# Clarification of the details of the proof of Cayley Hamilton theorem in commutative algebra

I am trying to understand this proof of the Cayley Hamilton theorem from commutative algebra by Atiyah Mcdonald. So I am reading the following power point slides which gives more details but there is still some details I am unsure of: https://raphaelponge.files.wordpress.com/2022/10/chap2.pdf

Proposition (Cayley-Hamilton Theorem; Proposition 2.4) Note that $$A$$ is commutative and has a unit.

Suppose that $$M$$ is a finitely generated $$A$$-module and $$\mathfrak{a}$$ is ideal of A. Let $$\phi: M \rightarrow M$$ be an A-module endomorphism such that $$\phi(M) \subseteq \mathfrak{a} M$$. Then $$\phi$$ satisfies an equation of the form, $$\phi^n+a_1 \phi^{n-1}+\cdots+a_n=0, \quad a_i \in \mathfrak{a}$$ Remarks

• We identify $$A$$ with its image in $$\operatorname{End}_A(M)=\operatorname{Hom}_A(M, M)$$.
• The above equality holds in $$\operatorname{End}_A(M)$$, which is an A-module.

Proof of Proposition $$2.4$$.

• Let $$x_1, \ldots, x_n$$ be generators of $$M$$. Then: $$\text { (*) } \quad \phi x_j=a_{1 j} x_1+\cdots+a_{n j} x_n, \quad a_{i j} \in \mathfrak{a} .$$
• Let $$B$$ be the sub-ring of $$\operatorname{End}_A(M)$$ generated by $$\phi$$ and $$A$$. This is a commutative ring.
• Set $$a=\left[a_{i j}\right] \in M_n(\mathfrak{a})$$ and $$b=\phi I_n-a \in M_n(B)$$. Note that $$M_n(B)$$ acts on $$M^n$$. Then $$(*)$$ means that $$b x=0 \quad \text { with } x=\left[\begin{array}{c} x_1 \\ \vdots \\ x_n \end{array}\right] \text {. }$$

Proof of Proposition $$2.4$$ (continued).

• Let $$c$$ be the cofactor matrix of $$b$$. As $$B$$ is a commutative ring, we have
• As $$b x=0$$, we get $$c b=\operatorname{det}(b) I_n .$$ $$0=c b x=\operatorname{det}(b) x=\left[\begin{array}{c} \operatorname{det}(b) x_1 \\ \vdots \\ \operatorname{det}(b) x_n \end{array}\right]$$
• As $$x_1, \ldots, x_n$$ generate $$M$$, this gives $$\operatorname{det}(b)=0$$ in $$\operatorname{End}_A(M)$$.
• Here $$b=\phi I_n-a$$ with $$a \in M_n(\mathfrak{a})$$. Expanding the equation $$\operatorname{det}\left(\phi I_n-a\right)=0$$ shows there are $$a_1, \ldots, a_n$$ in a such that $$\phi^n+a_1 \phi^{n-1}+\cdots+a_n=0 .$$
• Equivalently, if $$P(\lambda)=\operatorname{det}\left(\lambda I_n-a\right)$$ is the characteristic polynomial of $$a=\left[a_{i j}\right]$$, then $$P(\phi)=0$$.

So my question is: what is the significance of $$B$$ being a commutative ring? Based on the proof, I assume that $$c b=\operatorname{det}(b) I_n .$$ will not hold if $$B$$ is not commutative. Why is this the case?

Also is $$M^n$$ a $$M_n(B)$$ module?

I have not done linear algebra over rings before so I am not sure how much theorems from linear algebra over $$R$$ and $$C$$ carry over to linear algebra over rings. Also is $$M^n$$ a $$M_n(B)$$ module?

• You should add that the general assumption is that any ring $A$ is commutative and has a unit element (the first sentence on the first slide). Feb 4 at 11:28

The determinant $$\det(b)$$ can be defined only when the entries of $$b$$ are in some commutative rings. In particular, we need the commutativity of $$B$$ to define the cofactor matrix. To exemplify, consider the $$2\times 2$$ multiplication $$\begin{pmatrix}S&-Q\\-R&P\end{pmatrix} \begin{pmatrix}P&Q\\R&S\end{pmatrix}= \begin{pmatrix}SP-QR&SQ-QS\\-RP+PR&-RQ+PS\end{pmatrix}.$$ The RHS is far from being proportional to a scalar matrix if $$B$$ is not commutative!
For the second question, yes, $$M_n(B)$$ acts on $$M^n$$ and makes it a module by a multiplication of matrices. However, in this proof of Cayley-Hamilton theorem (sometimes called "the determinant trick"), we only consider $$M$$, not $$M^n$$ and we just borrow a matrix notation and use it as a computational tool, as far as I understand.