Degree of splitting field for $f(x) = x^4 - x^2 + 4$ over $\mathbb{Q}$ I started for finding the roots of the polynomial (4 in total) which took the forms 
$$ \pm \sqrt{\frac{1}{2}\left( 1 \pm i \sqrt{15} \ \right)} $$
I figured that adjoining the positive square root gives us the negative square root, so at most, only $  \sqrt{\frac{1}{2}\left( 1 + i \sqrt{15} \ \right)} = z$ and $\sqrt{\frac{1}{2}\left( 1 - i \sqrt{15} \ \right)} = z'$ need to be adjoined to $\mathbb{Q}$ to get $f(x)$ to split. 
Where I'm stuck on is how to confirm that these are the two roots that need to be adjoined, and if they are, how to calculate the degree of the splitting field over $\mathbb{Q}$.  $f(x)$ looks to me to be irreducible (though I've no idea how to convincingly show it; guidance on this would be much appreciated) the tower theorem would give me that (where $K$ is the splitting field of $f(x)$) $$[K:\mathbb{Q}] = 4[\mathbb{Q}(z, z'): \mathbb{Q}(z)] $$
But I'm not sure what to do here either.
Thanks for reading!
 A: First of all we find some relations between the roots $\alpha_1=\sqrt{\frac{1}{2}(1+i\sqrt{15})},$ $-\alpha_1$, $\alpha_2=\sqrt{\frac{1}{2}(1-i\sqrt{15})}$ and $-\alpha_2$, where $\alpha_1$ and $\alpha_2$ has positive real part.
We have that $\alpha_1 \alpha_2 = \sqrt{\frac{1}{4}(1-(i\sqrt{15})^2)} = \sqrt{\frac{1}{4}16} = 2 \in \mathbb{Q}.$ Also $(\alpha_1+\alpha_2)^2=5 \ \Rightarrow \ \alpha_1+\alpha_2=\sqrt{5}.$ Furthermore, $(\alpha_1+\alpha_2)(\alpha_1-\alpha_2)=\alpha_1^2-\alpha_2^2 = i\sqrt{15} \ \Rightarrow \ \alpha_1-\alpha_2 = i\sqrt{3}.$
As you noted, the splitting field $\mathbb{K}$ of $f$ is $$\mathbb{K}=\mathbb{Q}(\alpha_1,-\alpha_1,\alpha_2,-\alpha_2)=\mathbb{Q}(\alpha_1,\alpha_2).$$
Now, since $\alpha_1 \alpha_2 \in \mathbb{Q}$, we have that $\alpha_2=\frac{2}{\alpha_1} \in \mathbb{K}$. Therefore $\mathbb{K}=\mathbb{Q}(\alpha_1)$ since $\mathbb{K}$ is a field and $\alpha_2$ could be written as a rational multiple of the inverse of $\alpha_1.$ Here, with your work, we can conclude directly that $$[\mathbb{K}:\mathbb{Q}] = 4.$$
If you want to prove that $f$ is irreducible over $\mathbb{Q}[x]$, I would do it like this. We know that in $\mathbb{K}[x]$ is $$f(x)=(x-\alpha_1)(x+\alpha_1)(x-\alpha_2)(x+\alpha_2).$$ Clearly, none of the roots of $f$ is in $\mathbb{Q}$ so if $f$ were reducible in $\mathbb{Q}[x]$ it won't be factored with a linear factor. So focus on the possible factorizations as a product of quadratic polynomials. There are $3$ possibilities:


*

*\begin{align*}
f(x)&=[((x-\alpha_1)(x+\alpha_1)][(x-\alpha_2)(x+\alpha_2)]\\&=(x^2-\frac{1}{2}(1+i\sqrt{15}))(x^2-\frac{1}{2}(1-i\sqrt{15})).
\end{align*}

*\begin{align*}
f(x)&=[(x-\alpha_1)(x-\alpha_2)][(x-\alpha_1)(x+\alpha_2)]\\&=(x^2-(\alpha_1+\alpha_2)x+\alpha_1\alpha_2)(x^2+(\alpha_1+\alpha_2)x+\alpha_1\alpha_2)\\&=(x^2-\sqrt{5}x+2)(x^2+\sqrt{5}x+2)
\end{align*}

*\begin{align*}
f(x)&=[(x-\alpha_1)(x+\alpha_2)][(x+\alpha_1)(x-\alpha_2)]\\&=(x^2-(\alpha_1-\alpha_2)x-\alpha_1\alpha_2)(x^2+(\alpha_1-\alpha_2)x-\alpha_1\alpha_2)\\&=(x^2-i\sqrt{3}x-2)(x^2+i\sqrt{3}x-2)
\end{align*}
None of the factorizations above are valid in $\mathbb{Q}[x]$ since has coefficients not in $\mathbb{Q}.$ Hence, $f$ is irreducible over $\mathbb{Q}[x]$ and is $[\mathbb{K}:\mathbb{Q}]=4.$
