Here $\zeta_n$ denotes the primitive n-th root of unity.

These days I am learning field theory. According to my lecture, for a radical extension we consider the splitting field of $x^n-a$ where $a$ is not a n-th power in $\Bbb Q$. We have a tower $\Bbb Q(\zeta_n,\sqrt[n]{a})/\Bbb Q(\zeta_n)/\Bbb Q$. And we have an injective homomorphism from $Gal(\Bbb Q(\zeta_n,\sqrt[n]{a}))/\Bbb Q(\zeta_n))$ to the additive group $\Bbb Z/n\Bbb Z$.

We have learnt that the $F$-algebra homomorphism from a field $F(\alpha)$ to another field $K$ is in bijection to the roots of the minimal polynomial of $\alpha$ over $F$ in $K$. Here we have all the n roots in $\Bbb Q(\zeta_n,\sqrt[n]{a})$. So I think if we can prove that $x^n-a$ is irreducible over $\Bbb Q(\alpha)$, we can prove that the upper layer of extension has degree $n$ and hence the Galois group $Gal(\Bbb Q(\zeta_n,\sqrt[n]{a}))/\Bbb Q(\zeta_n))$ is the whole $\Bbb Z/n\Bbb Z$. But I am not able to find a proof so far. So could someone please tell me if it is true? How to prove that? Any reference would also be appreciate. Thanks a lot.

  • $\begingroup$ What exactly do you want to prove? Note that $x^n-a$ is in general not irreducible over $\mathbb Q(\zeta_n)$ even if it is irreducible over $\mathbb Q$. $\endgroup$ – MooS May 31 '17 at 13:17
  • $\begingroup$ @MooS Thanks for tell me that. But may I please ask for (if there is ) a general method to determine the galois group of radical extension? Thanks a lot. $\endgroup$ – non-abelian group of order 9 May 31 '17 at 13:21
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    $\begingroup$ For the Galois group of $x^n-a$ over $\mathbb{Q}$ see this question. $\endgroup$ – Dietrich Burde May 31 '17 at 13:43
  • $\begingroup$ Sorry, @DietrichBurde, I should have looked at your answer to that question before giving my much more partial answer below. $\endgroup$ – Lubin May 31 '17 at 18:57
  • $\begingroup$ The Kummer extension idea is that If $x^m-b$ is irreducible over $K(\zeta_m)$ then $Gal(K(\zeta_m,\sqrt[m]{b})/K(\zeta_m))$ contains for each $k$ an automorphism $\sigma_k(\sqrt[m]{b}) = \zeta_m^k \sqrt[m]{b}$ so that $$Gal(K(\zeta_m,\sqrt[m]{b})/K(\zeta_m))= \mathbb{Z}/m\mathbb{Z}$$ Here you need to find the least power of $c=\sqrt[n]{a}$ such that $c^m \in \mathbb{Q}(\zeta_n)$, and obtain that $x^m-c^m$ is irreducible over $\mathbb{Q}(\zeta_n,\zeta_m)=\mathbb{Q}(\zeta_n)$, so that $$Gal(\mathbb{Q}(\zeta_n,c)/\mathbb{Q}(\zeta_n))=\mathbb{Z}/m\mathbb{Z}$$ $\endgroup$ – reuns May 31 '17 at 21:34

There is an “of course” reason why your conjecture fails, and a more serious reason.

The “of course” reason is that if $a$ is not an $n$-th power in $\Bbb Q$, but is an $m$-th power for $m$ a proper divisor of $n$, then $[\Bbb Q(\sqrt[n]a\,):\Bbb Q]\le n/m$. Just think of the case where $a$ isn’t a fourth power, but is a square of a rational number. So I think your hypothesis wants to be that for all $m|n$ with $m>1$, $a$ is not an $m$-th power of a rational.

But here is a serious counterexample to your conjecture: consider $\Bbb Q(\zeta_8,\sqrt[8]2\,)$. You know that $\zeta_8=(1+i)\big/\sqrt2$, so that $\sqrt2\in\Bbb Q(\zeta_8)$. But it’s not a square there, so $[\Bbb Q(\zeta_8,\sqrt[8]2\,):\Bbb Q(\zeta_8)]=4$. To be explicit, the factorization of $X^8-2$ over $\Bbb Q(\zeta_8)$ is $$ X^8-2=(X^4-\zeta_8-\zeta_8^{-1})(X^4+\zeta_8+\zeta_8^{-1})\,. $$


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