# Question about integral element and faithful module.

The following is a part of proposition $$5.1$$ in An Introduction to Commutative Algebra by Atiyah&Mcdonald. First let $$A\subset B$$ be commutative rings with identity and $$b\in B$$.

If there is a faithful $$A[b]$$-module $$M$$ such that $$M$$ is finitely generated as an $$A$$-module, then $$b$$ is integral over $$A$$.

I cannot see how the faithfulness of $$M$$ is used, so could anyone please give a counterexample when $$M$$ is not faithful? The proof in the book uses Cayley-Hamilton with $$\phi$$ to be the right multiplication by $$b$$. Thank you.

Without faithfulness, all what we can conclude is that multiplication by $$\beta = b^n + a_1 b^{n-1} + \cdots + a_0$$ is always zero. The element $$\beta$$ itself doesn't have to be zero.
For a counterexample without faithfulness, extend the action of $$\mathbb Z$$ on itself to $$\mathbb Z[x]$$ such that $$x \cdot a = 0$$ for all $$a \in \mathbb Z$$. Note that $$x$$ is an indeterminate and cannot be the root of a polynomial over $$\mathbb Z$$.