Understanding Jurgen Neukirch proof that $A[b_1, \dots, b_n]$ finitely generated $\implies$ it's integral over $A$. > ... Conversely, assume that the $A$-module $A[b_1, \dots, b_n]$ is finitely generated and that $w_1, \dots, w_r$ is a system of generators. THen for any element $b \in A[b_1, \dots, b_n]$, one finds that $$ bw_i = \sum\limits_{j=1}^r a_{i,j} w_j, i=1, \dots, r, a_{i,j} \in A \\\ $$ From (2.3) [$AA^* = A^*A = \det(A)I$], we see that $\det(bI - (a_{i,j}))w_i = 0, i=1,\dots, r$, and since **$1$ can be written $1 = c_1 w_1 + \dots + c_r w_r$, the identity $\det(bI-(a_{i,j})) = 0$** gives us a monic equation for $b$ with coefficients in $A$. I don't understand the bolded part. How does $1$ being writable as a sum of the generators imply that $\det(bI - (a_{i,j})) = 0$ ? Thank you!

Multiply $\det(bI-(a_{ij}))w_i=0$ by $c_i$ for every $i$ (note BTW that the $i$ in $a_{ij}$ is a placeholder, and not the same $i$ as in $w_i$), giving $$c_iw_i\det(bI-(a_{ij})) = 0.$$ Add those up, use the fact that $\sum c_iw_i=1$, and you are done.