Let $A$ be an integral domain and $K$ its field of fractions. Let $L/K$ be a finite separable extension of degree $n$. Let $B$ be the integral closure of $A$ in $L$. Then, by a theorem, there exists a basis $\lbrace e_1,\ldots, e_n \rbrace \subset B$ of $L$ over $K$. Since $L/K$ is separable of degree $n$, then there are precisely $n$ monomorphisms $\sigma_i:L \to \bar{K}$ fixing $K$ pointwise, where $\bar{K}$ is a fixed algebraic closure of $K$. Let $M:= (\sigma_i(e_j))_{1\leq i,j \leq n}$. I wonder why $\det M \neq 0$. I need a proof not depending on Galois Theory because i didn`t study it yet.
Thanks in advance.