Cayley-Hamilton Theorem

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 $\phi(M) \subseteq \mathfrak a M$.

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 $a_i \in \mathfrak a$.

Corollary: The Classical Cayley-Hamilton Theorem
With $A$ as above, let $N=(a_{ij})$ be an $n\times n$ matrix with entries in $A$.

Let $I_n$ denote the $n \times n$ identity matrix.

Let $p_N \left({x}\right)$ be the determinant $\det \left({x \cdot I_n - N}\right)$.

Then $p_N \left({N}\right) = \mathbf 0$ as an $n \times n$ zero matrix. That is,


 * $ N^n + b_{n-1} N^{n-1} + \cdots + b_1 N + b_0 = \mathbf 0$

where the $b_i$ are the coefficients of $p_N(x)$.

Corollary: Nakayama's Lemma
Let $A$, $M$ be as above, and $\mathfrak a$ an ideal of $A$ such that $M = \mathfrak a M$.

Then there is $a \in \mathfrak a$ such that $1 + a \in \operatorname{Ann}_A(M)$, the annihilator of $M$.

Hence if $\mathfrak a \subseteq \operatorname{Jac}(A)$, the Jacobson radical of $A$ then $M = 0$.

Proof
Let $m_1,\ldots,m_n$ be a set of generators for $M$.

Then for each $i$, $\phi(m_i) \in \mathfrak aM$, say:


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

for $i = 1,\ldots,n$. Thus for each $i$:


 * $\displaystyle \sum_{j \mathop = 1}^n \left[ \delta_{ij}\phi - a_{ij} \right]m_i = 0\qquad (1)$

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

Now let $\Delta = (\phi\delta_{ij} - a_{ij})$ and let $\operatorname{adj}(\Delta)$ be the adjugate matrix of $\Delta$.

Recall Cramer's Rule:


 * $\operatorname{adj}(\Delta)\cdot\Delta=\Delta\cdot\operatorname{adj}(\Delta)=\det(\Delta)\cdot I_n$

Multiplying through by $\operatorname{adj}(\Delta)$ in $(1)$ and applying Cramer's rule we see that:


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

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

But $\det\left( \phi\delta_{ij}-a_{ij} \right)$ is a monic polynomial in $\phi$ with coefficients in $\mathfrak a$, so we have an equation of the required form.

Proof of The Classical Cayley-Hamilton Theorem
Taking $\phi = N$ in the proof of Theorem 1 we see that $N$ satisfies:


 * $p_N(x) = \det\left( x\cdot I_n - N\right) = 0$

Take $\mathfrak a$ to be the ideal generated by the entries of $N$.

Proof of Nakayama's Lemma
Take $\phi$ to be the identity in Theorem $1$, and $a = a_0 + \cdots + a_{n-1}$.

Then $a \in \mathfrak a$ and $1 + a \in \operatorname{Ann}_A(M)$ as required.

For the second part, notice that if $\mathfrak a \subseteq \operatorname{Jac}(A)$, then by Characterisation of Jacobson Radical $1+a$ is a unit in $A$.

Thus a unit $(1+a) \in A$ annihilates $M$ and necessarily $M = 0$