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.
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 $$ X^8-2=(X^4-\zeta_8-\zeta_8^{-1})(X^4+\zeta_8+\zeta_8^{-1})\,. $$