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Hello I would like to calculate all subfields of the field in which $f$ splits in linear factors (I just don't know the right english word for this field :) )

$f:=X^6-2$ is irreducible with Eisenstein criterion. Define $a:=2^\frac{1}{6}$ and $\xi:=\exp(\frac{2\pi i}{6})$. Then $$X^6-2=(X-a)\cdot\ldots\cdot(X-a\xi^5).$$

Now $[\mathbb{Q}(a):\mathbb{Q}]=6$, since $f$ is the minimal polynomial of $a$.

and $[\mathbb{Q}(a)(\xi):\mathbb{Q}(a)]=2$, since $a\in\mathbb{R}$, so $X^2-X+1$ is still the minimal polynomial of $\xi$.

Therefore $[\mathbb{Q}(a,\xi):\mathbb{Q}]=12=ord(Gal(f,\mathbb{Q}))$, since $K:=\mathbb{Q}(a,\xi)$ is the field in which $f$ splits in linear factors.

Now calculate the automorphisms:

First there are automorphisms:

$$\phi:\mathbb{Q}(a)\rightarrow\mathbb{Q}(\xi a), \phi|_\mathbb{Q}=id_\mathbb{Q}, \phi(a)=\xi a$$

and the extension

$$\varphi:K\rightarrow K, \varphi|_\mathbb{Q}=id_\mathbb{Q}, \varphi(a)=\xi a, \varphi(\xi)=\xi.$$

There is also an extension with $\overline{\varphi}(\xi)=\overline{\xi}$, since $\xi$ and $\overline{\xi}$ are roots of the minimal polynomial of $X^2-X+1$, but one can get this by using the product of $\varphi$ and the following automorphism

$$\psi:K=\mathbb{Q}(a)(\xi)\rightarrow\mathbb{Q}(a)(\overline{\xi}),\psi|_\mathbb{Q}=id_\mathbb{Q}, \psi(a)=a, \psi(\xi)=\overline{\xi},$$

since $X^2-X+1$ is still the minimalpolynomial of $\xi$ and $\overline{\xi}$ over $\mathbb{Q}(a)$.

So $G:=Gal(f,\mathbb{Q})=\{id,\varphi,\ldots,\varphi^5,\psi,\psi\varphi,\ldots,\psi\varphi^5\}$.

So one gets the subgroups of $G$ of order $2,3,4$ or $6$.

But now I am having problems to calculate the subfield. For example for $\psi\varphi^3$ one has:

$$[K:K^{<\psi\varphi^3>}]=ord(<\psi\varphi^3>)=2$$ and $$[K^{<\psi\varphi^3>}:\mathbb{Q}]=6$$.

($K^{<\psi\varphi^3>}$ denotes the usual fixgroup) Now the usual approach I know would be to find an linear combination $c$ of elements of the fixgroup and show by using the degree formula that $K^{<\psi\varphi^3>}=\mathbb{Q}(c)$ but I cannot find such $c$. So what is the usual approach to solve this?

Thanks for your help!

user408858
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  • See here for the "usual approach to solve this". – Dietrich Burde Mar 21 '18 at 19:11
  • It's called the "splitting field". – saulspatz Mar 21 '18 at 19:11
  • Could $c$ be some $\xi^j a$? – Angina Seng Mar 21 '18 at 19:11
  • I don't see why this is an answer to my question about determining the subfields of the splitting field. As far as I can see the link only considers the splitting field itself? – user408858 Mar 21 '18 at 21:50
  • Well I could solve it by using $a_1+a_2$ for the maps $\psi\varphi^i$ where $a_1,a_2$ are roots and using $a^2+\xi$ for $\varphi^3$, $a^3+\xi$ for $\varphi^2$ and of course $K^{<\varphi>}=K^{<\varphi^5>}=\mathbb{Q}(\xi)$. Especially $\xi a+\xi^2a$ is fixed by $\psi\varphi^3$ and for example $a+\xi a$ is fixed by $\psi\varphi$ doing the jobs. – user408858 Mar 22 '18 at 13:26

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