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Given $\alpha\in \Bbb C$ trascendental , and such that $|\alpha|=1$ (I don't know if this is necesary but I need only this case). Then I have to prove that the polynomial $x^3-\alpha \in \Bbb Q(\alpha)[x]$ is irreducible.

I want to prove this because this is a way to prove that the angle given by $\alpha$ cannot be trisected with rule and compass.

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Isn't this a duplicate of your earlier question, 241726? –  Gerry Myerson Nov 21 '12 at 4:52

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The roots of $\,x^3-\alpha\,$ are $\,\alpha^{1/3}\,,\,\alpha^{1/3}w\,,\,\alpha^{1/3}w^2\,\,\,,\,w:=e^{2\pi i/3}\,$ . If any of this is in $\,\Bbb Q(\alpha)\,$ then

$$\alpha^{1/3}w^k=\frac{p(\alpha)}{q(\alpha)}\in\Bbb Q(\alpha)\,\,,\,p(x),q(x)\in\Bbb Q[x]\,\,,\,0\leq k\leq 2\Longrightarrow$$

$$\Longrightarrow\alpha=\frac{p(\alpha)^3}{q(\alpha)^3}\Longrightarrow \alpha q(\alpha)^3-p(\alpha)^3=0$$

If we choose the fraction $\,p(\alpha)/q(\alpha)\in\Bbb Q(\alpha)\,$ reduced, the above gives us a contradiction (either $\,\alpha\,$ isn't transcendent or else the polynomial it satisfies in the last equality above is the zero polynomial...)

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Show that the polynomial is irreducible in $\mathbb{Q}[\alpha]$. It follows by Gauss lemma that it's irreducible in the fraction field. This is since $\mathbb{Q}[\alpha]\cong \mathbb{Q}[X]$, which is a UFD.

Your result then follows easily, since assuming your polynomial isn't irreducible, it has a root, so

$\sqrt[3]{\alpha} = a_1\alpha^{e_1}+\ldots + a_n\alpha^{e_n},\;\; a_i\in \mathbb{Q}$.

Now this gives an algebraic relation for $\alpha$ over $\mathbb{Q}$.

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If it's not irreducible, it has a root, so $$\root3\of\alpha={f(\alpha)\over g(\alpha)}$$ for some polynomials $f,g$. Cube, clear fractions to get an algebraic equation for $\alpha$, contradiction.

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