Finite length $A$-algebras are finitely generated? Let $k$ be a field and $M$ a module over a (associative, unital) finite-dimensional $k$-algebra $A$.
The length of $M$ is the unique length of a composition series for $M$.
How does $M$ having finite length imply that $M$ is finitely generated as an $A$-module?
I know that there are similar statements for modules over artinian rings (and that finite-dimensional algebras are artinian rings etc.) but I can't at the moment see how to show this directly even though I've been told it's apparently easy.
 A: Let $0 = M_0 \subsetneq M_1 \subsetneq \dotsb \subsetneq M_t = M$ be a composition series for $M$, so every factor $M_i / M_{i-1},\ 1 \leq i \leq t$ is simple and so $M_{i-1}$ is maximal in $M_i$.
For every $1 \leq i \leq t$ choose $x_i \in M_i \setminus M_{i-1}$, then $M_i$ is the module generated by $M_{i-1} \cup \{x_i\}$, which is also the module generated by $X_{i-1} \cup \{x_i\}$ where $X_{i-1}$ is a generating set for $M_{i-1}$ ($0$ is generated by the empty set).
By induction $X_t =  \{x_1, \dotsc, x_t\}$ generates $M_t = M$, so in particular $M$ has a generating set of cardinality $l(M)$.
A: Hint:Firstly remark that having length $1$ implies the module is finitely generated. Since it is an irreducible module over an associative algebra.
Suppose the result true for length $n$ module, let $M$ a module of length $n+1$,$N_0\subset N_1...\subset N_n\subset M$ a composition sequence of length $n+1$,shows that the length of $N_n$ is $n$ and the length of $M/N_n$ is 1. Conclude recursively.
A: A module of finite length is noetherian, so it is finitely generated.
