Proving $End_A(V) $ is a local ring whose Jacobson -radical is a nil-radical ideal

Question

Consider the following question asked in my quiz on Algebraic Geometry.

Let A be a ring and V be an indecomposible A-module which is artinian and well as noetherian. Then $End _AV$ is a local ring whose Jacobson -radical is a nil - radical ideal.

Attempt: Let f belongs to $ End_A V$ and Then there exists $m \in \mathbb{N}$ and $m'\in \mathbb{N}$ such that $Ker f^n = Ker f^m$ for all $n \geq m $ and $Img f^n = Img f^{m'}$, $n \leq m'$ . I can write $V = Ker f^m \oplus Img f^{m'}$ . Now, V is indecomposible implies that either $ Kerf^m =0$ (1)or $ Img f^{m'} =0$(2).

If (1) is true then $f^m$ is injective. If (2) is true then f is nilpotent.

But still I am not sure how should I proceed to prove what is asked.

Can you please help with outlines?

Answer 1

An injective endomorphism on an Artinian module is an isomorphism. So you have shown that all elements are nilpotent or units.

If $x$ is nilpotent, then $1-x$ is a unit

By a well-known characterization of local rings this would mean that the ring is local. Needless to say its maximal ideal is the set of all nilpotent elements, and hence it is nil.

(Actually I found another old post which covers the two points above too: here).

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Alternatively there is an even more sledgehammery approach follows using an interesting theorem that endomorphism rings of modules of finite length are semiprimary. For noncommutative rings in general it may not be local because the indecomposability hypothesis was not used. To show it is local, we have to explain why the semisimple ring $R/J(R)$ has no nontrivial idempotents, and therefore is a division ring.

If $R/J(R)$ had a nontrivial idempotent, it would lift to a nontrivial idempotent of $R$. But a nontrivial idempotent of $R=End(V_A)$ by definition decomposes $V$ into two nonzero halves. Since this has been ruled out, $R/J(R)$ is a division ring, hence $R$ is a local semiprimary ring.