Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

up vote 5 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.

share|improve this answer
2  
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
2  
+1 for the solution without isomorphisms and quotients. –  azimut Mar 14 '13 at 15:33
add comment

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

share|improve this answer
2  
What are the maps in the sequence? –  Fredrik Meyer Mar 14 '13 at 13:58
1  
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
4  
The maps are $x \mapsto (x,-x)$ and $(x,y) \mapsto x+y$. –  Martin Brandenburg Mar 14 '13 at 14:05
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.