# Prove that $A[b_1,...,b_n]$ is a finitely generated A-module if $b_i's$ are integral over A

This question was left as an exercise in my class on commutative algebra and I am not sure about it's solution. So, I am posting it here.

Question: if $$b_1,..., b_n \in B$$ are integral over A, then show that $$A[b_1,...,b_n]$$ is a finitely generated A-module.

Attempt: Definition of finitely generated module is : The left R-module M is finitely generated if there exist $$a_1, a_2, ..., a_n$$ in M such that for any x in M, there exist $$r_1, r_2, ..., r_n$$ in R with $$x = r_1a_1 + r_2a_2 + ... + r_na_n$$.

Here, $$b_i's$$ are integral over A => $$b_i$$'s are solution of $$x^n + a_{n-1} x^{n-1} +...+ a_0=0$$ where all $$a_i's$$ are in A.

Now, use the above definition of integral over A I have proved that all $$x\in A[b_1,...,b_n]$$ can be represented in terms of $$a_i's\in A$$ and in powers of $$b_i's$$ my generating set being {$$b_1,...,b_n$$}, but that was not neat ie I couldn't prove this exactly that for any x there exists $$A_i 's \in A$$ such that $$x= A_1 b_1 +...+ A_n b_n$$ .

So, does that makes my proof wrong or it is fine? If it is wrong then kindly tell correct way to prove it.

• Each $b_j$ is such that $f_j(b_j)=0$ for some $f_j\in A[x]$ monic. Take $N=\sup_j \deg f_j$, then the finitely generated $A$-module $M=\sum_{j\le n} \sum_{l\ne N} A b_j^l$ is easily seen to be a ring, ie. $M=A[b_1,\ldots,b_n]$. May 27, 2022 at 0:38

It is not true in general that if $$b_1,\dots,b_n \in B$$ are integral over $$A$$, then $$\{b_1,\dots,b_n\}$$ generates $$A[b_1,\dots,b_n]$$ as an $$A$$-module.
To prove the result, recall that given an extension of rings $$R \subseteq S$$ and $$s \in S$$, $$s$$ is integral over $$R$$ if and only if $$R[s]$$ is a finite $$R$$-module.
We now prove the result by induction on $$n$$:
If $$n=0$$, then $$A[\emptyset]=A$$ is of course finite over $$A$$.
Now suppose $$n > 1$$ and that the result holds for $$n-1$$. $$b_n$$ is integral over $$A$$, so a fortiori over $$A[b_1,\dots,b_{n-1}]$$, hence $$A[b_1,\dots,b_n]=A[b_1,\dots,b_{n-1}][b_n]$$ is finite over $$A[b_1,\dots,b_{n-1}]$$. By induction hypothesis, $$A[b_1,\dots,b_{n-1}]$$ is finite over $$A$$. By the transitivity of finite ring extensions, the result follows.