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Show that $\mathbb{Q}(\sqrt[4]{2}, i)$ is also the splitting field of $x^4 + 2$ over $\mathbb{Q}$.

I solve it as $x^4+2=0$

then $x^4=-2$

$\implies \mathbb Q(\sqrt[4]{-2}, \sqrt{-2})$

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Form of roots: $z^4 = -2 \Rightarrow z_k = 2^{\frac{1}{4}}e^{(\frac{pi+2pk}{4})i}$ for $k=0,1,2,3$.

More clearly: Let $\alpha = 2^{\frac{1}{4}} \omega$, where $ \omega = \frac{-\sqrt{2}+i\sqrt{2}}{2}$. And one can read off the roots by plugging in values of $k$ above.

Explanation: It suffices to have $\alpha,i$ since if we have $\alpha \Rightarrow$ we have $\alpha^2 = \sqrt{2}, \alpha^4 = 2, \frac{\alpha^2}{\alpha^4}=\frac{\sqrt{2}}{2}$. Now if we have $i$ then we are can get all of the roots by simply multiplying and taking powers.

Hence, the splitting field of $f(x)=x^4-2$ is : $\mathbb{Q}(2^{\frac{1}{4}},i)$.

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  • $\begingroup$ i do it like what you said and K=0 i have (2)^.25 exp(pi/4) and k=1 i have (2)^.25 exp(pi2/4) =(2)^.25 *i so,Q((2)^.25 *i) is that = to Q((2)^.25 ,i) $\endgroup$ – user146264 May 1 '14 at 5:41
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Note that the fourth roots of $-2$ are $\alpha=\sqrt[4]{2}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)$, and $-\alpha$, $i\alpha$, and $-i\alpha$.

These roots are all in $\mathbb{Q}(\sqrt[4]{2},i)$, so the splitting field $F$ is a subfield of $\mathbb{Q}(\sqrt[4]{2},i)$.

To show that $F$ is all of $\mathbb{Q}(\sqrt[4]{2},i)$, it is enough to show that $\sqrt[4]{2}$ and $i$ are both in $F$.

To show $i$ is in $F$ is easy, for $i=\frac{\alpha i}{\alpha}$.

To show that $\sqrt[4]{2}$ is in $F$, note that $\alpha^2=\sqrt{2}(-i)$. Since $i\in F$, it follows that $\sqrt{2}\in F$. But since $\alpha$, $i$, and $\sqrt{2}$ are in $F$, it follows that $\sqrt[4]{2}$ is in $F$.

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  • $\begingroup$ i am little pit confuse since i see from split field the extension will be 8 but from x^4+2=(x-a)(x+a)(x^2-x-a) that mean we have four extension dimension $\endgroup$ – user146264 May 3 '14 at 2:49
  • $\begingroup$ The splitting field does have degree $8$ over $\mathbb{Q}$. The comment above does not contradict that. I am assuming that your $a$ is my $\alpha$. Already $\mathbb{}(\alpha)$ has degree $4$ over $\mathbb{Q}$, and it is not all of our splitting field. $\endgroup$ – André Nicolas May 3 '14 at 3:02

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