# Endomorphismrings of maximal submodules.

The question I am interested in answering is the following:

Suppose that for a pair of $d$-dimensional modules $M$ and $N$ over a $k$-algebra ($k$ a field) $R$ we have that $\dim_k \rm{Hom}_R(X,M)\leq \dim_k \rm{Hom}_R(X,N)$ for all finite dimensional $R$-modules $X$. This gives a partial order on the space of $d$-dimensional modules called the hom-order, denoted by $M \leq _{\rm{hom}} N$ .Given a maximal submodule $M_1$ of $M$ does there exist a maximal submodule $N_1$ of $N$ such that $M_1\leq _{\rm{hom}} N_1$?

Now, a general fact in the case $M \leq _{\rm{hom}} N$ is that $\dim_k \rm{End}_R(M)\leq \dim_k \rm{End}_R(N)$. Moreover, the hom-order is insensitive to common direct summands, so one may as well suppose that $M$ and $N$ have no non-zero common direct summands. In that case $\dim_k \rm{End}_R(M) < \dim_k \rm{End}_R(N)$. This then suggests that given $M_1$ (maximal) one should look at maximal submodules of $N$ with dimension of endomorphism ring maximal among maximal submodules.

So my question here is: Can one say something sensible of the dimension of endomorphism rings of maximal submodules in general? In examples I have noticed that the iso-classes of maximal submodules of $N$ with maximal (in the sense of dimension) endomorphism ring is finite. Is this property perhaps a closed property (in the above setting*)?

(*) That is, maximal submodules of $N$ with maximal endomorphism ring. It is not true in general, e.g. $k$ field with $\rm{char}\; k =0$, let $R=k[x,y]/(x^2,y^2)$ then $\rm{rad }\; R=(x,y)$ has a 1-parameter family of maximal submodules given by $R/(x+\lambda y)$ where $\lambda \in \mathbb{P}^1(k)$.

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Which assumptions of the $k$-algebra do you make or allow? – Natalie May 4 '12 at 17:32
Finite dimensional k-algebra or if you like finitely generated k-algebra where k is an artinian ring but then one considers lengths of hom-spaces over k(or the center of R). – Tore Forbregd May 4 '12 at 20:09