Splitting Field Computations (13.4.1-4)

Dummit and Foote Abstract Algebra, section 13.4, exercises 1-14:

Determine the degree of the splitting field over $\mathbb{Q}$ for the following polynomials:
$x^4-2$
$x^4+2$
$x^4+x^2+1$
$x^6-4$

Proof: $x^4-2$: Letting $α$ be a solution to this polynomial, we see the elements $α$,$iα$,$-α$, and $-iα$ are the four distinct solutions. Therefore we have the splitting field contains $α$ and $i$ and also a field containing $α$ and $i$ contains the splitting field, so the splitting field is precisely $\mathbb{Q}(α,i)$. We shall show $x^4-2$ is irreducible over $\mathbb{Q}[x]$ by first observing it has no roots in $\mathbb{Z}$ and also does not decompose into two quadratics: $$x^4-2=(x^2+ax+b)(x^2+cx+d)$$ $$bd=-2⇒d=-2/b$$ $$ad+bc=0⇒c=2a/b^2$$ $$b+d+ac=0⇒a^2=-b(b^2-2)/2$$ $$a+c=0⇒a=2a/b^2⇒a=0,b^2-2=0⇒b \not ∈ \mathbb{Z}$$ Thus $[\mathbb{Q}(α)~:~\mathbb{Q}]=4$, and since $\mathbb{Q}(α)⊂\mathbb{R}$ we have $x^2+1$ irreducible over $\mathbb{Q}(α)$, so the computed degree is $8$.

$x^4+2$: Again letting $α$ be a root we have the splitting field is $\mathbb{Q}(α,i)$. We shall show $x^4+2$ is irreducible over $\mathbb{Q}(i)$ by observing it doesn't have a root (else the splitting field would be $\mathbb{Q}(i)$ despite the fact that $α^2=\pm \sqrt{2}i \not ∈ \mathbb{Q}(i)$) and by computations similar as above it doesn't decompose into quadratics unless there exists $b$ such that $b^2-2=0$, despite $\pm \sqrt{2} \not ∈ \mathbb{Q}(i)$. Thus the degree is $8$.

$x^4+x^2+1$: After treating this as a quadratic in $x^2$ and applying some algebra, we come to the factorization in $\mathbb{C}[x]$ $$x^4+x^2+1=(x-1/2-\sqrt{3}/2i)(x-1/2+\sqrt{3}/2i)$$ $$(x+1/2-\sqrt{3}/2i)(x+1/2+\sqrt{3}/2i)$$ Therefore the splitting field is precisely $\mathbb{Q}(\sqrt{-3})$ and the degree is $2$.

$x^6-4$: We observe $x^6-4=(x^3-2)(x^3+2)$. Letting $α$ be a solution to $x^3-2$ and $\zeta$ being a primitive third root of unity, we see $α$, $\zeta α$, and $\zeta^2 α$ are the three solutions. As well, $-α$, $-\zeta α$, and $-\zeta^2 α$ are the solutions to $x^3+2$. Therefore the splitting field is $\mathbb{Q}(α,\zeta)$. Letting $α$ be the positive real solution, since $\zeta$ is not real and of degree two we must have the degree over $\mathbb{Q}$ is $2*3=6$.