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Show that $\Bbb Q(\sqrt[p]{a}, \omega )=\Bbb Q(\sqrt[p]{a}+ \omega )$, where $\omega=e^{2\pi i/p}$ and $a$ is prime. For simplicity let's call the left hand field $K$, and the right hand field $R$. I know that the strategy is to show inclusion both ways to get the inequality. I also know that $K$ is the splitting field of the polynomial of $\sqrt[p]{x}-a$, I have proven this.

I can show that $R \subseteq K$ since I know that $\omega, \sqrt[p]{a} \in K$ implying that $\omega+\sqrt[p]{a} \in K$ implying $R \subseteq K$.

I have trouble showing the opposite inclusion. I know the goal is to show that $\omega, \sqrt[p]{a} \in R$ knowing that $\omega+\sqrt[p]{a} \in R$. I tried multiple times to show this, mainly playing around with different polynomial and using the binomial theorem but I can't figure anything out for this.

Any hints or help to show this is appreciated. Thanks in advance

user218512
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    One option would be to determine the Galois group of $K$, then show that no non-trivial automorphism of $K$ fixes $\sqrt[p]{a}+\omega$. – carmichael561 Apr 11 '16 at 04:06
  • @carmichael561 unless I have made an error I have calculated the Galois group to be $S_p$. Then showing that only the identity automorphism fixes $\sqrt[p]{a}+\omega$ implies that $K\subseteq R$. I am new to Galois theory so maybe it's obvious and I don't see it, but why does this fact imply the inclusion? – user218512 Apr 11 '16 at 04:36
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    I think the Galois group is smaller than $S_p$. If I remember correctly, it's generated by the maps $\sqrt[p]{a}\mapsto \omega\sqrt[p]{a}$ and $\omega\mapsto \omega^2$. – carmichael561 Apr 11 '16 at 04:39
  • @carmichael561 I feel a bit confused about this but I'll go to sleep and look at it tomorrow and see what I come up with. I was most likely too hasty or making too many assumptions and will see it tomorrow. Thank you! – user218512 Apr 11 '16 at 04:47

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This is a classical primitive-element argument. Let $v=\sqrt[p]{a}$ and $u=v+\omega$. The minimal polynomial of $v$ is $V=X^p-a$ while the minimal polynomial of $\omega$ is $W=X^{p-1}+X^{p-2}+\ldots +1$. Next, consider the pertubated polynomial $F=W(u-X)$.

The roots of $F$ are $u-\omega=v,u-\omega^2,u-\omega^3,\ldots,u-\omega^{p-1}$.

The roots of $V$ on the other hand are $v,v\omega,v\omega^2,\ldots,v\omega^{p-1}$.

If $r$ is any common root to $F$ and $W$, we can write $r=u-\omega^i=v\omega^j$ for some indices $i$ and $j$ between $1$ and $p-1$. It follows that $v(1-\omega^j)=\omega-\omega^i$, and if $j\neq 1$ we could write $v=\frac{\omega-\omega^i}{1-\omega^j}\in{\mathbb Q}[\omega]$ which is impossible because $v$ has degree $p$ while $\omega$ has degree $p-1$. So $j=1$ and $r=v$. We have therefore shown that the only common root to $F$ and $V$ is $v$. So the GCD of those two polynomials is $X-v$. But since those two polynomials have coefficients in ${\mathbb Q}[u]$, so does their GCD (this is the "invariance of GCD" property and follows from the Euclidean algorithm). So $v\in{\mathbb Q}[u]$ which is exactly what you need.

Ewan Delanoy
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