$F$ is a field and $\alpha$ is any element. Then $[F(\alpha) :F(\alpha^3)]=1 \mbox{ or } 3$.

Question

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?

Answer 1

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.