I am reading the book Algebras, Rings and Modules, volume 1, by M. Hazewinkel and at the page 193 there is a proof about why the integral closure of a ring in a separable finite extension L over $k$ is finitely generated. I can't understand that point:
Let $ {W_1,...,W_n} $ a basis of $L$ over $k$. Then there is a dual basis $ {W_1^*,...,W_n^*}$ such that $ Sp(W_i,W_j^*)=δ_{ij}$ where $Sp$ is the trace. We can write $ W_J^*=X_1^jW_1+...+X_n^jW_n$ and for each $j = 1, ..., n$ this yields a system of linear equations: $$ \sum_{i=1}^n Sp(W_iW_k)X_i^J=Sp(W_kW_j^*)$$ $ (k=1,...,n)$ . Can anyone explain me how that system turns out? Thanks in advance.
Not sure about what you do not understand. The map $\DeclareMathOperator{\tr}{tr}(x,y)\mapsto \tr(xy)$ is a non-degenerate bilinear form on the $K$-vector space $L$,because the extension $L/K$ is separable, hence the basis $(W_i)_{1\le i\le n}$ has a dual basis $(W_i^*)_{1\le i\le n}$. (As I haven't read the book, I have no idea why the trace form is denoted $\mathrm{Sp}$.)
The system of equations consists in writing the definition of a dual basis, so as to find the values of the coordinates $X_i^j$ of each of the vectors $W_j^*$ in the basis $(W_i)_{1\le i\le n}$ . So the system of equations really is: $$\sum_{i=1}^n \tr(W_iW_k)X_i^j= \delta_{jk}\qquad (1\le j,k\le n).$$