There are enough Galois extensions? For me enough means that every finite extension of a certain field is included in a Galois extension of that field, formally: "Given a field $k$, and a finite extension $E/k$, there exists a Galois extension $L/k$ such that $K\subset E \subset L$?"
This statement is false, since every extension of a non-separable extension is non-separable (i.e non-Galois). So I thought to restrict to the case of finite separable extension, and the final question became: "Given a field $k$, and a finite separable extension $E/k$, there exists a Galois extension $L/k$ such that $K\subset E \subset L$?"
Since the normal closure is always defined (it is also finite), if $k$ is perfect (i.e. any algebraic extension is separable) the statement is always true. Any finite separable extension of $k$ can be extended to its normal closure which is a Galois extension.
But I'm getting bumped with the non-perfect case, how can I prove that the normal closure is separable? On the other hand, looking for a counterexample, I would say that the extension $F_p(T)[\sqrt[p]{T}]/F_p(T)$, which is not separable, is normal. But I failed in proving it.
Thank you for the help.
Best Answer
A finite extension of fields is normal if and only if it is the splitting field extension of some polynomial. By the primitive element theorem, a separable extension $E/L$ is simple, i.e. $E=L(\alpha)$ for some $\alpha\in L$. Since $E/L$ is separable, the minimal polynomial of $\alpha$ over $L$ is separable. The Galois closure of $E/L$ is the splitting field of this polynomial, which is normal (since it is a splitting field) and separable (since it is the splitting field of a separable polynomial).