Let $M$ be a finitely generated $R$-module, $\mathfrak a \lhd R$ an ideal and $\phi:M\to M$ an $R$-linear map such that $\phi(M)\subseteq \mathfrak a M$. Then $\phi^n+a_1 \phi^{n-1}+\cdots+a_n=0$.
(Atiyah/MacDonald, Proposition 2.4, page 21)
The proof goes as follows:
Let $x_1,…,x_n$ be a set of generators of $M$, then each $\phi(x_i)\in\mathfrak a M$ can be written as $\phi(x_i)=\sum_{j=1}^n a_{ij} x_j$ with some $a_{ij}\in\mathfrak a$, i.e.
$$\sum_{j=1}^n(\delta_{ij}\phi-a_{ij})x_j=0$$
By multiplying on the left by the adjoint of the matrix $(\delta_{ij}\phi-a_{ij})$ it follows that $\det(\delta_{ij}\phi-a_{ij})$ annihilates each $x_i$, hence is the
zero endomorphism of $M$. Expanding out the determinant, we have an equation
of the required form.
I understand the first part, but after "by multiplying on the left by the adjoint" I am not really sure what is happening anymore. Would somebody be so kind and shed some light on this?
Best Answer
Once you have: $$ \sum_{j=1}^n(\delta_{ij}\phi-a_{ij})x_j=0 $$ you can rewrite this as $(I\phi-A)X=0$, where $A=(a_{ij})$, $I=(\delta_{ij})$, and $X=(x_1,\dots,x_n)^T$. Since $\text{det}(I\phi-A)I=[\text{adj}(I\phi-A)](I\phi-A)$, multiplying by $\text{adj}(I\phi-A)$, whose entries are all in $\text{End}_R(M)$, gives: $$ \text{det}(I\phi-A)x_1=\cdots=\text{det}(I\phi-A)x_n=0 $$ and $\text{det}(I\phi-A)=0\in\text{End}_R(M)$.