1

I wonder if my proof of the above statement is correct. I did not find my proof in the linked answers, so now I am not sure if it is correct because it seems to be a little bit too easy.

Let be $f \colon M \to M$ a surjective endomorphisms of an $R$-module $M$. Then $M$ is an $R[x]$-module by letting $x$ act as $f$. Consider the ideal $I=(x)$. Then $IM=M$ holds since $f$ is surjective and $x \in I$. We can apply Nakayama's lemma because $M$ is a finitely generated $R[x]$-module. It follows that there exists an element $g \equiv 1 \ (\mathrm{mod}\ I)$ such that $gM=0$. Because $g \equiv 1 \ (\mathrm{mod}\ I)$ means that $g$ has constant term equal to $1$, it follows $n = 0$ for every $n \in \ker f$.

Surjective endomorphisms of finitely generated modules are isomorphisms

Surjective endomorphisms of finitely generated modules are injective

Heraklit
  • 387
  • After deducing that $g$ has constant term equal 1,u deduce that $f$ is invertible, i.e $f$ is an isomorphism, , with the inverse being a polynomial in $f$. Remark that the proof is exactly the same as the one found in the first link u provided! – belkacem abderrahmane Nov 03 '22 at 14:30
  • This is exactly what I found to be a bit cumbersome. Isn't it simpler as described in more detail in the answer below, to simply insert an element $n$ from the kernel into $g$ and see that $n=0$ follows. – Heraklit Nov 03 '22 at 14:39

1 Answers1

1

You can write $g$ as $1+h\cdot x$ for some $h \in R[x].$ If $m\in M$ belongs to the kernel of $f,$ note that $0=(1+h\cdot x)m=m+h\cdot f(m)=m.$ Hence, $f$ is an injection, so an isomorphism. This is the way I have usually seen the proof finished. You can see it as Theorem 2.4 in Matsumura's Commutative Ring Theory. You might want to check Vasconcelos' original paper, "On finitely generated flat modules," Trans. Amer. Math. Soc. 138 (1969) 505-512.

rschwieb
  • 153,510
Chris Leary
  • 2,993
  • @rschwieb - Thanks for the edit. It's amazing how what you think you are writing and what you actually write can be so different. – Chris Leary Nov 03 '22 at 17:12
  • Do you use Chrome? If you do and haven't already, I recommend the google scholar extension. It can generate such citations in a second or two. – rschwieb Nov 03 '22 at 17:16
  • @rschwieb - I do ue chrome, but have not used google scholar yet. I'll have to give it a go. – Chris Leary Nov 03 '22 at 17:47