# There are enough Galois extensions?

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.

• Just pick a primitive element of $E/k$ and take a splitting field of its minimal polynomial. May 25, 2011 at 10:59
• Your extension $F_p(T)[\sqrt[p]{T}]/F_p(T)$ is normal because it is the splitting field over $F_p(T)$ of the polynomial $X^p-T$. May 25, 2011 at 11:24

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).
• Thank you very much for your answer. If I'm not wrong your construction is exactly the same construction of the normal closure of an extension of field (i.e. the "smallest" normal extension of $L$ containing the field $E$, in your notation), and you point out that it is also separable because of the separability of the minimum polynomium of the primitive element. I'm right? May 25, 2011 at 12:06