Let $F$ be splitting Field of $x^3 -7 \in \mathbb{Q} [x]$ and let $G$ be the splitting field of $$ (x^3-7) (x^2-3) \in \mathbb{Q}[x]$$
Show that $F \not = G$
Show that $F \not = G$
The roots of $x^3-7$ are
$$\sqrt[3]{7},(\sqrt[3]{7}) w_3^1,(\sqrt[3]{7}) w_3^2 $$
The roots of $x^2-3$ are
$$ \sqrt{3},\sqrt{-3} $$
F by definition is
$$F = \mathbb{Q} \left (\sqrt[3]{7},(\sqrt[3]{7}) w_3,(\sqrt[3]{7}) w_3^2 \right)= \{ f_1+f_2\sqrt[3]{7} +f_3 (\sqrt[3]{7}) w_3 +f_4(\sqrt[3]{7}) w_3^2 ;f_1,f_2,f_3,f_4\in \mathbb{Q} \}$$
Now, call $H$ the splitting field of $x^2-3 \in \mathbb{Q}[x]$ over $\mathbb{Q}$ So, $$H = \mathbb{Q}(-\sqrt{3},\sqrt{3}) = \mathbb{Q}(\sqrt{3}) = \{h_1+h_2 \sqrt{3} : h_1,h_2 \in \mathbb{Q}\} $$
$G$ is a bigger splitting field getting the basis from combinations of $H$ and $F$
so they are
$$\sqrt{3}, \sqrt[3]{7},(\sqrt[3]{7}) w_3,(\sqrt[3]{7}) w_3^2, \sqrt{3}\sqrt[3]{7},\sqrt{3}(\sqrt[3]{7}) w_3,\sqrt{3}(\sqrt[3]{7}) w_3^2 $$
$G \not \subset F$ [$\therefore$ Not equal]
Its probably easiest just to say $\sqrt[3]{7} \not \in F $ but $\sqrt[3]{7}$ in $G$ and I might have done something wrong in the long way