0

Here is the question I want to answer:

Let $M$ be a finite $R$-module. Show that if $f \in \operatorname {End_R(M)}$ is surjective then it is also injective.

Hint: Let $R[X] \cong R^{[1]}$ and make $M$ into an $R[X]$-module by setting $Xm = f(m)$ for $m \in M.$ Show that $XM = M$ and apply Nakayama's Lemma.

My question is:

1- I do not understand why setting $Xm = f(m)$ for $m \in M$ will make $M$ into an $R[X]$-module. Could anyone clarify this to me, please?

2- Also, I do not see how I will use surjectivity in the proof that $XM = M$. Could any one elaborate this to me please?

user26857
  • 52,094

2 Answers2

0

Saying $M$ is an $R$-module is same as giving an abelian group $M$ along with a ring homomorphism $\psi: R\rightarrow \operatorname{End}_\mathbb Z (M)$. Since $f\in End_R(M)$, by the universal property of polynomial algebra $\exists! \ \Psi :R[X]\rightarrow \operatorname{End}_\mathbb Z (M)$ which extends $\psi$ and takes $X\mapsto f$. This makes $M$ an $R[X]$ module.

Note every $m\in M$ is of the form $f(m')=X.m'$ for some $m'\in M$.

user6
  • 4,092
  • 1
    it is better to explain to me what was the O.P. and the author of the answer doing here https://math.stackexchange.com/questions/2314566/surjective-endomorphisms-of-finitely-generated-modules-are-injective. Also what about my second question. –  Jan 26 '21 at 04:33
  • Here's a basic module theory exercise. Fix a ring $R$. Let $R-Mod$ be the category of left $R$-modules. Let $C$ be the category of all abelian groups together with a ring homomorphism from $R$ to the endomorphism ring (you should figure out what the morphisms are). Then $R-Mod$ and $C$ are isomorphic categories – user6 Jan 26 '21 at 04:39
  • I think I understand now what you said but how is this a proof of the question? –  Jan 26 '21 at 08:07
  • Where is your proof for $XM = M$? –  Jan 26 '21 at 08:22
  • How did we used that $M$ is finitely generated in that proof? –  Jan 26 '21 at 08:22
0

Let $R$ be a commutative ring with an $R$-module $M.$ Given any $R$-linear map $f : M \to M,$ consider the map $\cdot : R[X] \times M \to M$ defined by $r \cdot m = rm$ and $X \cdot m = f(m).$ By hypothesis that $f$ is $R$-linear, it follows that $X \cdot (m + m') = f(m + m') = f(m) + f(m') = X \cdot m + X \cdot m',$ hence $\cdot$ gives $M$ an $R[X]$-module structure that can be extended linearly to find $p(X) \cdot m$ for any element $p(X)$ of $R[X].$ (Check for yourself that the other properties work out.)

Given that $f : M \to M$ is surjective, it follows that for any element $m$ of $M,$ there exists an element $m'$ of $M$ such that $m = f(m') = X \cdot m'.$ Consquently, we have that $M \subseteq XM$ (here, we are suppressing the $\cdot$ notation for brevity). On the other hand, it is clear that $XM \subseteq M$: for any element $m$ of $M,$ we have that $X \cdot m = f(m)$ is in $M.$ We conclude that $XM = M.$

Last, we claim that $M$ is a finitely generated $R[X]$-module. By hypothesis that $M$ is a finitely generated $R$-module, there exist elements $m_1, \dots, m_n$ of $M$ such that every element $m$ of $M$ can be written as $m = r_1 m_1 + \cdots + r_n m_n$ for some elements $r_1, \dots, r_n$ of $R.$ By viewing the $r_1, \dots, r_n$ as constant polynomials of $R[X],$ it follows that $m = r_1 \cdot m_1 + \cdots + r_n \cdot m_n$ so that $M$ is a finitely generated $R[X]$-module. Consequently, by Nakayama's Lemma (applied to the finitely generated $R[X]$-module $M$ and the principal ideal $(X)$ of $R[X]$), there exists an element $p(X)$ of $(X)$ such that $(1_R + p(X)) M = 0.$ Can you finish the proof from here?

Dylan C. Beck
  • 4,811