TITLE: Question about a proof on Atiyah Macdonald
QUESTION [8 upvotes]: 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.

REPLY [12 votes]: 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?
See the wikipedia page for more on adjoint/adjugate matrices.
http://en.wikipedia.org/wiki/Adjugate_matrix