Irreducible polynomial/Splitting field

[mathmari]

Hey!

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

1. Split $f(x)$ into a product of first degree polynomials in $\mathbb{C}[x]$.
   
2. Show that $f(x)$ is an irreducible polynomial of $\mathbb{Q}[x]$.
   
3. Find the splitting field $E$ of $f(x)$ and the degree of the extension $[E:\mathbb{Q}]$.

I have done the following:

1. $f(x)=(x^2-(4i)^2)(x^2+(4i)^2)=(x-2\sqrt{i})(x+2\sqrt{i})(x-2\sqrt[3]{i})(x+2 \sqrt[3]{i})=(x-2 e^{\pi i/4})(x+2 e^{\pi i /4})((x-2 e^{\pi i/6})(x+2e^{\pi i/6})$
   
   Is it correct?? (Wondering)
2. $f(x)$ 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:
     $f(x)=(x^2-(4i)^2)(x^2+(4i)^2)$
     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:
     $f(x)=(x-2\sqrt{i})\left [(x+2\sqrt{i})(x^2-16)\right ] \\ =(x-2\sqrt{i}) (x^3-16x+2\sqrt{i} x^2-32\sqrt{i})$
     But the coefficients are not in $\mathbb{Q}$, So, it cannot be written in that way.
   Is it correct?? (Wondering)
3. 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?? (Wondering)$Irr(e^{\pi i/4}, \mathbb{Q})=x^4+1$
   
   $[\mathbb{Q}(e^{\pi i/4}):\mathbb{Q}]=4$
   
   How can I continue to find $[\mathbb{Q}(e^{\pi i/4}, e^{\pi i/6}): \mathbb{Q}]$?? (Wondering)
   

 

[Fallen Angel]

Hi,

I think is everything OK.

Remember that $[\Bbb{Q}(e^{\pi i/6},e^{\pi i / 4}) \ : \ \Bbb{Q}]=[\Bbb{Q}(e^{\pi i / 4}) (e^{\pi i/6})\ : \ \Bbb{Q}(e^{\pi i /4})][\Bbb{Q}(e^{\pi i / 4}) \ : \ \Bbb{Q}]$

So you have to compute $Irr(e^{\pi i / 6},\Bbb{Q}(e^{\pi i / 4}))$

 

[mathmari]

Remember that $[\Bbb{Q}(e^{\pi i/6},e^{\pi i / 4}) \ : \ \Bbb{Q}]=[\Bbb{Q}(e^{\pi i / 4}) (e^{\pi i/6})\ : \ \Bbb{Q}(e^{\pi i /4})][\Bbb{Q}(e^{\pi i / 4}) \ : \ \Bbb{Q}]$

So you have to compute $Irr(e^{\pi i / 6},\Bbb{Q}(e^{\pi i / 4}))$

$e^{\pi i / 6}$ is a root of $x^6+1$, right?? (Wondering)

But how could I find $Irr(e^{\pi i / 6},\Bbb{Q}(e^{\pi i / 4}))$ ?? (Wondering)

 

[Fallen Angel]

Try to factor $x^{6}+1$.

Is the same idea when you factor it over $\Bbb{Q}$, but this time the coefficients of your polynomials are in $\Bbb{Q}(e^{ \pi i/ 4})=\{a+be^{ \pi i/4} \ : \ a,b\in \Bbb{Q}\}$.

This carries a lot of work, but I have no a better idea. :(