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Determine the minimal polynomial of $e^{2\pi i/5}$ in $\mathbb{Q}$

Obviously $x^5$ is one of the polynomials such that $p(e^{2\pi i /5})=0$. However I am wondering if there is any polynomials with degree less than $5$.

If we think $e^{2\pi i /5}$ as a rotation counterclockwise $2\pi/5$, then $x^2$ means rotating $4\pi/5$ and $x^3$ means rotating $6\pi/5$. Once we have $6\pi/5$, we may can have $q'(x)=x^3+ax^2+bx$ so that $q'(e^{2\pi/5})$ be a rotation of $\pi$ or $0$. Then $q'(x)=x^3+ax^2+bx+c$ satisfies $q(e^{2\pi/5})=0$.

I guess the minimal polynomial should have degree $3$. Is it right?

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    https://en.wikipedia.org/wiki/Cyclotomic_polynomial – 2017-02-04

2 Answers 2

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No, 3 is not the answer.

Actually $x^5$ does not annihilate $e^{2\pi i/5}$, since $(e^{2\pi i/5})^5=e^{2\pi i}=1$. Try instead $x^5-1$. Can this polynomial be factored?

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    The degree will be 4? – 2017-02-04
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    $x^4+x^3+x^2+x+1$? – 2017-02-04
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    @Kenneth.K Yes, that is correct. – 2017-02-04
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$P(x)=x^4+ x^3+x^2+x+1$ is such that $P(\zeta_5)=0$ and is irreducible over $\Bbb{Q}$. Two ways to prove it:

Brute force: if it were reducible let's look at its possible factors; no rational root (the only possible ones are $\pm 1$) so there is no first degree. The only possible factorisation is therefore

$$P(x)=(x^2+ax+b)(x^2+cx+d)$$

We have $b=d=\pm 1$. Assume it is $+1$. We have $a+c=1$, $ac=-1$ and therefore $a,c=(1\pm\sqrt{5})/2$.

If $b=d=-1$ we still have $a+c=1$ however $ac=3$ and therefore $a,c=(1\pm\sqrt{13})/2$.

Eisenstein applied to $Q(x)=P(x+1)=((x+1)^5-1)/x$