I am trying to follow the construction of the Auslander-Reiten quiver of the Kronecker algebra, given by Barot in his book Introduction to the Representation Theory of Algebras.

There, he defines a morphism $f: M \to N$ between $A$-modules $M$, $N$ to be radical, if for every morphism $g: N \to M$ it holds true that $\xi = 1_M - g \circ f$ is an isomorphism.

Now, in the proof of lemma 6.2, he states:

[...] Since morphisms between non-isomorphic indecomposables are always radical [...]

I am trying to figure out, why this is the case. I found out that $\ker(\xi)$ cannot be $M$, as this would mean that $g$ is a retraction and hence $N \cong M \oplus \ker(g)$ which yields $M=0$ or $M \cong N$, both of which can not hold by assumption.

However, I think that I mixed up the terms irreducible module and indecomposable module and hence I am not sure how to proceed from here (if it is even possible).

  • $\begingroup$ I don't think you confused the terms: you stated that if $N$ decomposes as a direct sum then one of those factors is $0$, which is what indecomposable means. $\endgroup$ Dec 3, 2019 at 19:43
  • $\begingroup$ Captain Lama, this is not the problematic part. However, this only suffices to show that $\ker(\xi) \neq M$. How can I proceed from here to showing that $\xi$ is an isomorphism? For irreducible modules it would follow directly at this point. $\endgroup$ Dec 3, 2019 at 23:01

1 Answer 1


Barot is using Example 3.23 of the same book, which is the following:

enter image description here

By a spectroid he means a $k$-linear category $\mathcal{C}$ such that

  1. $\mathcal{C}(x,y)$ is finite dimensional;
  2. $\text{End}_{\mathcal{C}}(x,x)$ is local for each $x\in\mathcal{C}$;
  3. objects are pairwise non-isomorphic.

For your ends, he is considering the category whose objects are indecomposable $A$-modules.

Now, the proof of this example is not given, or at least I couldn't find it, in the book. However, the same result is in appendix A3 of Assem-Simson-Skowronski:enter image description here

and below is the proof of (b):enter image description here

If you don't have access to this book, the result (I.1.3) is the following result.

Lemma Let $A$ be a $k$-algebra, then the following are equivalent for $a\in A$:

  • $a\in\text{rad}\,A$;
  • $a$ is in the intersection of all left maximal ideals of $A$;
  • for any $b\in A$, the element $1-ab$ has a two sided inverse;
  • for any $b\in A$, the element $1-ab$ has a right inverse;
  • for any $b\in A$, the element $1-ba$ has a two sided inverse;
  • for any $b\in A$, the element $1-ba$ has a left inverse.

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