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I have a problem on splitting field as follows:

Determine the degree of the splitting field of the polynomial $x^6-7$ over:

(a) $\mathbb{Q},$

(b) $\mathbb{Q(\alpha)},$ where $\alpha$ is primitive 3rd root of unity,

(c) $\mathbb{F_3},$ (field with 3 elements).

I try to prove (a) in the following way: Since $x^6-7=(x^3-\sqrt7)(x^3+\sqrt7),$ then the splitting field $K$ for $x^6-7$ must contain the splitting fields for $(x^3-\sqrt7)$ and $(x^3+\sqrt7).$ The roots of $(x^3-\sqrt7)$ are $7^{1/6}, 7^{1/6}\xi, 7^{1/6}\xi^2,$ and the roots of $(x^3+\sqrt7)$ are $-7^{1/6}, -7^{1/6}\xi, -7^{1/6}\xi^2,$ where $\xi$ is a primitive sixth root of unity. Is this reasoning correct?

How do I solve parts (b) and (c)? Thanks in advance!

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As $\sqrt[6]{7}$ is a zero of the polynomial and $\sqrt[6]{7} \not \in \mathbb{Q}$ we adjoin it to $\mathbb{Q}$ to obtain $\mathbb{Q}(\sqrt[6]{7})$. Now in $\mathbb{Q}(\sqrt[6]{7})$ the polynomial factors into $(x - \sqrt[6]{7})(x + \sqrt[6]{7})(x^2 + \sqrt[6]{7}x + \sqrt[3]{7})(x^2 - \sqrt[6]{7}x + \sqrt[3]{7})$. Now the roots of the quadratic factors are $\xi\sqrt[6]{7},\xi^2\sqrt[6]{7},-\xi\sqrt[6]{7},-\xi^2\sqrt[6]{7}$, where $\xi$ is the third root of unity. As $\xi \not \in \mathbb{Q}(\sqrt[6]{7})$. Adjoining we obtain that $\mathbb{Q}(\sqrt[6]{7}, \xi)$ is the splitting field of the polynomial over $\mathbb{Q}$. Now we have:

$$[\mathbb{Q}(\sqrt[6]{7}, \xi) : \mathbb{Q}]=[\mathbb{Q}(\sqrt[6]{7},\xi) : \mathbb{Q}(\sqrt[6]{7})][\mathbb{Q}(\sqrt[6]{7}) : \mathbb{Q}] = 2 \cdot 6 = 12$$

This is true as the minimal polynomial of $\xi$ over $\mathbb{Q}(\sqrt[6]{7})$ is $x^2 + x + 1$ and the minimal polynomial of $\sqrt[6]{7}$ over $\mathbb{Q}$ is $x^6 - 7$. Therefore the degree is $12$.

For the second part as $\mathbb{Q}(\xi) = \mathbb{Q}(\xi^2)$ and using the previous part we have that the splitting field is again $\mathbb{Q}(\sqrt[6]{7}, \xi)$, but this time the degree is $6$.

For the third part we have that $x^6 - 7 = (x+1)^3(x+2)^3$ in $\mathbb{F}_3$, so the splitting field is infact $\mathbb{F}_3$ and the degree is of order $1$.

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  • $\begingroup$ @Bernard What was I thinking? I must have confused with the fact that $\xi$ is a zero of $x^3 - 1$. $\endgroup$ – Stefan4024 Oct 1 '16 at 16:57
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    $\begingroup$ @AyshaA. We're working in $\mathbb{F}_3$, which is infact isomorphic to $\mathbb{Z}_3$. $1=-2$ and $-1$ are triple roots in that field. $\endgroup$ – Stefan4024 Oct 3 '16 at 7:48
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    $\begingroup$ @AyshaA. I'm worried that I didn't get your point to be fair. But here's what I do. Obviously $1$ is a root of $x^6-7$, so divide $x^6-7$ by $x-1$ to get $x^5 + x^4 + x^3 + x^2 + x + 1$. $1$ is a root of this polynomial too, so dividing by $x-1$ yields $x^4 + 2x^3 + x + 2$. Again dividing by $x-1$ we have $x^3 + 1$. Now $-1$ is a root of this, so dividing by $x + 1$ we have $x^2 + 2x + 1$ and dividing again we obtain $x+1$. So therefore $1$ is a root three times, while $-1$ is also three times. $\endgroup$ – Stefan4024 Oct 3 '16 at 8:09
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    $\begingroup$ @AyshaA. Note that the minimal polynomial of $\sqrt[6]{7}$ in $\mathbb{Q}(\xi)$ is $x^6-7$. And the wanted number is the degree of the minimal polynomial in the field. $\endgroup$ – Stefan4024 Oct 6 '16 at 10:50
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    $\begingroup$ @AyshaA. You can't do what you did in the last sentence. For example $\mathbb{Q}(\xi)$ isn't a subfield of $\mathbb{Q}(\sqrt[3]{7})$. $\endgroup$ – Stefan4024 Oct 6 '16 at 10:52

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