Suppose $E$ is a finitely generated modul over a commutative ring $R$ with unity. Let $e_1,\ldots,e_n$ be generators of $E$, $I$ an ideal in $R$ and $u\in\Hom_R(E)$, such that:
$$
u(E)\sbe\Big\{\sum a_je_j:a_j\in I\Big\}~.
$$
Then there is a monic polynomial $p(x)=x^n+p_1x^{n-1}+\cdots+p_{n-1}x+p_n\in R[X]$ such that $p_k\in I^k$ and $p(u)=0$.
$\proof$
Since $e_1,\ldots,e_n$ are generators of $E$ and $u(E)\sbe IE$, there are $a_{jk}\in I$ such that for all $k=1,\ldots,n$:
$$
u(e_k)=\sum_l a_{kl}e_l
$$
Denoting by $1_E\in\Hom_R(E)$ the identity map, this says that
$$
\forall k=1,\ldots,n:\quad\sum_l(\d_{kl}u-a_{kl}1_E)e_l=0~.
$$
Definie a matrix $M\in\Ma(n,R[u])$ by
$$
M\colon=(\d_{jk}u-a_{jk}1_E)_{j,k=1}^n
=\left(\begin{array}{cccc}
u-a_{11}&-a_{12}&\ldots&-a_{1n}\\
-a_{21}&u-a_{22}&\ldots&-a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
-a_{n1}&-a_{n2}&\ldots&u-a_{nn}
\end{array}
\right)~.
$$
Then by means of matrix multiplication these relations can be written as:
$$
M\cdot(e_1,\ldots,e_n)^t=(0,\ldots,0)^t
$$
Since the unital ring $R[u]$ is commutative, the adjugate matrix $M^{ad}\in\Ma(n,R[u])$ of $M$ and $M$ (cf. e.g. wikipedia) satisfy the following relation:
$$
M^{ad}M=\det(M)diag\{1_E,\ldots,1_E\}~.
$$
It follows that
$$
(0,\ldots,0)^t
=M^{ad}\cdot(0,\ldots,0)^t
=M^{ad}M\cdot(e_1,\ldots,e_n)^t
=\det(M)diag\{e_1,\ldots,e_n\}
$$
i.e. $\det(M)\in R[u]\sbe\Hom_R(E)$ maps all generators $e_1,\ldots,e_n$ of $E$ to $0$, hence $\det(M)=0$ and thus the polynomial $p\colon=\det(\d_{jk}X-a_{jk})\in R[X]$ satisfies $p(u)=0$. Finally the assertion: $p_k\in I^k$ follows from the definition of the determinant.
$\eofproof$