Let $k$ be a field, $A$ a $k$-algebra and $X$ a finite dimensional indecomposable $A$-module. Then $\text{End}\_A (X)$ is a local ring. Let $m$ be its maximal ideal. Can we say anything about the $k$-dimension of $\text{End}\_A (X)/m$? It always equals $1$ if $k$ is algebraically closed. Can we bound it if $k$ is not algebraically closed?



Dimension of division rings coming from indecomposable modules
Even if $A$ is a finite dimensional $k$-algebra, there may be no bound on the dimension of $\text{End}\_A(X)/m$.


Let $Q$ be the Kronecker quiver (i.e., the quiver with two vertices and two arrows from the first vertex to the second), and let $A=kQ$ be its path algebra over $k$.


Suppose $k(\alpha)$ is a simple algebraic field extension of $k$. Let $X$ be the representation of $Q$ over $k$ with $k(\alpha)$ at both vertices, the first arrow acting as the identity, and the second arrow acting as multiplication by $\alpha$. Then $\text{End}\_A(X)\cong k(\alpha)$.


So as long as $k$ has simple algebraic extensions of unbounded degree, there is no bound on $\dim\_k\left(\text{End}\_A(X)/m\right)$ for finite dimensional indecomposable $A$-modules $X$.


