# Question about a proof on Atiyah Macdonald

I have a question about a step of a proof in Atiyah Macdonald. It's the proposition 2.4.

Let M be a finitely generate A-module, let a be an ideal of A, and let $\phi$ be an A-module endomorphism of M such that $$\phi \left( M \right) \subset aM$$ Then $\phi$ satisfies an equation of the form \eqalign{ & \phi ^n + a_1 \phi ^{n - 1} + ... + a_n = 0 \cr & \text{ where }\ \ a_i \in a \cr}

I don't understand why the determinant annihilates each $x_i$ because I did not understand the step of the adjoint matrix.

-
Put $b=(b_{ij}):=(\delta_{ij}\phi-a_{ij})$. This is a matrix with coefficients in $A[\phi]$. Let $c=(c_{ij})$ be the adjugate of $b$. Then $$0=\sum_i\ c_{ki}\ \sum_j\ b_{ij}\ x_j=\sum_j\ \sum_i\ c_{ki}\ b_{ij}\ x_j=\sum_j\ \delta_{kj}\ \det(b)\ x_j=\det(b)\ x_k.$$ – Pierre-Yves Gaillard Sep 17 '11 at 15:45
Let $\pi:A^n\to M$ be the $A$-linear map characterized by $\pi(e_i)=x_i$, and $\Phi:A^n\to A^n$ the $A$-linear map given by the matrix $(a_{ij})$. We have $\pi\circ\Phi=\phi\circ\pi$, and thus $\pi\circ f(\Phi)= f(\phi)\circ\pi$ for any $f$ in $A[X]$. In particular, if $\chi$ is the characteristic polynomial of $\Phi$, Cayley-Hamilton implies $\chi(\phi)\circ\pi=0$, and the surjectivity of $\pi$ yields $\chi(\phi)=0$. – Pierre-Yves Gaillard Sep 17 '11 at 15:59
On such determinant tricks see also here. – Bill Dubuque Sep 17 '11 at 18:35

For a square matrix $A$ with adjoint (or adjugate) matrix adj$(A)$, we have that $\textrm{adj}(A) A=A \textrm{adj}(A) = \textrm{det} (A) I$, where $I$ is the identity matrix.
Thus, in your case, let $A$ be the matrix such that $A_{ij} = \delta_{ij} \phi - a_{ij}$. It acts on $M^n$. Then, if $x = (x_1,\ldots,x_n)$, we have that $A x = 0$. Therefore, $\det(A) x=\textrm{adj}(A) A x =0$. This is what you were looking for, right?
In this case, the determinant of $A$ can be seen as an endomorphism of $M$. – Gooz Sep 19 '11 at 6:33