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?

  • $\begingroup$ 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). $\endgroup$ Commented Feb 4, 2023 at 11:28

1 Answer 1


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.

  • $\begingroup$ I think the proof uses the associativity of the action of $M_n(B)$ on $M^n$, which is almost equivalent to proving that it is indeed an action (so the module structure). $\endgroup$ Commented Jun 10 at 19:22

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