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:
\(\ds \adj \Delta \cdot \Delta\) | \(=\) | \(\ds \Delta \cdot \adj \Delta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \det \Delta \cdot \mathbf I_n\) |
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.