$p \in Q[x]$ has as a root a fifth primitive root of unity, then every fifth primitive root of unity is a root of $p$.

I'm extremely stuck. Can't figure it.

The conjugate is easy: let $w$ be a primitive root of unity, then $w^{-1}$ will also be a root, that's easy. But I'm missing $w^2$ and $w^3$. Why would they be also roots of p?

• What do you know about polynomials? – Igor Rivin Feb 26 '16 at 22:26
• I know that if x is root of P n times, then it will be root of P when you derive it up to P times, but not p+1 times I know that if P | Q, then P|Q share at least one root. – joaquinlpereyra Feb 26 '16 at 22:31
• How do you know that $1/w$ is a root? – Igor Rivin Feb 26 '16 at 22:32

Since $\;p(x)\in\Bbb Q[x]\;$ and it vanishes on $\;\zeta:=e^{2\pi i/5}\;$ , the minimal polynomial of $\;\zeta\;$ over the rationals (also known as a cyclotomic polynomial) also divides $\;p(x)\;$ , and thus

$$\Phi_5(x)\,|\,p(x)\implies p(x)=\overbrace{(x^4+x^3+x^2+x+1)}^{=\Phi_5(x)}\cdot h(x)\;,\;\;h(x)\in\Bbb Q[x]$$

Yet

$$\;x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}\implies$$

the roots of $\;\Phi_5(x)\;$ are precisely the four primitive roots of unity of order five, and thus these roots are also roots of $\;p(x)\;$

• @G. Sassatelli Thank you very much for your edit. It is clearer now. – DonAntonio Feb 26 '16 at 22:56
• You're welcome. I thought of a typo. – user228113 Feb 26 '16 at 22:58
• Thanks for this! My question is how would you go about this without using the knowledge you have about cyclotomic polynomial. This wasn't covered in my algebra course :( – joaquinlpereyra Feb 27 '16 at 20:13

Because $x^{5} - 1 = (x - 1) (x^{4} + x^{3} + x^{2} + x + 1)$, $w$ will be a root of f = $x^{4} + x^{3} + x^{2} + x + 1$. I think there is no alternative to proving that $f$ is irreducible in $\mathbb{Q}[x]$, which can be done with Eisenstein's criterion.

• Let $x=y+1.$ Then $f=y^4+5 y^3+10 y^2+10 y+5$ which meets Eisenstein's Criterion. – DanielWainfleet Feb 26 '16 at 23:25
• Exactly the point I had quoted from the Wikipedia. – Andreas Caranti Feb 26 '16 at 23:26

Let $R$ be the GCD of $P$ and $X^4+X^3+X^2+X+1$ (considering both polynomials as elements of $\mathbb Q[X]$.

It happens (see here for proof) that $R$ is also the GCD of $P$ and $X^4+X^3+X^2+X+1$ as elements of $\mathbb C[X]$. Since $(X-w)(X-\bar{w})$ divides both, $R$ has degree at least 2.

But $X^4+X^3+X^2+X+1$ is irreducible over $Q$ (Eisenstein can help you with that). Hence $R=X^4+X^3+X^2+X+1$.

Hence $X^4+X^3+X^2+X+1$ divides $P$ and the claim is proved.

You know $p$ factors as $(x-\omega)(x-\bar{\omega}) q(x)$, right? So the question is really "Can $q$ be just the polynomial $1$?" Well, if it were, would the coefficients of $x$ and $1$ be rationals? If yes, then the claim is false. :) So you have to show that they cannot be.

So now $q$ has to be at least linear ... and you can show that there's no (rational) linear thing you can multiply by those first two factors to make the coeffs all rational.

So $q$ must contain a quadratic with a pair of conjugate roots, say $\alpha$ and $\bar{\alpha}$, i.e. $$q(x) = (x-\alpha)(x-\bar{\alpha})r(x)$$ for some rational polynomial $r$.

Can you show that the product $(x-\omega)(x-\bar{\omega})$ will have rational coeffs only if $\alpha = \omega^2$ or $\omega^3$ (assuming that $\alpha$ and its conjugates are NOT roots of $r$)?

• In last paragraph, “the product” means $(x-\omega)(x-\bar\omega)q(x)$, yes? – Lubin Feb 26 '16 at 22:41
• Clarified (I hope). – John Hughes Feb 26 '16 at 23:26

The fifth roots of $1$ form a cyclic group of order $5$, so any of them, different from $1$, is primitive. If $w$ is one of the primitive roots, the others are $\omega^2$, $\omega^3=\omega^{-2}$ and $\omega^4=\omega^{-1}=\bar{\omega}$.

It is clear that $(x-\omega)(x-\omega^2)(x-\omega^3)(x-\omega^4)=x^4+x^3+x^2+x+1$. The minimal polynomial of $\omega$ is a factor of this degree $4$ polynomial, so it must have degree $2$ or $4$ (because a degree $3$ polynomial has a real root).

Thus we have to exclude that $\omega$ has degree $2$. If we assume this, then its minimal polynomial must be $(x-\omega)(x-\bar\omega)=x^2-(\omega+\bar\omega)x+1$, which implies $q=\omega+\bar\omega$ is rational.

Therefore also $(x-\omega^2)(x-\omega^3)$ would have rational coefficients and, indeed, $$\omega^2+\omega^3=\omega^2+\frac{1}{\omega^2}= \left(\omega+\frac{1}{\omega}\right)^2-2=q^2-2$$ Thus $$x^4+x^3+x^2+x+1=(x^2-qx+1)(x^2-(q^2-2)x+1)$$ The coefficient of the term of degree $1$ in the product can be easily computed as $$-q-(q^2-2)$$ so we have $-q^2-q+2=1$, so $q^2+q-1=0$. This is a contradiction, because $\sqrt{5}$ is not rational.

Therefore this polynomial divides $p$, as soon as $p(\omega)=0$.