Order of Galois group of $X^n-a$ over $\mathbb{Q}$

Question

The order of a Galois group, $G$, of a polynomial of the type $X^n-a$ over $\mathbb{Q}$, with $a, n \in \mathbb{N}$ is $|G|=n\cdot\varphi(n)$ when $n$ and $\varphi(n)$ are coprime (although I'm not sure about this condition). For instance, if $f(X)=X^5-2$, the Galois group has $5 \cdot\varphi(5)=5 \cdot 4 = 20$ elements. This comes in handy when deriving the corresponding Galois group.

Can this expression be generalized for the case when $n$ and $\varphi(n)$ are not coprime? If so, when does $a$ play a role in the number of elements?

I have checked that if $n$ divides 8, then the right expression seems to be $|G|=\frac{1}{2}n \cdot \varphi(n)$, but I don't know how to proceed with other cases (eg, $4, 6, 10$, etc).

Answer 1

As noted, the splitting field is $\mathbb{Q}(\zeta_n, \sqrt[n]{a})$ where $\zeta_n$ is a primitive $n$-th root of unity. Now it's known that $[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\varphi(n)$ so $\varphi(n)\mid |G|$.

For the remaining part, we need to know a bit more about $a$. For example if $a$ is a $n$-th power already, then the remaining extension is trivial so $a$ does play a role. In fact, the claim about $n$ and $\varphi(n)$ being coprime is not actually sufficient due to this: the polynomial $x^2-1$ is completely reducible so has trivial Galois group but $1 \neq \varphi(2) \times 2$.

Morally, the remaining condition to check is whether $a$ is a $d$-th power for any $d \mid n$. However, as @Jyrki points out, there are slight problems such as $2$ being a square in $\mathbb{Q}(\zeta_8)$ but not $\mathbb{Q}$. In general, $a$ is always a square in $\mathbb{Q}(\zeta_{4a})$ (and potentially even in $\mathbb{Q}(\zeta_a)$ e.g. if $a$ is prime congruent to $1 \bmod{4}$). Fortunately, it is only square roots where this problem occurs as noted in the comments.

Answer 2

"Can anyone refine @Matt B contribution?"

To the best of my knowledge, the most general result (for any field $K$) on the degree of $K(\sqrt [n] a)/K$ does not come from Kummer's theory, but from the following theorems:

1) Let $a\in K$ and $m,n$ two coprime integers. Then $X^{mn}-a$ is irreducible over $K$ iff both $X^m-a$ and $X^n-a$ are. This allows to reduce the problem to prime power exponents.

2) Let $p$ be a prime. Suppose that $X^p-a$ is irreducible over $K$, and let $\alpha$ be a root. Then: (i) If $p\neq2$, or $p=2$ and char$(K)=2$, $a$ is not a $p$-th power in $K(\alpha)$; (ii) If $p=2$ and char$(K)\neq 2$, $\alpha$ is a square in $K(\alpha)$ iff $-4a$ is a $4$-th power in $K$. The proof goes through the "Kummer detour".

3) Assume that $a$ is not a $p$-th power in $K$. Then: (i) If $p\neq2, X^{p^n}-a$ is irreducible over $K$ for any $n$ ; (ii) If $p=2$ and char$(K)=2, X^{2^n}-a$ is irreducible over $K$ for any $n$ ; (iii) If $p=2, char(K)\neq 2$ and $n\ge 2, X^{2^n}-a$ is irreducible over $K$ iff $-4a$ is not a $4$-th power. The proof relies of course on (1).

