# 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?

• The splitting field is $\mathbb{Q}(\zeta_{36},\sqrt{\zeta_3+1/2})$ a quadratic extension of $\mathbb{Q}(\zeta_{36})$, its Galois group should be $Z/(9)^\times \times \text{Aff}_1(Z/(4))$ where $\text{Aff}_1(Z/(4))$ acts on the $m$ of $i^m \sqrt[4]{-3}$ and $Z/(9)^\times$ acts on the $l$ of $\zeta_9^l$ – reuns Oct 28 '18 at 5:16
• One more time. There are infinitely many rational numbers, so the tag finite-fields is out of place. – Jyrki Lahtonen Oct 28 '18 at 17:42

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 $$C_3 \times D_4$$.