Cayley–Hamilton theorem for modules

Remarks

Let $x_1, \ldots, x_n$ generate $M$ and write $\varphi(x_i) = \sum_j a_{ij} x_j$ with $a_{ij} \in I$. View $M$ as a module over $R[t]$ with $t$ acting as $\varphi$. The matrix $B = t I_n - (a_{ij})$ kills the column vector $(x_1, \ldots, x_n)$; multiplying by the adjugate of $B$ shows that $\det B$ kills every $x_j$, hence all of $M$. Expanding $\det B$ gives the polynomial, and its coefficient of $t^{n-i}$ is a sum of products of $i$ entries in $I$. For $I = R$ and $M = R^n$, with $\varphi$ given by a matrix $A$, one recovers the Cayley–Hamilton theorem $p_A(A) = 0$.

Going deeper

For a student

This is the engine behind the determinant trick and hence behind Nakayama's lemma. Taking $\varphi = \id_M$ when $IM = M$ gives an element $1 + a_1 + \cdots + a_n$ with $a_i \in I$ that annihilates $M$. It is also the standard tool to prove that an element is integral. If $x$ lies in a ring $S \supseteq R$ and $M$ is a faithful $R[x]$-module that is finitely generated over $R$, then $x$ satisfies a monic polynomial over $R$.