I have to find the splitting field $L$ of the polynomial $f(x)=x^6-9x^4+3x^2+5\in \mathbb{Q}[x]$.
The splitting field is an extension that contains all roots, so I calculated the roots of $f(x)$ which are $x_{1}=-1$, $x_{2}=1$, $x_{3}=-\sqrt{4+\sqrt{21}}$, $x_{4}=\sqrt{4+\sqrt{21}}$, $x_{5}=-i\sqrt{\sqrt{21}-4}$, $x_{6}=i\sqrt{\sqrt{21}-4}$ and the splitting field should be $L=\mathbb{Q}[\sqrt{4+\sqrt{21}},i\sqrt{\sqrt{21}-4}]$. Is this right?
Now I need to find the degree of L over $\mathbb{Q}$, $[L:\mathbb{Q}]$ and I'm stuck. Thanks in advance for any help.
Best Answer
Factorizing the polynomial gives $(x^4 - 8x^2 - 5)(x + 1)(x - 1)$.
Hence the splitting field $L/\mathbb Q$ is the same as the splitting field of $g(x) = x^4 - 8x^2 - 5 = (x^2 - 4)^2 - 21$.
If $\alpha \in L$ is a zero of $g(x)$, then we have $(\alpha - 4)^2 = 21$, hence $L$ contains the subfield $K = \mathbb Q[\sqrt{21}]$.
It remains to factorize the polynomials $x^2 - (4 \pm \sqrt{21})$ over the field $K$. This is the same as checking whether $4 \pm \sqrt{21}$ is a square in $K$.
We do this by calculating the norm $N_{K/\mathbb Q}(4 \pm \sqrt{21}) = -5$, which is not a square in $\mathbb Q$, hence $4 \pm \sqrt{21}$ are not squares in $K$.
Therefore, $L$ contains the quadratic extension $E = K[\sqrt{4 + \sqrt{21}}]$ of $K$, which has degree $4$ over $\mathbb Q$.
It remains to factorize $x^2 - (4 - \sqrt{21})$ over $E$, i.e. to check whether $4 - \sqrt{21}$ is a square in $E$.
We prove a simple lemma.
Proof: The if part is clear. For the only if part, suppose $b = (u + v\sqrt a)^2$, we then have $b = u^2 + v^2a + 2uv\sqrt a$. Since $1, \sqrt a$ form an $F$-basis of $F[\sqrt a]$, it follows that $uv = 0$. But $v = 0$ implies that $b$ is a square in $F$, so we have $u = 0$ and $b = v^2a$, hence $ab = (va)^2$ is a square in $F$.
Now in our problem, applying the lemma to $F = K$ and $a, b = 4 \pm \sqrt{21}$, we see that it suffices to check whether the product $(4 + \sqrt{21})(4 - \sqrt{21}) = -5$ is a square in $K$.
Since $K = \mathbb Q[\sqrt{21}]$, we apply again the lemma to $F = \mathbb Q$ and $a, b = 21, -5$, and the problem becomes to check whether $21 \times (-5) = -105$ is a square in $\mathbb Q$. As it certainly is not, we conclude that $4 - \sqrt{21}$ is not a square in $E$.
Thus the splitting field $L$ is the quadratic extension $E[\sqrt{4 - \sqrt{21}}]$ of $E$, which has degree $8$ over $\mathbb Q$.
Putting together, one obtains $L = \mathbb Q[\sqrt{21}, \sqrt{4 + \sqrt{21}}, \sqrt{4 - \sqrt{21}}]$. It is easy to verify that it can also be written as $L = \mathbb Q[\sqrt{-5}, \sqrt{4 + \sqrt{21}}]$.