About Kronecker's lemma I'm studying the topic of integrally closed domains and related stuff and while I was reading (again) the paper "Bezout rings and their subrings" I found a result that is called "Kronecker's lemma" in the proof of theorem 2.7 of the above paper. Unfortunately, the reference of the author leads to the book "Les systèmes d'idéaux" which it seems very hard to find on the Internet.
So I decided to make some searching on Google, but all the results link to  Kronecker's lemma of field extensions or to his lemma about the convergence of series.

From the context of the theorem given in the paper I know that this Kronecker's lemma is relative to integrally closed domains, but I can't find a exact statement of what this result says. So maybe someone knows where I can find the statement of this lemma.

UPDATE: A extra searching on this lemma lead me to what is called "Dedekind's Prague theorem", which seems to be a different version of the above lemma, but again I couldn't find any proper statement of this theorem. As it seems to be that there are other related and similar results involving integrally closed domains, I would like to know about some other sources where these topics and ideas are discussed. Thanks to user26857 I found the book "Integral Closure of Ideals, Rings, and Modules", but for sure there are some other books.
 A: First of all, I want to thank @user26857 for helping me to find the result to which Cohn was referring in his paper "Bezout rings and their subrings".
Theorem: Let $R$ be an integral domain and let $f,g\in R[x]$. If $f=\sum_{i=0}^n a_ix^i$ and $g=\sum_{j=0}^m b_jx^j$, then $a_ib_j$ is integral over the ideal $A_{fg}$ for all $0\le i\le n$ and $0\le j\le m$.
Proof: Let $i,j$ arbitrary and set $u=a_ib_j$. If $f=0$ or $g=0$ the result is trivially true. So, let's suppose that $f\neq 0\neq g$. Since $u\in A_fA_g$, then $(u)\subseteq A_fA_g$, so $(u)A_f^m\subseteq A_f^{m+1}A_g$. By Dedekind-Mertens lemma we have $A_f^{m+1}A_g=A_f^m A_{fg}$, then $(u)A_f^m\subseteq A_{fg}A_f^m$.
Let's set $J=A_f^m$, so we have $(u)J\subseteq A_{fg}J$. As $J$ is finitely generated, there are $\alpha_1,\ldots, \alpha_k$ such that $J=(\alpha_1,\ldots, \alpha_k)$. Since $u\alpha_r\in A_{fg}J$ we deduce that there are $c_{rs}\in A_{fg}$ such that $$u\alpha_r=\sum_{s=1}^k c_{rs}\alpha_s.$$
We define the matrix $M=(\delta_{rs}u-c_{rs})_{k\times k}$, where $d_{rs}$ is Kronecker's delta. If we take the vector $v=(\alpha_1,\ldots, \alpha_k)^{T}$, then by construction it follows that $Mv=0$, then $$\text{adj}(M)Mv=0$$ $$\det(M)I_kv=0$$ $$\det(M)v=0.$$ Therefore, $\det(M)\alpha_r=0$ for all $1\le r\le k$. Since not every $\alpha_r$ is zero, it follows that $\det(M)=0$. Finally, expanding $\det(M)$ gives us an equation that shows that $u=a_ib_j$ is integral over $A_{fg}$.