Addendum. I have just realized that the OP question was about the degree of the splitting field of $X^n-a$ (I was confused by the expression "Galois group of a polynomial"), so I must go on further, taking $\mathbf Q$ as the base field. The final answer depends of course on data given about the element $a$. Let us first concentrate on the $p$-primary case, i.e. $n=p^m$, with $p$ odd. Throughout, suppose that $a$ is not a $p$-th power in $\mathbf Q$, and write $\alpha_r$ for a $p^r$-th root of $a$. According to 3i), the extension $F=\mathbf Q(\alpha_m)$ admits a tower of subextensions $F_r=\mathbf Q(\alpha_r)$ of degree $p^r$ over $\mathbf Q , r\le m$. On the other hand, consider the cyclotomic field $L=\mathbf Q(\zeta_{p^m})$ and its canonical tower of cyclic subextensions $L_r=\mathbf Q(\zeta_{p^r}), r\le m$. For $p$ odd, the extensions $F$ and $L_1$ are linearly disjoint, and $a$ is not a $p$-th power in $L_r$ for any $r$ (this comes from linear disjointness for $r=1$, and from the cyclicity of $L_r/L_{r-1}$ for $r>1$). So we can apply again 3i) to construct the tower of extensions $F_r.L_r, r\le m$, which culminates at the splitting field $F_m.L_m$, of degree $p^m\phi (p^m)$.

The passage from a $p$-primary degree to a composite degree $n$ is easy as long as $n$ is odd ./.

Answer 3

A very incomplete answer: assume that $a$ is neither a square nor a cube and $n=6$.

Then the splitting field is $\mathbb{Q}(a^{1/6})(i\sqrt{3})$ which does have degree $6*2=12$.

Answer 4

For $n\geq 2$, the splitting field $E$ of $f(x):=x^n-a\in \mathbb{Q}[x]$ over $\mathbb{Q}$ is $\mathbb{Q}(\zeta, \alpha)$, where $\zeta$ is an $n$-th primitive root of unity and $\alpha$ is a root of $f(x)$.

If $\mathbb{Q}(\zeta)\cap \mathbb{Q}(\alpha)=\mathbb{Q}$, then the Galois group $G_{\mathbb{Q}}(E)$ is isomorphic to the semidirect product $G_{\mathbb{Q}(\zeta)}(E)\rtimes G_{\mathbb{Q}(\alpha)}(E)$. Moreover, $G_{\mathbb{Q}(\zeta)}(E)$ is isomorphic to a subgroup of $(\mathbb{Z}_n, +)$ and $G_{\mathbb{Q}(\alpha)}(E)$ is isomorphic to a subgroup of $(\mathbb{U}_n, \cdot)$ (the group of invertible elements of $\mathbb{Z}_n$).

In case $f(x):=x^5-2$, by a simple degree argument, we have $\mathbb{Q}(\zeta)\cap \mathbb{Q}(\sqrt[5]{2})=\mathbb{Q}$.

Moreover, $G_{\mathbb{Q}(\sqrt[5]{2})}(E)\cong \mathbb{U}_5$ due to Translation Theorem:

($\mathbb{Q}(\zeta)/\mathbb{Q}$ is a Galois extension, hence $\mathbb{Q}(\zeta, \sqrt[5]{2})/\mathbb{Q}(\sqrt[5]{2})=E/\mathbb{Q}(\sqrt[5]{2})$ is a Galois extension and $G_{\mathbb{Q}(\sqrt[5]{2})}(E)\cong G_{\mathbb{Q}(\sqrt[5]{2})\cap \mathbb{Q}(\zeta)}(\mathbb{Q}(\zeta))=G_{\mathbb{Q}}(\mathbb{Q}(\zeta))\cong \mathbb{U}_5$ ).

Since $\vert G_{\mathbb{Q}(\zeta)}(E)\vert\geq 2$ and since $G_{\mathbb{Q}(\zeta)}(E)$ is a subgroup of a cyclic group of order $5$, we get $\vert G_{\mathbb{Q}(\zeta)}(E)\vert =5$. Hence, $G_{\mathbb{Q}(\zeta)}(E)\cong \mathbb{Z}_5$.

Finally, $G_{\mathbb{Q}}(E)\cong \mathbb{Z}_5\rtimes \mathbb{U}_5$ and $\vert G_{\mathbb{Q}}(E)\vert =20$.