I am trying to show that:Let $R$ be commutative with unity. if $N$ be a finitely generated $R-$module, then the following conditions are equivalent:
(a) $N$ is a sum of simple modules.
(b) $N$ is a direct sum of simple modules.
(c) Given any submodule $M \subset N,$ there exists a unique submodule $M' \subset N$ such that $N = M \oplus M'. $
For $(a) \implies (b),$ I said the following:
Assume $N$ is a sum of simple modules, we want to show that $N$ is a direct sum of simple modules i.e. $N = \oplus_{i=1}^n N_i$ where $N_i$ are simple modules. So to prove this, it is enough to show that $N_i \cap N_j = \phi.$ Assume towards contradiction that $M$ is a submodule of $N_i \cap N_j,$ then $M$ is a submodule of both $N_i, N_j$ but this is not possible because $N_i, N_j$ are simple which means they have no nontrivial submodule.
For $(b) \implies (c),$ I said the following:
Suppose that $N = \oplus_{i=1}^n N_i$ where $N_i$ are simple modules and suppose that $M$ is a submodule $N,$
$(1)$ if $M$ is simple then $M$ will be one of those $N_i.$ Let $M = N_1$ but then I do not know how to complete. Any help will be appreciated in that case.
$(2)$ if $M$ is not simple, also I do not know how to handle that case and I will be beyond grateful for any help.
It is clear to me that I have not used the main assumption that $N$ is a finitely generated $R-$module but I do not know how to use it.
For $(c) \implies (a),$ I said the following:
Given any submodule $M \subset N,$ there exists a unique submodule $M' \subset N$ such that $N = M \oplus M'$ and so $N$ is clearly a sum of modules but how to show that they are simple?
Also, any help will be greatly appreciated!
Best Answer
I think your hypotheses lack some necessary precision. Is $R$ commutative? unital? Noetherian, left or right? Is $N$ a left $R$-module? I think you intended $R$ to be commutative but what I wrote below doesn't require it.
$a) \Rightarrow b)$ seems incorrect or at least incomplete. It is possible for a vector space to be a sum of three vector subspaces which aren't in direct sum, even though they are pairwise in direct sum. For instance, $K^2=K(0,1) + K(1,0) + K(1,1)$, and the lines have pairwise intersection zero, but the overall sum isn't direct.
I'd prove it this way: let $N=\sum_i{N_i}$ where $N_i$ is simple. Because $N$ is finitely generated, we can assume that the $N_i$ are finitely many, and we can thus name them $N_1,\ldots,N_m$. For each $i$, call $N_i$ new if its intersection with $\sum_{j < i}{N_j}$ is zero. As $N_i$ is simple, if $N_i$ is not new then $N_i \subset \sum_{j < i}{N_j}$. It is then easy enough to check that $N$ is the direct sum of the new $N_i$.
Uniqueness in $b) \Rightarrow c)$ seems false. For instance, in $K^2$, we have $K^2=K(0,1)\oplus K(1,0)=K(0,1)\oplus K(1,1)$. But let's see if we can show existence: write a finite sum $N=\bigoplus_{i \in I}{N_i}$ where the $N_i$ are simple. Let $M$ be a submodule of $N$.
Let $J \subset I$ be a maximal subset such that $\sum_{j \in J}{N_j}$ is in direct sum with $M$. Let's show that $N=M \oplus \sum_{j \in J}{N_j}$. Clearly, it's enough to show that $P = M \oplus \sum_{j \in J}{N_j}$ contains every $N_i$. If $i \in J$, that's clear. If not, we know that $P$ isn't in direct sum with $N_i$ (by maximality of $J$), that is, $P \cap N_i$ is nonzero. As $N_i$ is simple, it follows $N_i \subset P$.
Again, how about existence in $c)$ implying $a)$? I'll simply prove that every submodule of $N$ satisfies existence in $c)$ as well. Indeed, let $M_1 \subset M$ be submodules, let $M' \subset N$ be such that $M \oplus M'=N$. Let then $M''$ be such that $M' \oplus (M'+M_1)=N$. Let $\pi:N \rightarrow M$ be the projection with kernel $M'$. It's easy to check that $M = \pi(M'') \oplus M_1$.
Note also that every submodule of $N$ is the image of $N$ under some projection so is finitely generated. In particular, every sequence of submodules $(M_n)_n$ such that $M_n \subset M_{n+1}$ is stationary.
Let $(M_n)$ be a sequence of submodules such that $M_n \supset M_{n+1}$. Let $M_n=N_n \oplus M_{n+1}$. Then $\left(\bigoplus_{p \leq n}{N_p}\right)_n$ is a nondecreasing sequence of submodules of $N$ so is stationary, which implies that for $n$ large enough $N_n=0$, ie $M_n=M_{n+1}$ so $M_n$ is stationary.
From this, it follows that the set of nonzero submodules of $N$ has minimal elements (the simple submodules). Let $S$ be their sum, we can write $N=S \oplus S_1$ for some submodule $S_1$. Assume $S_1$ is nonzero: since every non-increasing/nondecreasing sequence of submodules of $S_1$ is stationary, $S_1$ contains a simple submodule $P$. But $P \subset S$ by definition of $S$, so $P \subset S\cap S_1 = \{0\}$, a contradiction. So $N=S$ is sum of simple modules.