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Let $E$ be an field extension of $F$ and $\alpha \in E$. Determine $[F(\alpha) : F(\alpha^3)]$.

I'm unsure how to approach this problem because I thought I would try to test some examples and see what I get. I first tried with $\alpha = \sqrt{2}$ so $\mathbb{Q(\sqrt{2}})$ and $\mathbb{Q(2\sqrt{2}})$ in which case $$[\mathbb{Q(\sqrt{2}}) : \mathbb{Q(2\sqrt{2}})]=1$$ And so I thought the answer would be $1$ however, another example using $\alpha = 2^{1/3}$ we get $\mathbb{Q(2^{1/3})}$ and $\mathbb{Q(2)}$ in which case $$[\mathbb{Q(2^{1/3})}:\mathbb{Q(2)}]=3$$ And so now i'm unsure what the answer is, any help is appreciated thanks.

TAPLON
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1 Answers1

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Hint $\alpha$ is a root of $X^3-\alpha^3$, therefore its minimal polynomial is a divisor of this.

N. S.
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  • Since $(X- \alpha)(X^2 + \alpha X + \alpha^2)$. Do I need to show that the second factor is irreducible then conclude that the first one is the minimal polynomial and so the degree is one? – TAPLON Apr 24 '18 at 03:23
  • @JustinStevenson Your polynomial needs to have coefficients in $F(\alpha^3)$, $x-\alpha$ may not. – N. S. Apr 24 '18 at 03:33
  • So then is $X^3-\alpha^3$ the minimal polynomial then since the two factors both don't have coefficients in $F(\alpha^3)$? – TAPLON Apr 24 '18 at 03:38
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    @JustinStevenson How do you know that the coefficients are not in $F(\alpha^2)$? Maybe they are. Or maybe there are not. Which means that there are few distinct possibilities. – N. S. Apr 24 '18 at 03:58
  • @JustinStevenson Since the minimal polynomial is a divisor of the above, its degree is at most 3. You already have an example of degree 1 and one example of degree 3. All you have left is to figure is 2 is possible or not. – N. S. Apr 24 '18 at 04:00
  • Extra hint: Think cyclotomic fields! – C Monsour Apr 24 '18 at 04:13
  • Okay so using the last hint, using the extension field of the 3rd root of unity we get an extension of degree 2 and so all of 1,2 and 3 are possible. – TAPLON Apr 24 '18 at 04:19