Atiyah and MacDonald, Proposition 2.4

Question

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.

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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?

Answer 1

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 $\text{det}(I\phi-A)=0\in\text{End}_R(M)$.

Answer 2

The "adjoint" here means "classical adjoint". For a matrix $A$, the classical adjoint is the matrix whose $i,j$ entry is $(-1)^{i+j} A_{ji}$, where $A_{ji}$ means the $j,i$ cofactor---the determinant of the matrix obtained by omitting row $j$ and column $i$. The point is that the product of the classical adjoint and $A$ is a diagonal matrix with $\mathrm{det}(A)$ on the diagonal.