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 $m_1, \ldots, m_n$ be a generating set for $M$.

Then for each $i$, $\map \phi {m_i} \in \mathfrak a M$, say:

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

for $i = 1, \ldots, n$.

Thus for each $i$:

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

where $\delta_{ij}$ is the Kronecker delta.

Now let $\Delta$ be the matrix defined as:

$\Delta := \paren {\phi \delta_{ij} - a_{ij} }$

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

Recall Cramer's Rule:

\(\displaystyle \adj \Delta \cdot \Delta\)

\(=\)

\(\displaystyle \Delta \cdot \adj \Delta\)

\(\displaystyle \)

\(=\)

\(\displaystyle \map \det \Delta \cdot \mathbf I_n\)

Multiplying through by $\adj \Delta$ in $(1)$ and applying Cramer's Rule:

$\displaystyle \sum_{j \mathop = 1}^n \map \det \Delta m_i = 0$

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

But $\map \det {\phi \delta_{ij} - a_{ij}}$ 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.