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I know that if I have a matrix $A,$ the characteristic polynomial is determinant of the matrix $(A-\lambda I)$ where, $\lambda$ is an eigenvalue and $I$ is an identity matrix, and the characteristic equation is the characteristic polynomial equated to zero.

Let $k$ be a field and $R = k[x]$, I want to find the characteristic polynomial of an $R$-module of order (product of its prime ideals or product of all elementary divisors/ invariant factors) is $(x-1)^3(x + 1)^2.$ but I do not know where is the matrix in my case? where is the eigenvalue?

Could anyone help me answer those questions, please?

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    To give a $k[x]$-module is, up to isomorphism, to give a $k$-vector space $V$ together with a linear operator $T:V \to V$ (the idea is that $x^n$ acts on $V$ by $T^n$). In spelling out this equivalence you should be able to see how to build $T$ to have the order you desire (hint: Cayley-Hamilton Theorem). – Geoff Apr 05 '21 at 04:28
  • @Geoff is there any use of smith normal form? –  Apr 05 '21 at 05:21
  • It depends on how you construct the matrix. If you do something like the first answer in this question, the answer is probably. – Geoff Apr 05 '21 at 05:54
  • @Geoff Are you saying that I have my characteristic polynomial and I am trying to find the matrix corresponding to it here in my case? –  Apr 05 '21 at 06:09
  • I can see how can I use Cayley Hamilton theorem, could you explain a little bit more to me please? @Geoff –  Apr 05 '21 at 06:11
  • what is $n$ in $T^n$ in my case, is it just $2,3$?@Geoff –  Apr 05 '21 at 06:13

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