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How to solve following: Let $\Bbb C(\omega)$ be a field of rational functions with one undetermined $\omega$ over the field of complex numbers $\Bbb C$. If $f(x)=x^5+\omega$, find

a) $\operatorname{gal}(f\mid\Bbb C(\omega))$,

b) $\operatorname{gal}(f\mid\Bbb R(\omega))$.

This is what I have tried: $$x^5+\omega=(x+a)(x^4-ax^3+a^2x^2-a^3x+a^4),$$ $$x^5+\omega=\prod_{i=0}^4(x-\varepsilon^i\sqrt[5]{\omega}),$$ where $a=\sqrt[5]{\omega}$, and $\varepsilon=\sqrt[5]{-1}$.

As $\varepsilon\in\Bbb C\subset\Bbb C(\omega)$, we have $K=\Bbb C(\omega)(\sqrt[5]{\omega})$ and $|K:Q(\omega)|=5$ (K is a field of roots of f(x)).

If we denote $\phi_i: \sqrt[5]{\omega}\rightarrow \varepsilon^i\sqrt[5]{\omega}$, $0\leq i\leq 4$, it is easy to conclude that all $\phi_i$ are automorphisms of $K$ over $\Bbb C(\omega)$. We have $\phi_i\phi_j=\phi_{i+j}$ and $\phi^i_1=\phi_i$, so $\phi_1$ generates $\operatorname{gal}(K\mid\Bbb C(\omega))$, and $\operatorname{gal}(K\mid\Bbb C(\omega))$ is $\Bbb Z_5$.

How to do the same in b)? I assume that $x^4-ax^3+a^2x^2-a^3x+a^4$ is irreducible over $\Bbb R(\omega)$, but I don't know how to prove it? Is my solution to a) correct?

Any help is welcome.

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up vote 1 down vote accepted

You have almost correctly handled part a). A (minor) problem is that $\varepsilon^2 a$ is not a root of the polynomial, because $$ f(\varepsilon^2 a)=(\varepsilon^2a)^5+\omega=(-1)^2a^5+\omega=2\omega\neq0. $$ This is easy to fix. I would denote by $a$ a fifth root of $-\omega$. If we further denote by $\zeta$ the primitive fifth root of unity, $\zeta=e^{2\pi i/5}$. Then the roots of $f(x)$ are $a\zeta^k,$ $k=0,1,2,3,4$, and the automorphisms map $\phi_k:a\to\zeta^ka$. They form, indeed, a cyclic group of order five.

The difference to part b) will be that your base field no longer contains the the fifth root of unity $\zeta$. As that is the ratio of two roots, $\zeta=a\zeta/a$, its has to be included in the splitting field. So this time the splitting field is $$ K=\mathbb{R}(\omega)[\zeta, a]. $$ As $\zeta\in\mathbb{C}$, it is quadratic over $\mathbb{R}$. Therefore we have $[K:\mathbb{R}(\omega)]=10$.

As a splitting field of a separable polynomial $K$ is Galois over $\mathbb{R}(\omega)$. So you are looking for 10 distinct automorphisms. I am leaving that search for you to do at this point (ask for help, if you need). I do give you the hint that an automorphism will be fully specified, if you know how it maps the elements $a$ and $\zeta$. There are five possibilities for $\phi(a)$ and two possibilitied for $\phi(\zeta)$, so it looks like all the combinations will occur. Your task is to identify the permutations of the roots given by the automorphisms. Then you will surely be able to identify the group as a subgroup of $S_5$.

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