8
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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 $$0=\sum_i\ c_{ki}\ \sum_j\ b_{ij}\ x_j=\sum_j\ \sum_i\ c_{ki}\ b_{ij}\ x_j=\sum_j\ \delta_{kj}\ \det(b)\ x_j=\det(b)\ x_k.$$ $\endgroup$
    – Pierre-Yves Gaillard
    Sep 17, 2011 at 15:45
  • $\begingroup$ 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 $\chi(\phi)=0$. $\endgroup$
    – Pierre-Yves Gaillard
    Sep 17, 2011 at 15:59
  • $\begingroup$ On such determinant tricks see also here. $\endgroup$
    – Bill Dubuque
    Sep 17, 2011 at 18:35

1 Answer 1

Reset to default
12
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ On such determinant tricks see also here. $\endgroup$
    – Bill Dubuque
    Sep 17, 2011 at 18:37
  • $\begingroup$ But wait, I´m only a little confused, the determinant it´s a function that receives a homomorphism, and here , he is saying that the determinant vanishes in the generators, that are not homomorphism. Sorry for ask :S $\endgroup$
    – Daniel
    Sep 17, 2011 at 18:44
  • $\begingroup$ In this case, the determinant of $A$ can be seen as an endomorphism of $M$. $\endgroup$
    – Gooz
    Sep 19, 2011 at 6:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .