Splitting field of $X^6-7X^4+3X^2+3$ over $\mathbb Q$ and $\mathbb F_{13}$

Question

I want to find the splitting field and the degree of the splitting field over $\mathbb Q$ and $\mathbb F_{13}$ for the polynomial $X^6-7X^4+3X^2+3$.

Over $\mathbb Q$ the polynomial factors as $(X-1)(X+1)(X^4-6X^2-3)$, but how can I find the splitting field of $(X^4-6X^2-3)$? For $\mathbb F_{13}$ I have no idea.

Answer 1

Observe that $Y^2-6Y-3$ splits over $\mathbb{F}_{13}$ because its roots are $$\frac{6\pm\sqrt{36+12}}{2}=3\pm4\sqrt{3}=3\pm4(4)=7\text{ and }12$$ The roots of $X^4-6X^2-3$ are obviously the square roots of the roots of $Y^2-6Y-3$. The element $12$ is already a square in $\mathbb{F}_{13}$ (its square roots are $5$ and $-5=8$) so the splitting field of $X^4-6X^2-3$ over $\mathbb{F}_{13}$, and hence the splitting field of $X^6-7X^4+3X^2+3$ over $\mathbb{F}_{13}$, is $$\mathbb{F}_{13}(1,-1,5,-5,\sqrt{7},-\sqrt{7})=\mathbb{F}_{13}(\sqrt{7})\cong\mathbb{F}_{13^2}$$ The idea of analyzing $Y^2-6Y-3$ should help you figure out the case of $\mathbb{Q}$ on your own.