Find the splitting field in $\mathbb{C}$ of $f(x) = x^4 + 1 \in \mathbb{Q}[x]$ 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.
 A: *

*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.
A: 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  $\mathbb Q$.
A: 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.
