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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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, 2013 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, 2013 at 12:47
  • $\begingroup$ A detail... $\mathbb{C}$ is not the algebraic closure of $\mathbb{Q}(\sqrt[3]{2})$. It's smaller. $\endgroup$ Nov 26, 2015 at 0:36
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$\eta : \mathbb{Q}(\sqrt[3]{2})\rightarrow \mathbb{C}$ given by $\eta (\sqrt[3]{2})=\sqrt[3]{2}. \omega $ is a $\mathbb{Q}$ homomorphism.

But then....

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