0

Let $R$ be a commutative Artinian ring with unity. Prove that the surjective endomorphisms of every $R$-module are isomorphisms.

I have failed to sketch the proof for this. I do not even have a source of reference.

Proof:

Since $R$ is semiperfect, it is decomposable into a direct sum of indecomposable local rings. This implies that every $R$-module is finitely generated.

mariam
  • 283
  • 1
    Are the modules assumed to be finitely generated? Otherwise, consider the endomorphism of $R^{\oplus \mathbb N}$ that deletes the first coordinate and shifts everything else one position to the left. – Ravi Fernando Aug 11 '20 at 23:29
  • 2
    If the module is finitely generated, then it will have finite length. This property holds for modules of finite length. – Angina Seng Aug 11 '20 at 23:50
  • All nonzero rings have modules which aren’t finitely generated, so your last deduction is outlandish. – rschwieb Aug 12 '20 at 02:46

1 Answers1

0

It is false as noted in the comments.

It is fixed if the module is finitely generated. For then it follows that $R$ and $M$ are Noetherian, and that problem appears many places on this site.

rschwieb
  • 160,592