Question

Find a splitting field of $x^4-4x^2-5$ over $\mathrm{\mathbb{Q}}$ and show it has dimension 4 over $\mathrm{\mathbb{Q}}$.

Short answer

$Q\left(\sqrt{5},i\right)$ is splitting field. It is proved that it has dimension 4 over $\mathrm{\mathbb{Q}}$.

Step-by-step solution

Definition of splitting field.

A splitting field of a polynomial in a field is a smallest field extension of that field over which the polynomial splitting or split into linear factor.

Find splitting field

Let$f\left(x\right)=x^4-4x^2-5$

Find the root of the polynomial as follows,

$$\begin{gathered} x^4-4x^2-5=0 \\ {x}^4=4x^2+5 \\ {\left(x^2-5\right)}^2-9=0 \\ x=\pm \sqrt{5},\pm i \end{gathered}$$

Hence the splitting field is $\mathrm{\mathbb{Q}}\left(\sqrt{5},i\right)$.

Then minimal polynomial of $\mathrm{\mathbb{Q}}\left(\sqrt{5},i\right)$is $x^2-5$ which is 2 dimension and

$\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(i\right)\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{5},i\right)$

Therefore

$\begin{array}{rcl}\left[\mathrm{\mathbb{Q}}\left(\sqrt{5},i\right):\mathrm{\mathbb{Q}}\right]& =& \left[\mathrm{\mathbb{Q}}\left(\sqrt{5},i\right):\mathrm{\mathbb{Q}}\left(i\right)\right]\left[\mathrm{\mathbb{Q}}\left(i\right):\mathrm{\mathbb{Q}}\right]\\ & =& 2·2\\ & =& 4\\ & & \end{array}$

Therefore the dimension of splititing field is 4 over $\mathrm{\mathbb{Q}}$.