Let $E$ be the splitting field of $f(x)=x^4-x^2-2$ over $\Bbb Q$. Find a basis for $E$.

Question

Let $E$ be the splitting field of $f(x)=x^4-x^2-2$ over $\Bbb Q$. Find a basis for $E$.

The polynomial $f$ has roots $\pm\sqrt2$ and $\pm i$. I have a theorem which states that if $a$ is a zero of an irreducible polynomial over a field $F$, then the set $\{1,a,\dots,a^{n-1}\}$ is a basis for $F(a)$ over $F$. Here $n$ is the dimension of the polynomial.

So since I have four roots for $f$ could I pick any one of them an construct a base for $E$? That is would $\{1,\sqrt2, 2, \sqrt2^3, \sqrt2^4 \}$ be a base for $E$ as well as $\{1,i,i^2,i^3,i^4\}$?

Answer 1

The splitting field is not $\mathbb{Q}(\sqrt{2}\,)$, because this extension doesn't contain $i$.

Neither it is $\mathbb{Q}(i)$, because this extension doesn't contain $\sqrt{2}$.

You need to do a two step extension: the splitting field is $\mathbb{Q}(\sqrt{2},i)$ and apply the dimension formula (and its proof) to get that a basis is $$ \{1,\sqrt{2},i,i\sqrt{2}\} $$

Answer 2

Neither of these is a basis (the sets aren't $\mathbb{Q}$-linearly independent, since e.g. $\sqrt{2}^2 - 2 = 0$). The problem is that your theorem is about an extension of the form $F(\alpha)/F$, but the extension in question is neither $\mathbb{Q}(i)$ nor $\mathbb{Q}(\sqrt{2})$.

Answer 3

$f(x)=x^4-x^2-2=(x^2-2)(x^2+1)$

$K=\frac{Q[x]}{\langle x^2-2\rangle}\cong \Bbb{Q}[\sqrt{2}]$

$L=\frac{K[x]}{\langle x^2+1\rangle}\cong \Bbb{K}[i]$

$\Bbb{Q}\overset{\{1, \sqrt{2}\}}\to K\overset{\{1, i\}}\to L$

$[L:\Bbb{Q}]=[K:\Bbb{Q}]\times [L:K]=4$

Splitting field $\mathbb{Q}[\sqrt{2}, i]$ and has basis$\{1, \sqrt{2}\}\times \{1, i\} =\{1, \sqrt{2}, i,\sqrt{2}i\}$