Question

Find the spititing field of .

(a)Over$\mathbb{Q}$ .

(b)Over $ℝ$.

Short answer

(a)Splitting field over$\mathbb{Q}$is localid="1660218516331" $\mathbb{Q}(\sqrt[4]{2},i)$

(b)Splitting field over$ℝ$ is $ℝ\left(i\right)$.

Step-by-step solution

Definition of Splitting field

A Splitting field of a polynomial with coefficient in the smallest field extension of that field over which the polynomial splits into linear factors.

Step-2: Find the Splitting field 

Consider a polynomial$x^n-2$. Now let$f\left(x\right)=x^4-2$.

Find the root of$f\left(x\right)=x^4-2$as follows,

$\begin{array}{c}{x}^4-2=0\\ (x^2+\sqrt{2})(x^2-\sqrt{2})=0\\ (x+i\sqrt[4]{2})(x-i\sqrt[4]{2})(x+\sqrt[4]{2})(x+\sqrt[4]{2})=0\end{array}$

This gives

$x=\pm \sqrt[4]{2},i\sqrt[4]{2}$

.

(a)

Since the roots of a polynomial are$x=\pm \sqrt[4]{2},i\sqrt[4]{2}$. Thus the Splitting field of$f\left(x\right)$over$\mathbb{Q}$is,$\mathbb{Q}(\sqrt[4]{2},i)$

$\begin{array}{c}\mathbb{Q}(\pm \sqrt[4]{2},i\sqrt[4]{2})=\mathbb{Q}(\sqrt[4]{2},i\sqrt[4]{2})\\ =\mathbb{Q}(\sqrt[4]{2},i)\end{array}$

Equality holds as$i\in \mathbb{Q}(\sqrt[4]{2},i\sqrt[4]{2})$and$i\sqrt[4]{2}\in \mathbb{Q}(\sqrt[4]{2},i)$. Hence the Splitting field overlocalid="1660218531138" $\mathbb{Q}$is$\mathbb{Q}(\sqrt[4]{2},i)$.

(b)Step 3: Find the Splitting field

Since the root of the polynomial$f\left(x\right)$are

$x=\pm \sqrt[4]{2},i\sqrt[4]{2}$

As $\sqrt[4]{2}\in ℝ$. Thus the Splitting field of $f\left(x\right)$ over $ℝ$ is$ℝ\left(i\right)$ .