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Let $R$ be a unitary commutative ring, let $I$ be an ideal of $R$ and let $M$ be a finitely generated $R$-module. Suppose that $\phi$:$M$$\to$$M$ is a homomorphism such that $\phi$($M$)$\subseteq$$IM$. I wish to prove only using induction that there exist $a_i$, 1$\le$$i$$\le$$n$, in $I$ with $\phi^n+a_1\phi^{n-1}+...+a_n1_M=0$. When $M$ has only one generator, I can verify this fact. But I wonder how to do the inductive step.

P.S. Apparently all textbooks have presented the same proof for this theorem using the notion of determinant. But I think it must have a proof based on induction.

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  • $\begingroup$ I like this question, but you didn't specify what $n$ is. Do you want to prove that for each $\phi$ such an $n$ exist (with induction on the number of generators of $M$) or do you assume that $n$ is the number of generators of $M$, or even something else (e.g. the number of generators over $R$ of $M/IM$)? $\endgroup$ – Vincent Dec 20 '17 at 13:08
  • $\begingroup$ $n$ is assumed to be the number of generators of $M$ @Vincent $\endgroup$ – user514216 Dec 20 '17 at 13:25

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