For $f(x) = x^4 -x ^2 - 2 = (x^2-2)(x^2+1)$ in $\Bbb Q[x]$, show that $\Bbb Q(\sqrt{2}, i)$ is the splitting field.

Question

For $f(x) = x^4 -x ^2 - 2 = (x^2-2)(x^2+1)$ in $\Bbb Q$, show that $\Bbb Q(\sqrt{2}, i)$ is the splitting field.

I can see that the field is $\{a+b\sqrt{2} + i(c+d\sqrt{2}): a,b,c,d \in \Bbb Q\}$ and that $f(x)$ splits in it, but I don't know how to show that it's the smallest field extension that $f(x)$ splits in.

Anyone have any ideas?

Answer 1

The splitting field of $f$ is generated over $\Bbb Q$ by the roots $\{±\sqrt 2, i\}$ of $f$. Therefore it is $\Bbb Q(\sqrt 2, i)$.

Indeed, if $f$ splits over $K \supset \Bbb Q$, then every root $r_j$ of $f$ belongs to $K$, so that $\Bbb Q(r_1,\dots,r_n) \subset K$.

Answer 2

Maybe the answer is hiding behind the definition of the splitting field of a polynomial $f$. That is, given an extension $\mathbb{L} / \mathbb{K}$ and a polynomial $f \in \mathbb{K}[x]$, $\mathbb{L}$ is a splitting field of $f$, if $\mathbb{L}$ is generated over $\mathbb{K}$ by the roots $\{α_{1},...,α_{k}\}$ of $f$, in other words, $\mathbb{K}(α_{1},...,α_{k}) = \mathbb{L}$.

If you check the definition of a field after adjoining a number of roots you'll get your answer.