# For a projective cover $(\sigma, P)$ of a module $M$, $P$ is indecomposable implies $M$ is indecomposable

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.

• 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. – Zeek May 29 at 0:07
• 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 – 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.

|-------EDIT--------|

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.

• 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 – Cos May 29 at 3:16
• @cos What is "open sum"? I'm just defining $M_1 = f^{-1}(N_1)$ and likewise for $M_2$ – Badam Baplan May 29 at 3:20
• Sorry I got confused, think I already understand your proof but still understanding some details. I really aprecciate your throught this one @Badam Baplam – Cos May 29 at 3:31
• @Cos Sure, feel free to ask if something is unclear. – Badam Baplan May 29 at 3:37
• @Cos i updated and gave you some exercises to think about! – Badam Baplan May 30 at 16:14