I am trying to find a splitting field and its degree of $x^4+x^2-1$ over $\mathbb{Q}$
I make a substitution $x^2=u$ then I get $u^2+u-1=0$ and get 4 solutions
$u_{1}=\frac{-1+\sqrt{5}}{2}\Rightarrow x=\pm \sqrt{\frac{1}{2}(-1+\sqrt{5})}$
$u_{2}=\frac{-1-\sqrt{5}}{2}\Rightarrow x=\pm i\sqrt{\frac{1}{2}(1+\sqrt{5})}$
So I suppose I will be needing to adjoin $i$ and also, say $\alpha:=\sqrt{\frac{1}{2}(-1+\sqrt{5})}$ then I get also powers of $\alpha$
So I will need the following extension
$\mathbb{Q}(i,\alpha)$? with basis
$\{1,\alpha,\alpha^2,\alpha^3,i,i\alpha,i\alpha^2,i\alpha^3\}$ so the extension will have degree 8. Is this a sufficient argument?
Best Answer
Let $p(x)=x^4 +x^2 -1$. Let's first consider how $p$ factors over $\mathbb{Q}(\phi)$, where $\phi = \frac{1}{2}(1+\sqrt{5})$, the golden ratio. We know the minimal polynomial of $\phi$ is $x^2 - x -1$. As you can check, over $\mathbb{Q}(\phi)$ we have the factorization $$p(x) = ( x^2 +1- \phi )(x^2 + \phi), $$ If we now adjoin $\sqrt{\phi-1}$, we get the factorization $$p(x) = ( x +\sqrt{\phi-1} )(x-\sqrt{\phi-1})(x^2+\phi)$$ over $\mathbb{Q}(\phi, \sqrt{\phi-1}) = \mathbb{Q}(\sqrt{\phi-1})$. The extension $\mathbb{Q}(\phi) / \mathbb{Q}$ is quadratic and the extension $\mathbb{Q}(\sqrt{\phi-1})/\mathbb{Q}(\phi)$ was quadratic, so we have $[\mathbb{Q}(\sqrt{\phi-1}):\mathbb{Q}]=4$.
Note that that $\mathbb{Q}(\sqrt{\phi-1})$ is a subfield of the reals. Since the roots of $x^2+\phi$ are not real, $\mathbb{Q}(\sqrt{\phi-1})$ cannot be the splitting field of $p(x)$.
If we adjoin $\sqrt{-\phi}$, we get the factorization $$p(x) = ( x +\sqrt{\phi-1} )(x-\sqrt{\phi-1})(x + \sqrt{-\phi})(x-\sqrt{-\phi}),$$ over $F=\mathbb{Q}(\sqrt{\phi-1}, \sqrt{-\phi})$, which is finally the product of linear factors.
Now $F/\mathbb{Q}(\sqrt{\phi-1})$ is a quadratic extension, so $[F:\mathbb{Q}(\sqrt{\phi-1})]=2$. Therefore $[F:\mathbb{Q}]=8$.
Thus the splitting field of $x^4 +x^2 -1$ is over $\mathbb{Q}$ is $F=\mathbb{Q}(\sqrt{\phi-1}, \sqrt{-\phi})$, an extension of degree 8.