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I have already shown that if $F\subset K$ is a Galois extension, then for any intermediate field $L$, we have $L\subset K$ is a Galois extension. I then want to show that it's not necessarily true that $F\subset L$ is a Galois extension. I want to show this by example.

My approach was to find an irreducible polynomial $f$ over $\mathbb{Q}$, its splitting field $E$ (which is Galois since ${\rm char}(\mathbb{Q}) = 0)$ and then find an intermediate field $L$ such that $\mathbb{Q}\subset L$ is not normal. I tried to do this with $f(x) = x^3 +27x +6$ since I worked with its Galois group in a previous exercise, but I didn't succeed. Maybe it's doable with that polynomial and I just cannot see the solution. I then tried to do it with $g(x) = x^3 -2$ since its roots are "nicer" than those of $f(x)$, but again I didn't succeed. I'll just show what I did in the latter case, and maybe someone will spot a mistake or see how I should have proceeded.

So let's look at $g(x) = x^3 -2$. Let $\omega = e^{2\pi i/3}$, then the roots of $g$ are $x_1 = 2^{1/3}$, $x_2 = 2^{1/3}\omega$ and $2^{1/3}\omega^2$. We have $[\mathbb{Q}(x_1):\mathbb{Q}]= 3$, and the minimal polynomial of $x_2$ and $x_3$ over $\mathbb{Q}(x_1)$ is of degree $2$, so $[\mathbb{Q}(x_1,x_2):\mathbb{Q}(x_1)] = 6$. Let $E$ be the splitting field of $g$ over $\mathbb{Q}$, so $[E:\mathbb{Q}] = 6$. At this point I looked at $\mathbb{Q}(x_2)$ and tried to work out if it's a normal extension of $\mathbb{Q}$. Now I remind myself that since $g$ has $3$ distinct roots in $E$ we have $G = {\rm Gal}(E/\mathbb{Q})$ is a subgroup of $S_3$. Since $\mathbb{Q} \subset E$ is normal ($E$ is the splitting field of $g$) we have that $[E:\mathbb{Q}] = \left|G\right| = 6$, so in fact $G\cong S_3$. The only normal subgroup of $S_3$ is $A_3$, so maybe there is a solution hiding here somewhere?

I don't know how to proceed from here. The minimal polynomial of $x_2$ over $\mathbb{Q}$ is $g$ itself, so $[\mathbb{Q}(x_2):\mathbb{Q}] \neq 2$ which is what I wanted (the subgroups of $S_3$ of order $2$ are not normal). Can I adjoin $\omega$ to $\mathbb{Q}$ and get an intermediate field? Any tips, hints or complete answers will be appreciated, and of course if mistakes was made I would like to know.

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  • $\begingroup$ Pick $F \subset L$ any finite separable non-Galois extension, and call $K$ the normal closure. $\endgroup$ – Crostul May 31 '16 at 12:52
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Consider $\omega$ such that $\omega^3=1$, i.e. $\omega$ is a cube root of unity. The splitting field of $x^3-2=0$ over $\mathbb{Q}$ is $\mathbb{Q}(\omega,\sqrt[3]{2})$.

Also, $\mathbb{Q}(\sqrt[3]{2})\subset\mathbb{Q}(\omega,\sqrt[3]{2})$. But, $\mathbb{Q}(\sqrt[3]{2})$ is not a Galois extension. This is because, if it was Galois, then it would be a normal and separable extension. Hence, it must contain all the roots of $x^3-2=0$. But 2 roots of this equation are complex and this is clearly a real extension.

Alternatively, one can notice that $\mathbb{Q}(\sqrt[3]{2})$ is a Galois extension only if $|Aut(\mathbb{Q}(\sqrt[3]{2}))|=[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$

But $|Aut(\mathbb{Q}(\sqrt[3]{2}))|$=1 as the automorphisms fix $\mathbb{Q}$ and $\sqrt[3]{2}$, which is a root of the irreducible polynomial $x^3-2$ ca only be sent to $\sqrt[3]{2}$, as the other roots are not in the field (automorphisms send roots of irreducible polynomials to roots of the same polynomial).

Thus $\mathbb{Q}(\sqrt[3]{2})$ is not Galois.

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  • $\begingroup$ Thanks, this cleared it up for me. $\endgroup$ – Auclair May 31 '16 at 13:51

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