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Let $F$ be any field.

I have tried to find $[F(\alpha):F(\alpha^3)]$.

First, $x^3 - \alpha^3 $ is polynomial over $F(\alpha^3)$ having zero $\alpha$.

So $\mathrm{irr}(\alpha,F(\alpha^3)$ divides $x^3-\alpha^3$.

So $\deg(\alpha,F(\alpha^3))$ is 1,2, or 3.

Claim : $\deg(\alpha,F(\alpha^3)) \;|\; 3$, that is, $\deg(\alpha,F(\alpha^3)\neq 2$.

Could you help me?

user26857
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Seongqjini
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1 Answers1

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This is false. Let's denote $\beta=\root3\of2$, $F=\Bbb{Q}(\beta)$ and $\alpha=\omega\root3\of2$, where $\omega=e^{2\pi i/3}$. We then have $\alpha\notin F(\alpha^3)=F\subset\Bbb{R}$, and $\alpha^3=2\in F$. But $$\alpha^2+\alpha\beta+\beta^2=\frac{\alpha^3-\beta^3}{\alpha-\beta}=\frac{2-2}{\alpha-\beta}=0.$$ Therefore $\alpha$ is a zero of the polynomial $$ m(x)=x^2+\beta x+\beta^2\in F[x]\subseteq F(\alpha^3)[x]. $$ Hence $$ [F(\alpha):F(\alpha^3)]=2 $$ in this case.

Jyrki Lahtonen
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    Simpler example: $\omega^3 \in \mathbb Q$, but $\omega$ has degree $2$ over $\mathbb Q$. – Dustan Levenstein Aug 13 '16 at 14:21
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    LOL. A good one, @DustanLevenstein! My autopilot was dead set to use the fact that $\Bbb{Q}(\root3\of2)$ and $\Bbb{Q}(\omega\root3\of2)$ are not linearly disjoint over their intersection. – Jyrki Lahtonen Aug 13 '16 at 14:23