An easy trick to compute the Galois group of a product of high degree polynomials If $f(X)=(X^4+3)(X^9-1) \in \mathbb{Q}[X]$ is there a trick to find its Galois group?
For example, if $f$ has only factos of degree $3$ and $2$ then its easy calculated because it can only be a direct product of $S_{3}, A_{3}, S_{2}$ and $\{id\}$ using the discriminant Theorem.
Now with high orders it became impossible to use discrimant. 
Is there a easy way to solve it without enumerating all $f's$ roots and combining them with the automorphism?
 A: The splitting field of $x^4+3$ is $\mathbb{Q}(\zeta_4, \sqrt[4]{-3})$. Hence the splitting field of $(x^4+3)(x^9-1)$ is $L=\mathbb{Q}(\zeta_{36}, \sqrt[4]{-3})=\mathbb{Q}(\zeta_{36}, \sqrt{\zeta_3+1/2})$, evidently it has degree $24$ over $\mathbb{Q}$.
Let $\alpha = \sqrt{\zeta_3+1/2}$, then $\sigma\in G=\text{Gal}(L/\mathbb{Q})$ is determined by $(a,b) \in (\mathbb{Z}/36\mathbb{Z})^\times \times (\mathbb{Z}/4\mathbb{Z})$:
$$\sigma:\zeta_{36} \mapsto \zeta_{36}^a \qquad \alpha \mapsto i^b \alpha$$
Note that $(a,b)$ has to satisfy certain condition:
$$\zeta_3 \mapsto \zeta_3^a \qquad \alpha^2 = \zeta_3 + \frac{1}{2} \mapsto (-1)^b (\zeta_3 + \frac{1}{2})$$
therefore $$\zeta_3^a + \frac{1}{2} = (-1)^b (\zeta_3 + \frac{1}{2})$$
this implies $$\tag{1} a\equiv 1 \pmod{3} \iff b\equiv 0 \pmod{2} \qquad a\equiv 2 \pmod{3} \iff b\equiv 1 \pmod{2}$$
Therefore elements in $G$ can be regarded as those $(a,b)\in (\mathbb{Z}/36\mathbb{Z})^\times \times (\mathbb{Z}/4\mathbb{Z})$ that satisfies $(1)$, the group operation is given by
$$({a_1},{b_1})({a_2},{b_2}) = ({a_1}{a_2},{a_1}{b_2} + {b_1})$$
Sage says $G$ as an abstract group, is isomorphic to $C_3 \times D_4$.
A: Algorithms to compute Galois groups are described in the answers to a question posed on Math Overflow.
In the specific example which you gave, it may be possible to calculate the Galois group by other simpler means.
