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$B$ is a ring and $A$ is a subring of $B$.

Atiyah's proposition $5.1$ says that the following statements are equivalent.

1- $x \in B$ is integral over $A$

2-$A[x]$ is finitely generated $A$-module

3-$A[x]$ is contained in a subring $C$ of $B$ such that $C$ is a finitely generated $A-$module

4-There exists a faithful $A[x]-$module $M$ which is a finitely generated as an $A-$module.

Being finitely generated $A$ module implies the existence of a finite generating set $\{ x_1, x_2...\dots x_n\}$ and then we write the $$sx_i= \sum_{j=1}^{n} a_{ij}x_j$$ and then $$0= \sum_{j=1}^{n} (\delta_{ij} -x_i )a_{ij}x_j$$

Then I will get a matrix equation $Bx=0$ and multiplication by the adjoint of matrix $B$, I get $\operatorname{det}(B)(x_j)=0$ and hence $$\operatorname{det}(B)=\operatorname{det}(B)\cdot (1)=\sum_j \operatorname{det}(B) a_j x_j=0$$ and thus we have the integral equation $\operatorname{det}(B)$ for $x$.

Here are my questions

1-Where are we using the faithfulness of the $A[x]$ module in the implication $4) \rightarrow (1)$.

(2)-Also can I have a counterexample to see why it is necessary?

(3)-Since $x_i$ are a generating set, then sure the $x_i$'s are non-zero. Then the system of equations $Bx=0$ doesn't have a unique solution. Hence the matrix $B$ is not invertible. Hence $\operatorname{det}(B) \neq 0$. Is this proof also valid. If yes, then why none of the texts has presented this basic/easy proof?

permutation_matrix
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  • I went through https://math.stackexchange.com/questions/903360/integral-ring-extensions-and-finitely-generated-as-a-module and https://math.stackexchange.com/questions/1544407/equivalent-conditions-for-integral-element and https://math.stackexchange.com/questions/2288601/using-cayley-hamilton-and-faithful-modules-to-show-x-in-b-is-integral-over-a but none of them answers my question completely. – permutation_matrix Nov 13 '22 at 18:06
  • $(2)$ A counterexample is $\Bbb Z[1/2]$, which is not finitely generated. Here $A=\Bbb Z$. – Dietrich Burde Nov 13 '22 at 18:20
  • @DietrichBurde, I didn't ask for a counterexample incase it is not finitely generated. I am asking about the faithfulness. Not finitely generated has many for example $\mathbb{Z} \subset \mathbb{Z}[x]$. Isn't it? – permutation_matrix Nov 13 '22 at 18:23
  • I thought you had asked for a counterexample to $(2)$, which says $A[x]$ is a finitely generated $A$-module, sorry (you wrote "(2)-Also can I have a counterexample"). – Dietrich Burde Nov 13 '22 at 19:08
  • No Problem. Anyways the counterexample of polymial ring over integers works as I provided in the comment? – permutation_matrix Nov 13 '22 at 20:35

1 Answers1

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For your counter-example, take $A=\mathbb Z$ and $x$ transcendental, so that $A[x]$ is a polynomial ring. Now take $M=\mathbb Z$, which as an $A[x]$-module has $x$ acting as zero. $M$ is finitely generated as an $A$-module, but $x$ is not integral over $A$.

The faithfulness means that we can obtain information about $x$ by looking in $M$. If it is not faithful, then we can’t, as in the counter-example.

Andrew Hubery
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