# Galois Theory: Are separable and normal equivalent?

I have to proof the following Theorem

Theorem: Let $\mathbb{Q} \subseteq L$ be a normal field extension with $L \subseteq R$ and $[L:\mathbb{Q}]=2^\lambda$. Then every $\alpha \in L$ is constructible.

Hint: use the fact that $L$ is also separable and therefore $[L:\mathbb{Q}]$ is a Galois extension.

But why is $L$ also separable?

Thanks a lot in advance!

## 1 Answer

In characteristic 0, every extension is separable.

• Just to add to that: You get non-separable extension $E/F$ when there is an $\alpha\in E$ such that the minimal polynomial of $\alpha$ in $F[x]$ has derivative $0$. This happens in characteristic $p$ when said minimal polynomial is a polynomial in $x^p$, but it never in characteristic $0$. In fact, it can never happen if $F$ is a so-called perfect field, which means that either the characteristic of $F$ is $0$, or the characteristic of $F$ is $p$ and every element in $F$ has a $p$-th root. Any non-perfect field does admit a non-separable extension. – Arthur Jun 14 '17 at 10:44
• Okay and then in the above theorem, since the extension is normal, it is also algebraic, and the derivative of every minimal polynomial over $\mathbb{Q}$ is not equal to 0, because $\mathbb{Q}$ has characteristic 0. And that is the Definition of being a separable extension. Thank you very much! – vaoy Jun 14 '17 at 11:42