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I am trying to find the degree of the splitting field for $ x^3-5 $ over $\mathbb{Q}$. I have so far:

The splitting field will be $\mathbb{Q}(\sqrt[3]{5},u)$ where u is the 3rd root of unity. So I'm looking for the degree of $\mathbb{Q}(\sqrt[3]{5},u):\mathbb{Q}$.

$ x^3-5 $ is irreducible by Eisenstein's criterion so the degree of $\mathbb{Q}(\sqrt[3]{5}):\mathbb{Q}$ = deg($ x^3-5 $) = 3

I think the next step would be to find the degree of $\mathbb{Q}(\sqrt[3]{5},u):\mathbb{Q}(\sqrt[3]{5})$ and apply the tower law but I'm not sure if this is correct or how to do this.

Also would the degree of the splitting field over $\mathbb{Q}$ be the same as the degree of Gal($\mathbb{Q}(\sqrt[3]{5},u):\mathbb{Q}$)

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  • $\begingroup$ Computation! Computation! You should check that $-\frac12+\frac{\sqrt{-3}}2$ is a cube root of unity. $\endgroup$
    – Lubin
    Jan 3, 2016 at 5:58

2 Answers 2

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You know the total extension is either $3$ or $6$ because the degree of the splitting field of $p(x)$ must divide $\deg(p(x))!$. You have a degree $3$ sub extension, so this is just asking if the extension by the third root of five is the whole field. Obviously it's not, because that's a real field and the splitting field is not a real field.

To see intuitively why it's $2$ and not $3$, remember that if $u$ satisfies $x^3=1$, that polynomial is divisible by $x-1$ so it also satisfies $x^2-x+1$. This is an irreducible polynomial that generates the extension.

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  • $\begingroup$ Thanks! I'm still a bit confused though. By inverting the Tower law using that the degree I'm looking for is 6 would that not give that the degree of $\mathbb{Q}(\sqrt[3]{5},u):\mathbb{Q}(\sqrt[3]{5})$ is 2? If my understanding of degree is correct it is the number of items in the basis needed to form the extension which I would have thought would be u, u^2 and 1 so 3? $\endgroup$ Jan 2, 2016 at 17:26
  • $\begingroup$ @user2973447 I added an explonation $\endgroup$ Jan 2, 2016 at 17:32
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Let $\zeta$ be a third root of unity and $K = \mathbf{Q} ( \sqrt[3]{5}), \zeta)$. $K$ contains the cyclotomic field of $3$rd roots of unity and is generated over it by $\sqrt[3]{5}$. Thus the degree is at most 6. $\mathbf{Q} (\sqrt[3]{5})$ and $\mathbf{Q}(\zeta)$ are both subfields of degrees 3 and 2 respectively. Since the degrees are relatively prime, the degree must be divisible by $3*2 = 6$ and so $[K: \mathbf{Q}] = 6$.

An even more direct route is to just use the fact that it is a composite field of relatively prime degrees.

Your final query regarding the Galois group is a bit confusing to me. Gal represents a Galois group, so it doesn't make sense to talk about its degree. I am guessing you mean to say order of the group then. In that case, this is always true. By definition, $K/F$ is Galois extension if $|\textrm{Aut} (K/F) | = [K:F]$ and we denote $\textrm{Aut} (K/F)$ as $\textrm{Gal}(K/F)$

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  • $\begingroup$ Thanks! Can you explain why the degree of $\mathbf{Q}(\zeta)$ is 2 though please? I would have thought it would be 3 as $(\zeta)^3 = 1$ $\endgroup$ Jan 2, 2016 at 17:28
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    $\begingroup$ @user2973447 A primitive cubic root $\zeta$ of $1$ satisfies $\zeta^2+\zeta+1=0$. In general, the degree of a (primitive) $p$-th root of $1$ is $p-1$. $\endgroup$
    – egreg
    Jan 2, 2016 at 17:33

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