suplement $P'$ we have that $P'+\ker(\sigma)=P$ but as $\ker(\sigma)$ is superfluous , $P'=P$. So $\sigma(P)=\sigma(P')= M=M_{1} +M_{2}$ but I don't see how to get $M_{1}=M$ or $M_{2}=M$....

Also with my general hypothesis I got that $P$ is indecompsable in the way I mean it first is equivalent to be directly indecomposblae (direct sum) but I don't know if this helps.

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    $\begingroup$ I have recently found that what Kasch calls an indecomposable module is usually called a hollow module, namely one that cannot be expressed as a sum of proper submodules. $\endgroup$ – Zeek May 29 at 0:07
  • $\begingroup$ Exactly! You helped me by proving this for directly indescomposable according to Kasch. With the general hypothesis M indescomposable implies M indescomposable but not the other implication. @zeek $\endgroup$ – Cos May 29 at 2:42

More generally we have the following:

Let $f:M \twoheadrightarrow N$ a surjection of $R$-modules with superfluous kernel. If $M$ is indecomposable then $N$ is indecomposable.

Proof: Suppose for the sake of contradiction that $N = N_1 + N_2$ with $N_1, N_2 \not= N$. Set $M_i = f^{-1}(N_i)$, so $M_i$ is a proper submodule of $M$, and is superfluous because $M$ is indecomposable. Also note that $f(M_1 + M_2) = N_1 + N_2 = N$. Recall that a surjective morphism of modules $M \twoheadrightarrow N$ has superfluous kernel iff $f(M') \not= N$ for every submodule $M' \subsetneq M$, and thereby conclude that $M_1 + M_2 = M$. Thus $M$ is the sum of superfluous submodules, which is absurd.


Lemma Let $f: M \rightarrow N$ a surjective morphism of $R$-modules. The following are equivalent.

(1) $\ker(f)$ is a superfluous submodule of $M$.

(2) For any submodule $M'$ of $M$, $f(M') = N$ implies $M' = M$.

(3) For any module morphism $g: L \rightarrow M$, $fg$ surjective implies $g$ surjective.

Proof $(1) \implies (2)$ Let $M' \subseteq M$ with $f(M') = N$. Then for any $m \in M$ we can find $m' \in M'$ such that $f(m) = f(m')$, i.e. $m - m' \in \ker(f)$. This shows that $M' + \ker(f) = M$, and since $\ker(f)$ is superfluous, we conclude $M' = M$.

$(2) \implies (1)$ Suppose that $\ker(f) + M' = M$. Then $f(M') = f(\ker(f) + M') = f(M) = N$ and by assumption $M' = M$. Thus $\ker(f)$ is superfluous in $M'$ by definition.

$(3) \implies (2)$ If $f(M') = N$ then the composition $M' \subseteq M \rightarrow N$ is a surjection so the inclusion $M' \subseteq M$ is a surjection, aka $M' = M$.

$(2) \implies (3)$ Given $g: L \rightarrow M$ such that $fg$ is surjective, so $g(L)$ is a submodule of $M$ such that $f g(L) = N$, and by assumption $g(L) = M$, i.e. $g$ is surjective.

These concepts and facts have duals. I think it would be a nice exercise for you to work out the following.

A module is called uniform if every submodule is essential.

Exercise (A) Let $f: M \hookrightarrow N$ an injective morphism of $R$-modules. The following are equivalent: (1) $f(M)$ is an essential submodule of $N$.
(2) For every $g: N \rightarrow O$, $gf$ injective implies $g$ is injective.

Exercise (B) Let $f: M \hookrightarrow N$ be injective with essential image. If $M$ is uniform then $N$ is uniform.

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  • $\begingroup$ Are you saying $f^{-1}$ open sum? Thats not true or why you can set $M_{i}=f^{-1}(N_{i})$ for $i=1,2$?? @Badam Baplan $\endgroup$ – Cos May 29 at 3:16
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    $\begingroup$ @cos What is "open sum"? I'm just defining $M_1 = f^{-1}(N_1)$ and likewise for $M_2$ $\endgroup$ – Badam Baplan May 29 at 3:20
  • $\begingroup$ Sorry I got confused, think I already understand your proof but still understanding some details. I really aprecciate your throught this one @Badam Baplam $\endgroup$ – Cos May 29 at 3:31
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    $\begingroup$ @Cos Sure, feel free to ask if something is unclear. $\endgroup$ – Badam Baplan May 29 at 3:37
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    $\begingroup$ @Cos i updated and gave you some exercises to think about! $\endgroup$ – Badam Baplan May 30 at 16:14

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