How to show that if $M$ is an $A$ module of finite type and $L$ and $N$ are submodules such that $L+N$ and $L\cap N$ are of finite type, then $L$ and $N$ are of finite type?
I am having a real problem with this exercise, although it should be simple.
I can show that if $\pi : L+N\to N$ (or to $L$), then since $L+N$ is of finite type, there is a surjection $$\varphi : A^k \to L+N,~k\in\mathbb{Z},$$ so the composite $$\pi\circ \varphi : A^k \to N~~\text{or to $L$} ~(*)$$ defines a surjection.
I am not quite sure about the definition of finite type. I read somewhere that it relies on the existence of a surjection like $(*)$ but this can't be the case, since we didn't use $L\cap N$ has finite type.
Any hints, suggestions, answers?
You're correct on the definition of finite type (more commonly referred to as "finitely generated"). But your homomorphism $\pi$ may not be well-defined, since there may be an element $x\in L+N$ such that $x = l + n = l' + n'$ for $n\neq n'$.
Instead you should consider the exact sequence
$$ 0\to L\cap N \to L\oplus N\to L + N \to 0 $$
Here the map $L\oplus N\to L+N$ is just subtraction. The lemma that you would then want to prove is that given an exact sequence of the form
$$ 0\to A \to B\to C \to 0 $$
with $A$ and $C$ finitely generated, prove that $B$ is finitely generated. To do this, simply take a set of generators $\{a_1,...,a_n\}$ for $A$ and a set of generators $\{c_1,...,c_m\}$ for $C$. Then let $\{b_1,...,b_m\}$ be a set of elements of $B$ mapping to the $c_i$. It should be easy now to show that $\{a_1,...,a_n,b_1,...,b_m\}$ generate $B$.