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: enter image description here

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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.

  • $\begingroup$ Thanks! But I need to clarify one thing: $\text{End}(A^n)$ is the endomorphism ring of the abelian group $A^n$. Thus the elements of $\text{End}(A^n)$ preserve only the addition (they are not necessarily $A$-linear). If $f \in \text{End}(A^n)$, then $f(a) = f(\sum_{i=1}^{n} a_i e_i) = \sum_{i=1}^{n} f(a_i e_i) \neq \sum_{i=1}^{n} a_if(e_i)$. So how do you conclude that "An endomorphism of $A^n$ is $0$ iff it sends each $e_j$ to $0$...". I suppose it is a mistake. However this works if $f \in A'[\varphi]$, because all elements in $A'[\varphi]$ are $A$-linear. $\endgroup$ – user128245 Sep 23 '16 at 20:31
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    $\begingroup$ Ah, I assumed $\operatorname{End}(A^n)$ referred to the $A$-linear endomorphisms. You're correct that the argument still works because $A'[\varphi]$ is contained in the subring of $A$-linear endomorphisms. $\endgroup$ – Eric Wofsey Sep 23 '16 at 21:48

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