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Let $R$ be a commutative ring with $1$, $M$ be a finitely generated $R$-module, $\mathfrak{i}$ an ideal of $R$, and $\phi$ an $R$-homomorphism such that: $$1.\;\phi(M)\subseteq \mathfrak{i}M=M$$ i. Prove that for some $n$ and $a_i\in\mathfrak{i}:\;\phi^n+a_{n-1}\phi^{n-1}+\cdots+a_0=0$

As was suggested by the author, I proved this by adapting the proof of Cayley-Hamilton I know, using $\mathbf{A}\,\text{adj}\,\mathbf{A}=I\det \mathbf{A}$. My set-up was that since $\phi$ is an $R$-homomorphsim its action can be represented by a matrix $\mathbf{P}$ with entries in $R$, and since $\phi(M)\subseteq\mathfrak{i}M$ these entries must lie in $\mathfrak{i}$. Moreover $M\cong R^n/N$ for some $n$ since finitely generated, so this matrix does indeed have finite dimension.

  • is this enough justification to then carry on with the usual Cayley-Hamilton proof?

ii. deduce that there exists $r\in R$ such that $rM=0$ and $r-1\in \mathfrak{i}$.

  • plugging in $M$ (I interpret $a_0$ as the map $a_0\,\text{id}$), I get $Q+a_0M=0$ where $Q$ is a submodule of $M$. If we had $Q=M$ then $r=1+a_0$ gives me what I want, but I can't see why $Q$ should be the whole of $M$ (if it indeed should), so a little stuck here.

Thank you for any help.

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marked as duplicate by user26857 abstract-algebra May 26 '16 at 10:59

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