On pages 42-43 in Reid's book "Undergraduate commutative algebra" there is a proof of the Cayley - Hamilton theorem which reduces to proving that $\det{\Delta} \in A'[\varphi]$ and $\det{\Delta} = 0$.
I do not understand the following statement:
As I understand $(\det{\Delta}) \ e_j = 0, \ j=\overline{1,n} \ $ implies $\det{\Delta} = 0$ iff $\text{Ann}_{A'[\varphi]}(A^n) = (0) \iff A^n$ is faithful $A'[\varphi]$-module. Is it true? If so, how do I prove, that $A^n$ is faithful?
By definition, $A'[\varphi]$ is a subring of $\operatorname{End}(A^n)$. An endomorphism of $A^n$ is $0$ iff it sends each $e_j$ to $0$, and so in particular this applies to any element of $A'[\varphi]$. To put it another way, by definition $A'[\varphi]$ is a ring of functions on $A^n$ and the $A'[\varphi]$-module structure of $A^n$ is just evaluation of these functions, and so it is a faithful module because a function is determined by its values at every point.