Lemma 10.15.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Let $\varphi : M \to M$ be an endomorphism such that $\varphi (M) \subset IM$. Then there exists a monic polynomial $P = t^ n + a_1 t^{n - 1} + \ldots + a_ n \in R[T]$ such that $a_ j \in I^ j$ and $P(\varphi ) = 0$ as an endomorphism of $M$.

Proof. Choose a surjective $R$-module map $R^{\oplus n} \to M$, given by $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ for some generators $x_ i \in M$. Choose $(a_{i1}, \ldots , a_{in}) \in I^{\oplus n}$ such that $\varphi (x_ i) = \sum a_{ij} x_ j$. In other words the diagram

$\xymatrix{ R^{\oplus n} \ar[d]_ A \ar[r] & M \ar[d]^\varphi \\ I^{\oplus n} \ar[r] & M }$

is commutative where $A = (a_{ij})$. By Lemma 10.15.1 the polynomial $P(t) = \det (t\text{id}_{n \times n} - A)$ has all the desired properties. $\square$

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