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?
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$\begingroup$ Bound in terms of what? Any division $k$-algebra $D$ that is finite dimensional over $k$ is possible for $\text{End}_A (X)/m$, as you could just take $A=X=D$. $\endgroup$– Jeremy RickardCommented Jun 15, 2022 at 18:33
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$\begingroup$ For a fixed $A$. $\endgroup$– kevkev1695Commented Jun 15, 2022 at 18:39
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$\begingroup$ If you fix $A$ you will need $A$ to have infinite representation type to have any chance of unbounded dimension since you will needed infinitely many indecomposables. $\endgroup$– Benjamin SteinbergCommented Jun 15, 2022 at 19:14
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$\begingroup$ You might look at the notion of endo-wild algebras. They should give examples. I don't know enough to know if they exist over Q as the base ring $\endgroup$– Benjamin SteinbergCommented Jun 15, 2022 at 19:24
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1 Answer
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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 $X$.
answered Jun 16, 2022 at 6:30