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I'm asked to find the degree of the splitting field of $x^3-7$ over the rationals.

The roots are $\sqrt[3] 7e^{\frac{2\pi ik}{3}},\ k=0,1,2$. Explicitly, $$x_1=\sqrt[3] 7,\\ x_2=\sqrt[3] 7 \bigg(-\frac{1}{2}+i\frac{\sqrt 3}{2}\bigg),\\ x_3=\sqrt[3] 7 \bigg(-\frac{1}{2}-i\frac{\sqrt 3}{2}\bigg).$$

The splitting field is an extension that contains all roots. I see at least two essentially different possibilities of building it. First, I could adjoin $\sqrt[3] 7, i,\sqrt 3$. Such an extension will contain all roots. On the other hand, I could adjoin $\sqrt[3] 7, i\sqrt 3$. This extension also contains all the roots. I believe the two extensions have different degrees. So how should I understand which degree I should find to begin with? Which extension is the splitting field?

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    $\begingroup$ The splitting field should be the one with lowest degree of extension, shouldn't it? Which field of the two has a smaller degree? $\endgroup$ – Batominovski Jul 16 '18 at 18:39
  • $\begingroup$ @Batominovski For example in the Wikipedia definition, I don't see the requirement on the degree. $\endgroup$ – user437309 Jul 16 '18 at 18:41
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    $\begingroup$ Yes, the degrees of the two extension are different. So one of them is not the splitting field. You can think about it as follows. The number $x_2/x_1=(-1+i\sqrt3)/2$ is certainly in the splitting field. Because $\Bbb{Q}$ is a subset of the splitting field you can then deduce that $i\sqrt{3}$ must be in. But, it stops there, right? $\endgroup$ – Jyrki Lahtonen Jul 16 '18 at 18:42
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    $\begingroup$ On the wiki page, that is said in literally the first sentence of the page. "In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial splits or decomposes into linear factors." $\endgroup$ – Batominovski Jul 16 '18 at 18:42
  • $\begingroup$ @Batominovski I looked in the "Definition" section. Is it implied by some condition in the formal definition stated on wiki? $\endgroup$ – user437309 Jul 16 '18 at 18:43
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The splitting field $K$ of $x^3-7$ is $\mathbb Q(\sqrt[3]7, \omega)$, where $\omega=-\frac{1}{2}+i\frac{\sqrt 3}{2}$ is a primitive cubic root of the unity, a root of $x^2+x+1$. There is no need to decompose $\omega$ into $i$ and $\sqrt 3$.

It is true that $L =\mathbb Q(\sqrt[3]7, i, \sqrt 3)$ contains all the roots of $x^3-7$, but $L$ is not the smallest such field. That is $K$.

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You have $x^3-7$ and $\alpha=\sqrt [ 3 ]{ 7 }$ is a clear root of $x^3-7$. Then after factoring and applying the quadratic formula $($if needed$)$ one factors $x^3-7=(x-\alpha)(x-\alpha\zeta )(x-\alpha\zeta ^2)$ where $\zeta $ is a complex cube root of unity. $\zeta ^2+\zeta +1=0$ and $\zeta \notin\mathbb{R}$ hence $\notin\mathbb{Q}(\alpha)$, so splitting the field has the degree $3\cdot2=6$. In fact the splitting field is $\mathbb{Q}(\alpha,\zeta)$.

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