5
$\begingroup$

a Galois extension is defined as an extension which is separable and normal. I am looking for an example of an extension which is separable but not Galois. Apparently it is the case of$$\mathbb Q(\sqrt[3]{2})$$ but I don't really understand why. Can someone explain ? It is separable because $\sqrt[3]{2}$ is separable over $\mathbb Q$, but why is it not normal over $\mathbb Q$ ? Thanks.

$\endgroup$
7
$\begingroup$

Notice that the field $\mathbb Q(\sqrt[3]{2})$ is contained in $\mathbb R$, but the other two roots of the irreducible polynomial $x^3-2$ are complex. Thus, the field extension contains only one of the three roots of an irreducible polynomial, thus is not normal.

$\endgroup$
  • $\begingroup$ Thanks fot the quick answer. However I still don't really understand the link between the two characterizations of a normal extension ; namely $L$ is a normal extension of $K$ if one of the following equivalent conditions holds : (i) every $K$-homomorphism from $L$ to an algebraic closure $\Omega$ of $K$ is an automorphism of $L$ (ii) every irreducible polynomial of $K[X]$ which has at least one root in L, has all its roots in L. $\endgroup$ – Asinus Dec 2 '13 at 12:46
  • $\begingroup$ This would mean that if $\mathbb Q(\sqrt[3]{2})$ is not normal, there should be a $\mathbb Q$-homomorphism of $\mathbb Q(\sqrt[3]{2})$ to $\mathbb C$ which is not an automorphism of $\mathbb Q(\sqrt[3]{2})$. I just don't see what such a homomorphism would be like... $\endgroup$ – Asinus Dec 2 '13 at 12:47
  • $\begingroup$ A detail... $\mathbb{C}$ is not the algebraic closure of $\mathbb{Q}(\sqrt[3]{2})$. It's smaller. $\endgroup$ – João Dos Reis Nov 26 '15 at 0:36
2
$\begingroup$

$\eta : \mathbb{Q}(\sqrt[3]{2})\rightarrow \mathbb{C}$ given by $\eta (\sqrt[3]{2})=\sqrt[3]{2}. \omega $ is a $\mathbb{Q}$ homomorphism.

But then....

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.