Faithful modules in integral dependence

Question

In Atiyah-Macdonalds, Proposition 5.1, for a ring $B$ and a subring $A$, we have

There exists a faithful $A[x]$-module $M$ which is finitely generated as an $A$-module $\Rightarrow$ $x \in B$ is integral over $A$.

The proof uses the fact that $M$ is faithful. Namely, consider the endomorphism $\varphi$ of $M$ that is multiplication by $x$. Since $x M \subseteq M$, and $M$ is faithful, then $x^n + a_1 x^{n-1} + \cdots + a_n = 0$ for suitable $a_i \in A$.

I'm unsure about why in the last statement we uses that $M$ is a faithful module. Thanks for any insights!

Answer 1

If you don't use that $M$ is faithful, you'll only know that:

$$x^nm+a_1x^{n-1}m+\dots+a_nm=(x^n+a_1x^{n-1}+\dots+a_0)m=0$$ for every $m\in M$. And you can't conclude that $x^n+a_1x^{n-1}+\dots+a_0=0$

An equivalent definition of a faithful module is:

An $R$-module $M$ is faithful if for some $m\in M$, $r\cdot m\neq 0$ for all non-zero $r\in R$