Calculating Splitting field

Question

Find the splitting field of the polynomial and degree over $\mathbb{Q}$

$P(X)=X^4+2$.

The roots of $P(X)$ are $\sqrt[4]{2}\sqrt{i},\ -\sqrt{i}\sqrt[4]{2}, \ i\sqrt{i}\sqrt[4]{2},\ -i\sqrt{i}\sqrt[4]{2}$, therefore,

a splitting field is $K=\mathbb{Q}(\sqrt{i},\sqrt[4]{2})$.

I have problems to calculate the degree.

$[K:\mathbb{Q}]=[K:\mathbb{Q}(\sqrt{i})][\mathbb{Q}(\sqrt{i}):\mathbb{Q}]=[K:\mathbb{Q}(\sqrt{i})]*4$

My question is: What is the degree $[K:\mathbb{Q}(\sqrt{i})]$?

I think: $polmin(\sqrt[4]{2},\mathbb{Q}(\sqrt{i}))=X^4+2 $

Thanks you all.

Answer 1

The way you get to the splitting field is inconclusive. For example, how do you show that $\root4\of2$ is in there? This is a key step in my solution (see below), so it needs special attention.

Another obvious error is on the last line. Surely the minimal polynomial of $\root4\of2$ over any extension field of $\Bbb{Q}$ has to be a factor of $x^4-2$.

I try to remedy these maladies, and leave a few details about the steps to you.

Combine the following points/steps:

• $\sqrt i=(\pm)\frac{1+i}{\sqrt2}$. BTW, I kinda dislike the notation $\sqrt i$, as this hides the fact that $i$ has two square roots. It might be preferrable to refer to this number as a specific eighth root of unity. For example, the plus sign in the above formula: $\zeta_8=(1+i)/\sqrt2 =e^{i\pi /4}$.
• It easily follows from this that $\Bbb{Q}(\sqrt i)=\Bbb{Q}(i,\sqrt2)$.
• Let us denote one of the zeros of $x^4+2$ by $u=\zeta_8\root4\of2$. Then $u^2= i\sqrt2$, so $u^2\in \Bbb{Q}(i,\sqrt2)$. Therefore $[\Bbb{Q}(u,i,\sqrt2):\Bbb{Q}(i,\sqrt2)]\le2.$
• The other zeros of $x^4+2$ are $iu,-u$ and $-iu$, so the splitting field is $$K=\Bbb{Q}(u,i,\sqrt2)=\Bbb{Q}(u,i)=\Bbb{Q}(u,\sqrt2)$$ (don't know yet which way of looking at is most useful for the purposes of solving this problem).
• Because $\zeta_8\in K$, it follows that $\root4\of2\in K$.
• The intersection $\Bbb{Q}(i,\sqrt2)\cap\Bbb{R}=\Bbb{Q}(\sqrt2)$. This implies (how? why?) that $\root4\of2\notin\Bbb{Q}(i,\sqrt2)$.
• This implies (how? why?) that $u\notin \Bbb{Q}(i,\sqrt2)$. Therefore $[\Bbb{Q}(u,i,\sqrt2):\Bbb{Q}(i,\sqrt2)]\ge2.$

At this point you should be able to answer your question.