Existence of Galois Closure

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.

• Why do you think the $P_i$ having a root in common changes anything ? – reuns Oct 11 '18 at 22:21
• Look at the compositum of splitting fields – user10354138 Oct 11 '18 at 22:26
• At first a separable extension is an algebraic extension $F/K$ such that every $a \in F$ is separable over $K$ (there is a non-zero polynomial $h \in K[x]$ with $h(a) = 0$). It is a theorem that $K(\alpha)$ and $K(\alpha_1,\ldots,\alpha_n)$ are separable extensions of $K$ when the $\alpha_j$ are separable over $K$. Those theorems come before your question. – reuns Oct 11 '18 at 22:28
• @reuns Ah yes! A polynomial is separable when all of its irreducible factors are separable... and since it is true for all $P_i$ we are done. Thank you! – DesmondMiles Oct 11 '18 at 22:33
• The compositum of two normal extensions is normal (recall normality can be characterized as: for any homomorphism $\sigma$ to the algebraic closure does not leave the extension, and we also have $\sigma(L_1L_2)=\sigma(L_1)\sigma(L_2)$), and compositum of separable is separable (the $L_2$-minimal polynomial of a primitive element $\alpha$ of $L_1/K$ divides the $K$-minimal polynomial of $\alpha$). – user10354138 Oct 11 '18 at 22:59