What is the splitting field of $x^6+x^4-2x^2-2$ over $\mathbb{Q}$?

Question

What is the splitting field of $x^6+x^4-2x^2-2$ over $\mathbb{Q}$? And what is its degree? I think if $\alpha_1,\cdots,\alpha_6$ are the roots of this polynomial, then $\mathbb{Q}(\alpha_1,\cdots,\alpha_6)$ is the splitting field, and therefore its degree is 7 or 6, right?

Answer 1

First, find the roots of $p(x) = x^6+x^4-2x^2-2$: Since all exponents are even, we may put $y= x^2$ and then first find the roots of $q(y) = y^3+y^2-2y-2$. One root of $q(y)$ is easily guessed to be $-1$, so we see that $q(y) = (y+1)(y^2-2)$. Hence the roots of $q(y)$ are $-1,\sqrt2$ and $-\sqrt2$.

It follows that the roots of $p(x)$ have to be $\pm{\rm i}, \pm\sqrt[4]{2}$ and $\pm{\rm i}\sqrt[4]{2}$. Hence, the splitting field of $p(x)$ over $\mathbb Q$ is $\mathbb Q(\rm i,\sqrt[4]{2})$ (which clearly contains all the roots and is minimal with this property).

The degree is $[\mathbb Q(\rm i,\sqrt[4]2):\mathbb Q ]=8$, because of $[\mathbb Q(\sqrt[4]{2}):\mathbb Q] = 4$, $[\mathbb Q(\rm i,\sqrt[4]{2}): \mathbb Q(\sqrt[4]{2})]=2$ (since $\mathbb Q(\sqrt[4]{2})\subseteq \mathbb R$) and the degree formula for field extensions.