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Suppose $\mathcal C$ is a preadditive, Karoubi category with a zero object. What further assumptions on $\mathcal C$ are required to ensure that the endomorphism ring of an indecomposable object is a local?

By an object $X$ being indecomposable, I mean that in any biproduct decomposition $X \cong X_1 \oplus X_2$, one has $X_i \cong 0$ for some $i=1,2$.

Thanks for your help.

flag	asked Jul 22 2011 at 10:18 benjamin 23●2

Johannes Hahn · Accepted Answer · 2011-07-22 17:29:51Z UTC

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Hi benjamin.

You're asking if $\mathcal{C}$ is a so called Krull-Schmidt-category. There are several sufficient conditions know from ring theory if $\mathcal{C}$ is a module category. For example the argument George mentioned gives an affirmative answer if $\mathcal{C}$ is the category of finitely generated (left)modules over an artinian ring. You get another characterisation by considering semiperfect rings. A ring is semiperfect iff the category of finitely generated projective (left)modules is Krull-Schmidt. Examples of semiperfect rings include all left or right artinian ring as well as all finitely generated $R$-algebras where $R$ is a complete local ring or a discrete valuation ring.

This connection between semiperfect rings and Krull-Schmidt-categories is in fact somewhat stronger: One can show that $\mathcal{C}$ is Krull-Schmidt iff $End_\mathcal{C}(X)$ is semiperfect for all objects $X\in\mathcal{C}$.

answered Jul 22 2011 at 17:29

Johannes Hahn
1,725●6●12

	nice answer! however, being Krull-Schmidt does not look equivalent to endos of indecomposables being local (will call this having local indecomposables). With the standard definition (the one for which the characterization of $\mathcal{C}$ in terms of $End_\mathcal{C}(X)$ semiperfect for all $X$ holds) being Krull-Schmidt implies having local indecomposables. But it looks like you'll need to require the existence of direct sum decompositions into indecomposables, which is not a trivial matter. – Eduardo Pareja Tobes Jul 23 2011 at 15:28
	What you could say is something like $\mathcal{C}$ has local indecomposables iff the full subcategory of finite direct sums of indecomposables is a Krull-Schmidt category; but this is a little bit trivial. – Eduardo Pareja Tobes Jul 23 2011 at 15:29
	this is the answer i was looking for, since existence of indecomposables and of decompositions into indecomposables isn't important to me (reading the question again, i realise it wasn't clear). i've expanded on johannes' answer below. – benjamin Jul 23 2011 at 17:38
	details: a ring is called semiperfect if the quotient by its radical is semisimple and idempotents lift modulo the radical. thus any left artinian ring is semiperfect, in particular. (see e.g. Anderson & Fuller) $\cat C$ is Karoubi, so idempotents split; so $R=EndX$ has no non-trivial idempotents. $R$ is semiperfect, so $R/J$ has none either, and is moreover semisimple ($J=rad(R)$). thus the regular $R/J$-module is simple, so $J$ is a maximal left ideal of $R$. it follows that $J$ is the unique maximal ideal; thus $R$ is local. – benjamin Jul 23 2011 at 17:43
	@eduardo: actually, it seems semiperfect endomorphism rings is sufficient to guarantee a decomposition into indecomposables, if $\mathcal{C}$ is Karoubi. a ring $R$ is semiperfect iff there exists orthogonal idempotents $e_i$ summing to $1$ such that $e_i R e_i$ is a local ring (see e.g. Anderson & Fuller 26.7). Thus, writing $im(e_i)$ for a splitting of $e_i$, we have $X \cong \oplus_i im(e_i)$, and that $End im(e_i) = e_i R e_i$ is local. Local rings have no non-trivial idempotents, so it follows that $im(e_i)$ is indecomposable. – benjamin Jul 23 2011 at 17:57

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George C. Modoi · Answer 2 · 2011-07-22 16:58:14Z UTC

Suppose your additive category is an abelian one (for example a category of modules), and let $X$ be an object. Suppose $X$ is of finite length, that is there is a filtration $0=X_0\leq X_1\leq\cdots\leq X_n=X$, such that $X_i/X_{i-1}$ is simple for all $i=1,\cdots,n$. Then it is easy to show that every object is both artinian and notherian, and for every endomorphism $f:X\to X$ the conditions to be monomorphism, epimorphism and automorphism are equivalent. Therefore, the endomorphism ring of $X$ is local.
Thus in a slightly more particular case, a sufficient condition is every object is of finite length.

endomorphism rings of indecomposable objects

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