7
$\begingroup$

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} $ enter image description here

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

  • 0
    On such **determinant tricks** see also [here.](http://math.stackexchange.com/questions/25045/problem-on-finitely-generated-ideal/25048#25048)2011-09-17

1 Answers 1

10

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

  • 0
    In this case, the determinant of $A$ can be seen as an endomorphism of $M$.2011-09-19