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 $\phi \left({M}\right) \subseteq \mathfrak a M$.

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 $\mathbf N = \left({a_{ij} }\right)$ be an $n \times n$ matrix with entries in $A$.

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

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

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 $p_N \left({x}\right)$.

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 generating set for $M$.

Then for each $i$, $\phi \left({m_i}\right) \in \mathfrak a M$, say:


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

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

Thus for each $i$:


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

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

Now let $\Delta$ be the matrix defined as:
 * $\Delta := \left({\phi \delta_{ij} - a_{ij} }\right)$

Let $\operatorname{adj} \left({\Delta}\right)$ be the adjugate matrix of $\Delta$.

Recall Cramer's Rule:

Multiplying through by $\operatorname{adj} \left({\Delta}\right)$ in $(1)$ and applying Cramer's Rule:


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

Therefore $\det \left({\Delta}\right)$ 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 $\mathfrak a$.

Thus 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:


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

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 $1 + a \in \operatorname{Ann}_A \left({M}\right)$ as required.

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

Thus a unit $\left({1 + a}\right) \in A$ annihilates $M$ and necessarily $M = 0$.