Let $L/K$ be a finite separable extension of a field $K$. A Galois Closure $M$ of $L/K$ is defined as a minimal degree extension of $L$ for which $M/K$ is Galois.
I want to show that the Galois Closure of $L/K$ exists. Here is my approach:
Let ${e_1,\ldots,e_n}$ be a (finite!) basis for $L$ as a $K$-vector space. Each $e_i$ has a minimal (and separable) polynomial $P_i(x)$. Define $L_0$ to be $L$ with adjoined all of the roots of the $P_i$. It is easy to check that $L_0$ is a finite extension and that it is the splitting field of $A(x)=\prod P_i(x)$. However, if any of the $P_i$ have common roots, then $A(x)$ will not be separable, despite each $P_i$ being separable.
Any idea how to fix this or I have to think in a different way? Any help appreciated.