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

Question

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.

Answer 1

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.