I am having some trouble with finding the splitting field of this polynomial. I know that once I find the roots of these I can construct the field that I am looking for but, I am having problems in finding its roots. Could anyone help me?
Thanks a lot.
The roots of this polynomial are the primitive 8th roots of unity in $\mathbb{C}$ (i.e. $e^{\pi i/4},e^{3\pi i/4}, e^{5\pi i/4}$, and $e^{7\pi i/4}$).
Rewrite these roots using $e^{i\theta}=\cos\theta + i\sin\theta$ to conclude that the splitting field is contained in $\mathbb{Q}(\sqrt{2},i)$.
Next, compute $[\mathbb{Q}(\sqrt{2},i):\mathbb{Q}]$ and conclude you have found the splitting field.
When we are looking for splitting field of $$x^{4}+1$$ in $\mathbb C$ we are free to find the complex roots of this polynomial .
$$x^{4}+1=0$$ or$$x^{4}=-1$$ or,$$x^{4}=e^{i(2k+1)\pi}$$ or, $$x=e^{{i(2k+1)\pi}\over {4}} \ \ for\ \ k=0,1,2,3;$$
Now putting the values $k=0,1,2,3;$ and keeping in mind that $e^{i\theta }=cos \theta + i \ sin \theta $
the following values of $x$ are found :
$${{1}\over {\sqrt2}}+{{i}\over {\sqrt2}}$$, $${{-1}\over {\sqrt2}}+{{i}\over {\sqrt2}}$$, $${{-1}\over {\sqrt2}}+{{-i}\over {\sqrt2}}$$, $${{1}\over {\sqrt2}}+{{-i}\over {\sqrt2}}$$
Now from here, can you see that splitting field containing all these $4$ elements adjoined to $\mathbb Q$ is equal to $\mathbb Q(i,\sqrt2)$ which has degree $4$ over $\mathbb Q$.
You could also use quadratic completion to avoid complex numbers until the last step: $$ x^4+1=(x^2+1)^2-2x^2=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1) $$ which requires to adjoin $\sqrt2$.
Then in the next step one gets $$ x^2±\sqrt{2}x+1=(x±\sqrt{2}/2)^2+1/2=(x±\sqrt{2}/2(1+i))(x±\sqrt{2}/2(1-i)) $$ requiring $i$ as additional element in the field.
