# Cayley-Hamilton Theorem/Finitely Generated Modules

## Theorem

Let $A$ be a commutative ring with unity.

Let $M$ be a finitely generated $A$-module.

Let $\mathfrak a$ be an ideal of $A$.

Let $\phi$ be an endomorphism of $M$ such that $\map \phi M \subseteq \mathfrak a M$.

Then $\phi$ satisfies an equation of the form:

$\phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$

with the $a_i \in \mathfrak a$.

## Proof

Let $\set {m_1, \ldots, m_n}$ be a generator for $M$.

Then for each $i \in \set {1, \ldots, n}$, as $\map \phi {m_i} \in \mathfrak a M$, we can write:

$\ds \map \phi {m_i} = \sum_{j \mathop = 1}^n a_j m_j$

where $a_j \in A$ for $j \in \set {1, \ldots, n}$.

Thus for each $i \in \set {1, \ldots, n}$:

$(1): \quad \ds \sum_{j \mathop = 1}^n \sqbrk{\delta_{ij} \phi - a_{ij} \chi_M } m_i = 0$

where:

$\delta_{i j}$ is the Kronecker delta with values in $A$
$\chi_M$ is the identity mapping on $M$

Recall that the set of automorphisms $\Aut M$ is an $A$-module.

Recall also the ring of $n$-matrices $\map {\MM_{\Aut M} } n$.

Now let $\Delta \in \map {\MM_{\Aut M} } n$ defined as:

$\Delta := \paren {\phi \delta_{i j} - a_{i j} \chi_M}$

Then $\paren 1$ can be written as:

$\Delta \cdot \begin{bmatrix} m_1 \\ \vdots \\ m_n \end{bmatrix} = \mathbf 0_{ {\Aut M}^n}$

Let $\adj \Delta$ be the adjugate matrix of $\Delta$.

Recall Cramer's Rule with respect to $\map {\MM_{\Aut M} } n$:

 $\ds \adj \Delta \cdot \Delta$ $=$ $\ds \Delta \cdot \adj \Delta$ $\ds$ $=$ $\ds \map \det \Delta \mathbf I_n$ $\ds$ $=$ $\ds \begin{bmatrix} \det \Delta & 0 & \cdots & 0 \\ 0 & \det \Delta & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \det \Delta \\ \end{bmatrix}$

Multiplying the above equality by $\adj \Delta$ and applying Cramer's Rule:

$\ds \map \det \Delta m_i = 0$

for each $i \in \set {1, \ldots , n}$.

Therefore $\map \det \Delta$ annihilates each $m_i$ and is the zero endomorphism of $M$.

But $\map \det {\delta_{i j} \phi - a_{i j} \chi_M }$ is a monic polynomial in $\phi$ with coefficients in $\mathfrak a$.

Thus we have an equation of the required form.

$\blacksquare$

## Source of Name

This entry was named for Arthur Cayley and William Rowan Hamilton.