My attempt:
I'm performing the proof by induction on $t$. For the base case of induction I'm assuming that $u_1$ is an integral element over $A$, so $A[u_1]$ is a finitely generated $A$-module (I already tested this). Then I propose the induction hypothesis and to prove the thesis I'm writing
$$A[u_1,...,u_t]=A[u_1,...,u_{t-1}][u_t]$$
So I want to prove that $u_t$ is an integral element over $A[u_1,...,u_{t-1}]$ and this is where I have not been able to advance because I don't know very well how to proceed, my reasoning is as follows:
For $u_t$ to be an integer element there must exist a monic polynomial $F \in A[u_1,...,u_{t-1}][X]$ such that $F(u_t)=0$ but the only thing I know about $A[u_1 ,...,u_{t-1}]$ is that it is finitely generated by the inductive hypothesis and furthermore that $u_t$ is an integer element over $A$ by hypothesis.
I tried to propose the form of the polynomial $F(X)$ by taking its coefficients in $A[u_1,...,u_{t-1}]$ and since this last set is finitely generated then there are $p_1,...,p_n \in A[ u_1,...,u_{t-1}]$ such that every element $x \in A[u_1,...,u_{t-1}]$ is written as a linear combination of the $p_i$ with $i\in\{1,...,n\}$, that is to say that there exist $a_1,...,a_n \in A$ such that
$$x=\sum_{i=1}^{n}a_ip_i$$
Any suggestions you can give me to continue?
Recall the definition of an integral element. An element u is said to be integral over a ring A, if there exists a monic polynomial f(x) with coefficients in A such that f(u)=0.
Also recall how subrings are generated. A and u_1, u_2, ...,u_t is denoted by $$A[u_1,u_2,...,u_t]$$ This subring contains all polynomials in u_1, u_2, ...,u_t with coefficients from A.
For our base step. You are already using induction on t. So for the base case t=1, we assume u_1 is an integral element over A. Also assume A[u_1] is a finitely generated A-module.
Now for the induction step. Assume the statement is true for t-1. So $$A[u_1,...,u_{t-1}]$$ is also a finitely generated ring A.
As you already stated $$A[u_1,...,u_t]=A[u_1,...,u_{t-1}][u_t]$$ This means that $$A[u_1,...,u_t]$$ is obtained by adjoining u_t to $$A[u_1,...,u_{t−1}]$$ So for the statement to be true, u_t needs to be integral over $$A[u_1,...,u_{t−1}]$$