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 $\mathbb Z$.


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