Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

3 Answers

up vote 1 down vote accepted

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...)

share|improve this answer
add comment

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 $\mathbb{Q}$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.