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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!

nekodesu
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  • Here's a hint about why the last statement is useful. $M$ being a faithful $N$-module means that if $n \in N$ and $nm = 0$ for all $m \in M$, then $n=0$. Apply this to $N=A[x]$ and notice that we want to find some monic polynomial $f(X) \in A[X]$ with $x$ as a root, i.e. $f(x) =0$. – ggg Feb 07 '18 at 15:08
  • can I get an example to show that faithfulness is necessary? I want to understand by an example where there is a $A[x]$ module $M$ which is finitely generated $A$ yet $x$ is not integral. – permutation_matrix Nov 10 '22 at 12:29

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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 $r\in R$

cansomeonehelpmeout
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