I want to find a splitting field for : $x^4+3 \in \mathbb{Q}[x]$ and the degree of that extension , I would be thankful if you verify/correct what I have tried for this problem :
First I find the zeros of this polynomial over $\mathbb{C}$ : $$x_1=\sqrt[4]{3}(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} ) $$ $$x_2=\sqrt[4]{3}(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} ) $$ $$x_3=\sqrt[4]{3}(-\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} ) $$ $$x_4=\sqrt[4]{3}(\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} ) $$ So I have to join $\sqrt[4]{3}$ , $\sqrt2$ and $i$ to the field $\mathbb{Q}$. Hence the splitting field is $\mathbb{Q}(\sqrt2,i,\sqrt[4]{3})$
I know that $[\mathbb{Q}(\sqrt2,i):\mathbb{Q}]=4$, and I observe that $\mathbb{Q}(\sqrt2,i,\sqrt[4]{3}) = [\mathbb{Q}(\sqrt2,i)](\sqrt[4]{3})$ And that $$[[\mathbb{Q}(\sqrt2,i)](\sqrt[4]{3}):\mathbb{Q}(\sqrt2,i)]=4$$ because the minimal polynomial of $\sqrt[4]{3}$ over $\mathbb{Q}(\sqrt2,i)$ is $x^4-3$, which has degree 4. Hence $$[\mathbb{Q}(\sqrt2,i,\sqrt[4]{3}):\mathbb{Q}]=[[\mathbb{Q}(\sqrt2,i)](\sqrt[4]{3}):\mathbb{Q}(\sqrt2,i)] \cdot [\mathbb{Q}(\sqrt2,i):\mathbb{Q}]=4 \cdot 4=16 $$ And the basis for $\mathbb{Q}(\sqrt2,i,\sqrt[4]{3})$ is $$\mathcal{B}=\lbrace 1,\sqrt2,i,\sqrt[4]{3}, \sqrt[4]{3^2} , \sqrt[4]{3^3} , i\sqrt[4]{3} , i\sqrt[4]{3^2} , i\sqrt[4]{3^3} , \sqrt2 \sqrt[4]{3} , \sqrt2 \sqrt[4]{3^2} , \sqrt2 \sqrt[4]{3^3} , i \sqrt2 , i \sqrt2 \sqrt[4]{3} , i \sqrt2 \sqrt[4]{3^2} , i \sqrt2 \sqrt[4]{3^3} \rbrace$$