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The question asks to prove that if $M+N$ and $M \cap N$ are finitely generated modules, then M and N are also finitely generated.

I've tried to use basic definitions, but all failed.

I set some examples to study how I can find a generating set for M (or N) when I have the generators of $M+N$ and $M \cap N$ but then I realized that there's no simple way to do so because $3\mathbb{Z}+5\mathbb{Z}=\mathbb{Z}$ is a finitely generated $\mathbb{Z}$-module generated by $\{1\}$ and $3\mathbb{Z} \cap 5\mathbb{Z}=15 \mathbb{Z}$ is also a finitely generated $\mathbb{Z}$-module generated by $\{15\}$ but $M=3\mathbb{Z}$ is generated by $\{3\}$ which doesn't tell me anything conclusive.

I then set $M=\{(x,y,0): x,y \in \mathbb{R} \}$ and $N=\{(x,0,z): x,z \in \mathbb{R} \}$. Then $M+N=\mathbb{R}^3$ and $M \cap N = \{ (x,0,0): x \in \mathbb{R} \}$ both are finitely generated modules. The set $\{(0,1,1),(0,-1,1),(1,0,-1)\}$ is a generator for $M+N=\mathbb{R}^3$ and $\{(1,0,0)\}$ is a generator for $M \cap N$ but again there's no obvious way of obtaining generators for M or N.

Any insightful ideas will be appreciated.

EDIT: I have to add that this question is an exercise in the chapter 1 of the book I'm reading. Chapter 1 covers only basic definitions of modules, examples of modules, sub-modules, quotient modules and generating sets for modules. R-homomorphisms, exact sequences, Isomorphism theorems extended for modules, all are discussed in chapter 2. So I think I'm not allowed to use the materials covered in chapter 2 to solve exercises in chapter 1.

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up vote 6 down vote accepted

I assume that $M,N$ are submodules of a given $R$-module. Since $M+N$ is finitely generated, the same is true for $(M+N)/N \cong M/(M \cap N)$. Since $M \cap N$ is finitely generated, this implies that $M$ is finitely generated. By symmetry, also $N$ is finitely generated.

More explicitly (especially when you don't know quotients and isomorphism theorems yet): Let $\{m_i+n_i\}$ be a generating set of $M+N$. Let $\{u_j\}$ be a generating set of $M \cap N$. Then I claim that $\{m_i\} \cup \{u_j\}$ is a generating set of $M$ (in particular, if $M+N$ and $M \cap N$ are finitely generated, then the same is true for $M$). In fact, let $m \in M$. Then we can find $a_i \in R$ with $m = \sum_i a_i (m_i + n_i)$. It follows that $m - \sum_i a_i m_i \in M \cap N$, hence there are $b_j \in R$ with $m - \sum_i a_i m_i = \sum_j b_j u_j$, i.e. $m = \sum_i a_i m_i + \sum_j b_j u_j$, QED.

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In the given example ($M=3\Bbb Z,\ N=5\Bbb Z$), $1=2\cdot 3-5$ generates $\Bbb Z=M+N$, so $\{m_i\}_i=\{6\}$ now, and $\{u_j\}_j=\{15\}$ generates $M\cap N$. Then the generating set for $M$ we get by this procedure is $\{6,15\}$. – Berci Mar 14 '13 at 14:34
+1 for the solution without isomorphisms and quotients. – azimut Mar 14 '13 at 15:33

If think the exact sequence $0 \to M \cap N \to M \oplus N \to M + N \to 0$ should help to find a finite number of generators for $M \oplus N$ and thus for $M$ and $N$.

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What are the maps in the sequence? – Fredrik Meyer Mar 14 '13 at 13:58
Is such a sequence possible? I would like to know how to look at it this way... I can't guess what the second map would be. – rschwieb Mar 14 '13 at 14:02
The maps are $x \mapsto (x,-x)$ and $(x,y) \mapsto x+y$. – Martin Brandenburg Mar 14 '13 at 14:05

This question and its solutions would be a useful tool: $M$ finitely generated if submodule and quotient are finitely generated.

It gives you a lemma that if $A/B$ and $B$ are finitely generated, then so is $A$.

Then you would consider $\frac{M+N}{M}\cong\frac{N}{M\cap N}$ for inspiration.

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