Question

Show that $x^2-3$ and $x^2-2x-2$ are irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$ and have the same splitting field namely $\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$.

Short answer

$x^2-3$and $x^2-2x-2$ are irreducible.

Step-by-step solution

Definition of irreducible polynomial

Irreducible polynomial is a non constant polynomial that can not be factor into the product of two non constant polynomial

Showing that x2-3  is irreducible

Let$f\left(x\right)=x^2-3$

Find the root of this if they belong to$Q\left(x\right)$then$f\left(x\right)=x^2-3$is reducible otherwise it is irreducible.

Therefore the root of $f\left(x\right)=x^2-3$are

$\begin{array}{rcl}{x}^2-3& =& 0\\ x^2& =& 3\\ x& =& \pm \sqrt{3}\\ & & \end{array}$

As $\pm \sqrt{3}\notin Q\left(x\right)$ so $f\left(x\right)=x^2-3$ is irreducible in $Q\left(x\right)$.

So the split field of $f\left(x\right)=x^2-3$is$Q\left(\sqrt{3}\right)$

Showing that  x2-2x-2 is irreducible

Let$g\left(x\right)=x^2-2x-2$

Find the root of the polynomial$g\left(x\right)=x^2-2x-2$and if they belong to$Q\left(x\right)$then$g\left(x\right)=x^2-2x-2$is reducible if not then$g\left(x\right)=x^2-2x-2$is irreducible.

The roots of $g\left(x\right)=x^2-2x-2$are,

$$\begin{gathered} x^2-2x-2=0 \\ x=1\pm \sqrt{3} \end{gathered}$$

As $1\pm \sqrt{3}\notin Q\left(x\right)$. So$g\left(x\right)=x^2-2x-2$, is irreducible.

Now to show$Q\left(1\pm \sqrt{3}\right)=Q\left(\sqrt{3}\right)$

As$Q\left(\sqrt{3}\right)\subseteq Q\left(1\pm \sqrt{3}\right)\left(1\right)$

Since $1,1\pm \sqrt{3}\in Q\left(1\pm \sqrt{3}\right)$. So,

$$\begin{gathered} 1+\sqrt{3}-1=\sqrt{3}\in Q\left(\sqrt{3}\right) \\ Q\left(1\pm \sqrt{3}\right)\subseteq Q\left(\sqrt{3}\right)\left(2\right) \end{gathered}$$

From 1 and 2

$Q\left(1\pm \sqrt{3}\right)=Q\left(\sqrt{3}\right)$

This show that both the polynomial have spiliting field.