Irreducible polynomial/Splitting field Let $f(x)=x^4+16 \in \mathbb{Q}[x]$. 


*

*Split $f(x)$ into a product of first degree polynomials in $\mathbb{C}[x]$.

*Show that $f(x)$ is an irreducible polynomial of $\mathbb{Q}[x]$.

*Find the splitting field $E$ of $f(x)$ and the degree of the extension $[E:\mathbb{Q}]$.
I have done the following: 


*

*$f(x)=(x^2-4i)(x^2+4i)=(x-2\sqrt{i})(x+2\sqrt{i})(x-2\sqrt[3]{i})(x+2 \sqrt[3]{i})=(x-2 e^{\pi i/4})(x+2 e^{\pi i /4})((x-2 e^{\pi i/6})(x+2e^{\pi i/6})$


Is it correct?? 


*$f(x)$ is irreducible in $\mathbb{Q}$.
If it were not irreducible, then it could be written as a product of polynomials of $\mathbb{Q}[x]$ as followed:


*

*It can be written as a product of four first degree polynomials:
$f(x)=(x-2\sqrt{i})(x+2\sqrt{i})(x-2\sqrt[3]{i})(x+2 \sqrt[3]{i})$
But the coefficients are not in $\mathbb{Q}$, So, it cannot be written in that way.

*It can be written as a product of two second degree polynomials:
$f(x)=(x^2-4i)(x^2+4i)$ 
But the coefficients are not in $\mathbb{Q}$, So, it cannot be written in that way.

*It can be written as a product of a first degree and a third degree polynomial:
$f(x)=(x-2\sqrt{i})\left [(x+2\sqrt{i})(x^2+4i)\right ] \\ =(x-2\sqrt{i}) (x^3+4ix+2\sqrt{i}x^2+8i\sqrt{i})$
But the coefficients are not in $\mathbb{Q}$, So, it cannot be written in that way.
Is it correct?? 


*The splitting field is $E=\mathbb{Q}(\pm 2 e^{\pi i/4}, \pm 2 e^{\pi i/6})=\mathbb{Q}(e^{\pi i/4}, e^{\pi i/6})$


Is it correct?? 
$Irr(e^{\pi i/4}, \mathbb{Q})=x^4+1$
$[\mathbb{Q}(e^{\pi i/4}):\mathbb{Q}]=4$
How can I continue to find $[\mathbb{Q}(e^{\pi i/4}, e^{\pi i/6}): \mathbb{Q}]$?? 
 A: For $\;k=0,1\;$ :
$$\begin{align*}&\;\;\;\;\,i=e^{\frac\pi2i+2k\pi i}\implies \sqrt i\;\;\;\;=\;e^{\frac\pi4i+k\pi i}=\begin{cases}e^{\frac\pi4i}\;=\;\;\,\frac1{\sqrt2}(1+ i)\\{}\\e^{\frac{5\pi i}4}=-\frac1{\sqrt2}(1+i)\end{cases}\\{}\\
&-i=e^{-\frac\pi2i+2k\pi i}\implies \sqrt{-i}=e^{-\frac\pi4i+k\pi i}=\begin{cases}e^{-\frac\pi4i}=\;\;\,\frac1{\sqrt2}(1-i)\\{}\\e^{\frac{3\pi i}4}\;=-\frac1{\sqrt2}(1-i)\end{cases}\end{align*}$$
From here
$$x^4+16=(x^2-4i)(x^2+4i)=$$
$$=(x-\sqrt2(1+i))(x+\sqrt2(1+i))(x-\sqrt2(1-i))(x+\sqrt2(1-i)$$
If, for example, you multiply the first and third factor above you get
$$x^2-2\sqrt2\,x+4$$
and likewise with the second and fourth factors:
$$x^2+2\sqrt2\,x+4$$
This way, you can get the decomposition of the polynomial over $\;\Bbb R\;$ and check at once that there are no real, and thus also rational, roots.
With this you already have solved $\;(1)-(2)\;$
Now, for $\;(3)\;$ : observe that
$$r=\sqrt2(1+i)\implies r^2=2\cdot2i=4i\implies r^4=-16\implies \;[\Bbb Q(\sqrt2(1+i):\Bbb Q]=4$$
and a possible $\;\Bbb Q$-basis of this linear space is (with$\; w:=\sqrt2(1+i)=2\,e^{\frac\pi4i}$)
$$\left\{\;1\,,\,\,w\,,\,\,w^2=4i\,,\,\,w^3=4\sqrt2(-1+i)=-4\sqrt2(1-i)\;\right\}$$
Check that all the roots of $\;x^4+16\;$ are a linear combination of this basis, and you're done.
Of course, the two lines after "Now, for $\;(3)\;$" are superfluous as we already know that, but whatever.
