Question

Show that $\mathit{\mathbb{Q}}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\mathbf{i}\mathbf{)}$ is a splititing field of${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\sqrt{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{3}$ over$\mathit{\mathbb{Q}}\mathbf{\left(}\sqrt{\mathbf{2}}\mathbf{\right)}$ .

Short answer

$\mathbb{Q}(\sqrt{2}\pm i)$is the splititing field of$x^2-2\sqrt{2}x+3$.

Step-by-step solution

Definition of splititing field.

A splititing field of a polynomial with coefficient in a smallest field over which the polynomial split or split into linear facots.

Showing that ℚ(2±i)  is the splititing field of  x2−22x+3

Let$f\left(x\right)=x^2-2\sqrt{2}x+3$and splititing field of$f\left(x\right)$is$\mathbb{Q}(\sqrt{2},i)$over$\mathbb{Q}\left(\sqrt{2}\right)$.

First find the root of polynomial$f\left(x\right)=x^2-2\sqrt{2}x+3$.

$\begin{array}{c}{x}^2-2\sqrt{2}x+3=0\\ x^2-2\sqrt{2}+3=0\\ {(x-\sqrt{2})}^2+1=0\\ x=(\sqrt{2}\pm i)\end{array}$

Therefore the splititing field of$f\left(x\right)$is$\mathbb{Q}(\sqrt{2}\pm i)$. Now as

$i,\sqrt{2}\in \mathbb{Q}(\sqrt{2},i)$

So, their sum and different should also belong to$\mathbb{Q}(\sqrt{2},i)$. Therefore

$\sqrt{2}+i\in \mathbb{Q}(\sqrt{2},i)$

Also$\sqrt{2}-i\in \mathbb{Q}(\sqrt{2},i)$. So,

$\begin{array}{l}\sqrt{2}\pm i\in \mathbb{Q}(\sqrt{2}\pm i)\\ \mathbb{Q}(\sqrt{2},i)\subset \mathbb{Q}(\sqrt{2}\pm i)\left(1\right)\end{array}$

For reverse containment

$\begin{array}{l}\sqrt{2}\pm i\in \mathbb{Q}(\sqrt{2}\pm i)\\ \sqrt{2}\pm i+\sqrt{2}-i\in \mathbb{Q}(\sqrt{2}\pm i)\end{array}$

Simplify this

$\sqrt{2}\in \mathbb{Q}(\sqrt{2}\pm i)$

Also,

$\sqrt{2}\in \mathbb{Q}(\sqrt{2},i)$

Therefore,

$\mathbb{Q}(\sqrt{2}\pm i)\subset \mathbb{Q}(\sqrt{2},i)$

From 1 and 2

$\mathbb{Q}(\sqrt{2}\pm i)=\mathbb{Q}(\sqrt{2},i)$

.

$\mathbb{Q}(\sqrt{2}\pm i)$is the splititing field of$x^2-2\sqrt{2}x+3$.