Determining number of isomorhism classes of indecomposable modules of length $k$ Let $F$ be a field and $A$ be a finite dimensional $F$-algebra. We denote the number of isomorphism classes of indecomposable $A$-modules of length $k$, by $n_A(k)$ for all $k\in\mathbb{N}$.
I have to prove that:
(a) If $F$ is finite then $n_A(k)$ is finite for all $k$,
(b) If $F$ is infinite then $n_A(k)\leq |F|$ for all $k$.
For the proof of (a),
If $k=1$, then $n_A(1)$ is number of isomorphism classes of simple modules. We know that simple $A$-modules are determined by the maximal ideals of $A$. Since $A$ is finite, we have that $n_A(1)$ is finite. [Similar argument for $(b)$, proves that $n_A(1)\leq|F|$]
Let $\mathcal{N}_A(k)$ denotes the set of isomorphism class of indecomposable $A$-modules. We assume that $\mathcal{N}_A(k)$ is finite. Let $M\in\mathcal{N}_A(k+1)$. Let $0=M_0\subset M_1\subset\cdots\subset M_{k+1}=M$ be a composition series of $M$. Then $M/M_1\in\mathcal{N}_A(k)$. So we have finitely many choices for $M/M_1$. Also we have finitely many choices for $M_1$. I don't know how to proceed after this or if it'll be possible to solve the problem this way.
Also thought of using the result that
"If $A$ is right Artinian (here it is!), then there is a bijective correspondence between isomorphism classes of principal indecomposable $A$-modules and the isomorphism classes of simple $A/J(A)$-modules." But I couldn't find a way out.
Also I don't have any idea about towards $(b)$.
Any hint will be helpful. Thank you.
 A: Here's how one could answer (b).  In fact we'll prove the following (equivalent) statement:  If $F$ is infinite, then the set of isomorphism classes of modules of finite length has cardinality at most $|F|$.
Assume that $F$ is infinite, and let $M$ be a module of finite length over $A$.  Then there exists an epimorphism
$$
 A^k \to M \to 0,
$$
where $k$ is a non-negative integer.  This is just a way of saying that there exists a finite set $\{m_1, \ldots, m_k\}$ which generates $M$ as an $A$-module.  In fact, since $A$ is finite-dimensional over $F$, the existence of such an epimorphism is equivalent to the requirement that $M$ is of finite length.
This characterization of finite-length modules will give us the upper bound we are looking for.  Indeed, it implies that the set of isomorphism classes of finite length modules has cardinality bounded by that of the set of quotient modules of free modules $A^k$, with $k\geq 0$.  Now, fixing $k\geq 0$, a quotient of $A^k$ has the form $A^k/L$ for some $A$-submodule $L$ of $A^k$.  Thus quotients of $A^k$ are in bijection with submodules of $A^k$.  Any $A$-submodule of $A^k$ is in particular an $F$-subvector space of $A$ (viewed as a vector space).  The set of subvector spaces of $A^k$ of a given dimension $d$ is a projective variety called the Grassmannian $Gr(d,A^k)$. In particular, since $F$ is infinite,  $|Gr(d,A^k)|$ is at most $|F|$ (see below for another argument).
Therefore, the cardinality of the set of isomorphism classes of $A$-modules of finite length is bounded above by the cardinality of the union of all $|Gr(d,A^k)|$, where $d$ and $k$ are non-negative integers.  Since each $Gr(d,A^k)$ has cardinality bounded above by $|F|$, their (countable) union over all $d$ and $k$ will have cardinality bounded by $|F|$ as well.  This finishes the proof.

Here's an argument for the fact that $|Gr(d,A^k)|$ is at most $|F|$ when $F$ is infinite.  First, if $d=0$, then $|Gr(d,A^k)|=1$, and if $d > \dim A^k$, then $|Gr(d,A^k)|=0$.
Assume that $d>0$ and $d\leq \dim A^k$.  Any subvector space of dimention $d$ of $A^k$ is given by a basis.  Such a basis can be expressed as a $(\dim A^k \times d)$ matrix whose columns are linearly independent (each column is a basis vector, all expressed in some fixed basis).  The cardinality of the set of $(\dim A^k \times d)$ matrix is $|F|$.  Thus $|Gr(d,A^k)|$ is bounded above by $|F|$.

Similar ideas will prove (a): if $M$ is an $A$-module of length $k$, then there exists an epimorphism
$$
  A^k \to M \to 0
$$
(note that the $k$ in $A^k$ is the length of $M$).  Thus the number of isomorphism classes of $A$-modules of length $k$ is bounded above by the number of subvector spaces of $A^k$, which is finite.
