Question

Find and describe a splitting field of ${\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{1}$ over $\mathit{\mathbb{Q}}$.

Short answer

$\mathbb{Q}(\sqrt{2},i)$is the splitting field of$x^4+1$.

Step-by-step solution

Definition of splitting field

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

Showing thatQ(2,i)   is the splitting field of x4+1x4+1  

Let$L$ be the splitting field of the given polynomial. Now find the root of the polynomial $x^4+1$.

$\begin{array}{c}{x}^4+1=0\\ {(x^2+1)}^2-2x^2=0\\ (x^2+1+\sqrt{2}x)(x^2+1-\sqrt{2}x)=0\end{array}$

This implies

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

If$(x^2+1+\sqrt{2}x)=0$then

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

If$(x^2+1-\sqrt{2}x)=0$then

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

So the root of the given polynomial is

$\begin{array}{c}x=\pm \frac{1+i}{\sqrt{2}},\pm \frac{-1+i}{\sqrt{2}}\\ x=(\pm \sqrt{2}/2)\pm (\sqrt{2}/2)i\\ x=\mathbb{Q}\left(e^{i\pi /4}\right)\end{array}$

Since power of$e^{i\pi /4}$generates the other three root of$x^4+1$and also$i,\sqrt{2}\in L$.

Therefore

role="math" localid="1658905358452" $L\subset Q\left(e^{i\pi /4}\right)\left(1\right)$

But

$\begin{array}{l}Q\left(e^{i\pi /4}\right)=Q(i,\sqrt{2})\\ \left(Q\left(e^{i\pi /4}\right)\right)\subset L\left(2\right)\end{array}$

Thus from 1 and 2

$L=Q\left(e^{i\pi /4}\right)$

Hence the splitting field of $x^4+1$ is $Q(\sqrt{2},i)$.