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This is what I did to find the Galois group for $x^8+2$:

Splitting field: $$K = \Bbb{Q}(\zeta_8, \zeta_{16}2^{1/8})$$

Since $Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q}) \cong Aut( \langle\zeta_8\rangle) \cong \mathbb Z^{\times}_8 = \{1, 3, 5, 7\}$, we know that we have four subgroups in $Gal(Q(\zeta_8)/\Bbb{Q})$ (Let $\zeta_8 = \zeta$):

$$\sigma_1(\zeta) = \zeta$$

$$\sigma_3(\zeta) = \zeta^3$$

$$\sigma_5(\zeta) = \zeta^5$$

$$\sigma_7(\zeta) = \zeta^7$$

The polynomial of $\zeta_{16}2^{1/8}$ is $X^4 = \zeta^2_8 2^{1/2} = \sqrt{2}i$. So we have $X^4 - \sqrt{2}i = X^4 - \zeta^2_8(\zeta^a + \zeta^{-a})$

We know that for $\sigma \in Gal(K/\Bbb{Q})$, we have

$\sigma(\zeta_8) = \zeta^a_8$ for $a \in Z^{\times}_8$ and $\sigma(\zeta_{16}2^{1/8}) = \zeta_{16}\zeta^b_82^{1/8}$ for $a \in \mathbb Z^{\times}_8$ and $b \in \mathbb Z_8$.

We know that we have:

$(\zeta_{16}\zeta^b_82^{1/8})^4 = \sqrt{2}\zeta^4_{16}\zeta^{4b}_8 = \zeta^2_8(\zeta^a_8 + \zeta^{-a}_8)$.

But $\sqrt{2}\zeta^4_{16}\zeta^{4b}_8 = \zeta^2_8(\zeta^a_8 + \zeta^{-a}_8) \implies \sqrt{2}\zeta^{4b}_8 = \zeta^a_8 + \zeta^{-a}_8$, right? So this seems similar to my previous question.

I’m a bit surprised, because I thought they would be different, so I’m wondering if there is something that I’m missing but unaware of…

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If you have $i$ and $\sqrt 2$, then you also have $\zeta_8$ and then you also have $\zeta_{16}$ from $1+\zeta_8 =(2+\sqrt 2)\zeta_{16}$. So you have actually the same fields. – Hagen von Eitzen May 10 '13 at 16:46

1 Answer 1

up vote 3 down vote accepted

To elaborate on Hagen's comment:

The splitting field of $x^8+2$ is, as you note, $K = \mathbb{Q}(\zeta_8, (-2)^{1/8})$, while the splitting field of $x^8-2$ is $L = \mathbb{Q}(\zeta_8, 2^{1/8})$. Note that $(1+i)/\sqrt{2}$ is an eighth root of $-1$. Since $\big((-2)^{1/8}\big)^4 = i\sqrt{2}$ and $\zeta_8^2 = i$, we see that both $\sqrt{2}$ and $i$ are elements of $K$. Thus, so is $(-1)^{1/8}$.

Therefore, $(-1)^{1/8}(-2)^{1/8} = 2^{1/8}\in K$. So, $L\subseteq K$. Comparing degrees over $\mathbb{Q}$ or arguing as above, we see that $L = K$.

Note, it is not generally true that the splitting fields of $x^n-a$ and $x^n + a$ are the same field. Our proof depended on the fact that $\mathbb{Q}(\zeta_{16}) = \mathbb{Q}(\zeta_8,\sqrt{2})$.

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