# Why are morphisms between non-isomorphic indecomposable modules always radical?

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).

• 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. Dec 3, 2019 at 19:43
• 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. Dec 3, 2019 at 23:01

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

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:

and below is the proof of (b):

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.