The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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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