Lemma 10.16.2. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $\varphi : M \to M$ be an endomorphism. Then there exists a monic polynomial $P \in R[T]$ such that $P(\varphi ) = 0$ 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 $(a_{i1}, \ldots , a_{in}) \in R^{\oplus n}$ such that $\varphi (x_ i) = \sum a_{ij} x_ j$. In other words the diagram
\[ \xymatrix{ R^{\oplus n} \ar[d]_ A \ar[r] & M \ar[d]^\varphi \\ R^{\oplus n} \ar[r] & M } \]
is commutative where $A = (a_{ij})$. By Lemma 10.16.1 there exists a monic polynomial $P$ such that $P(A) = 0$. Then it follows that $P(\varphi ) = 0$. $\square$
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