# Cayley-Hamilton Theorem/Finitely Generated Modules

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

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:

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

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

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Thus for each $i$:

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

where $\delta_{i j}$ is the Kronecker delta.

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

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

This article, or a section of it, needs explaining.$\phi$ and $\delta_{i j}$ have changed places -- is this significant? Otherwise consistency in presentation is to be aimed for.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

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

Recall Cramer's Rule:

\(\ds \adj \Delta \cdot \Delta\) | \(=\) | \(\ds \Delta \cdot \adj \Delta\) | ||||||||||||

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

This article, or a section of it, needs explaining.Explanation needed as to what $\cdot$ means in this context, why it is significant to commute the factors of the expression, and how the final expression arises (probably explained once Cramer's rule is posted up.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

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

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

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

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