Cayley-Hamilton Theorem/Finitely Generated Module

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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 $\phi \sqbrk M \subseteq \mathfrak a M$.


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

Recall that the set of endomorphisms $\operatorname{End}_A(M)$ is an $A$-module with unity.

Let $R \subseteq \operatorname{End}_A(M)$ be the $A$-submudole generated by $\set {\phi^n : n \in \N}$.

That is:

$R = \set {a_0 + a_1 \phi + \cdots + a_k \phi^k : k \in \N, a_1,\ldots ,a_k \in A}$

Note that $R$ is a commutative ring with unity.

Recall also the ring of $n$-matrices $\map {\MM_R} n$.


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

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

where:

$\delta_{i j}$ is the Kronecker delta with values in $A$

Now let $\Delta \in \map {\MM_R} n$ defined as:

$\Delta := \paren {\phi \delta_{i j} - a_{i j} \mathbf 1_R }$

Then $\paren 1$ can be written as:

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


Let $\adj \Delta \in \map {\MM_R } n$ be the adjugate matrix of $\Delta$.

By Matrix Product with Adjugate Matrix:

\(\ds \adj \Delta \cdot \Delta\) \(=\) \(\ds \Delta \cdot \adj \Delta\)
\(\ds \) \(=\) \(\ds \map \det \Delta \mathbf I_n\)
\(\ds \) \(=\) \(\ds \begin{bmatrix}

\map \det \Delta & 0 & \cdots & 0 \\ 0 & \map \det \Delta & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \map \det \Delta \\ \end{bmatrix}\)

where $\map \det \Delta \in R$ denotes the determinant of $\Delta$.


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:

$\map \det \Delta = \mathbf 0_R$

On the other hand, $\map \det \Delta$ is a monic polynomial in $\phi$, since:

\(\ds \map \det \Delta\) \(=\) \(\ds \map \det {\delta_{i j} \phi - a_{i j} \mathbf 1_R }\)
\(\ds \) \(=\) \(\ds \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n \paren {\delta_{k \map \lambda k} \phi - a_{k \map \lambda k} \mathbf 1_R } }\) Definition 1 of Determinant
\(\ds \) \(=\) \(\ds \phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 \mathbf 1_R\)

where:

$\lambda$ goes over all permutations of $\set {1, 2, \ldots, n}$
$\map \sgn \lambda$ is the sign of the permutation $\lambda$
$a_0, a_1, \ldots, a_{n-1} \in \mathfrak a$

As $R \subseteq \operatorname{End}_A(M)$, this means:

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

$\blacksquare$


Source of Name

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

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