# Primitive $6^{th}$ root of unity

Why does any extension of $\mathbb{Q}$ that contains a root of $x^6+3$ have to contain a primitive $6^{th}$ root of unity? The roots of this equation are $\{\sqrt[6]{3}e^{\pi i/2},\sqrt[6]{3}e^{5\pi i/6},\sqrt[6]{3}e^{7\pi i/6},\sqrt[6]{3}e^{3\pi i/2},\sqrt[6]{3}e^{11\pi i/6},e^{13\pi i/6}\}$. A primitive $6^{th}$ root of unity has the form $\frac{1}{2}+\frac{\sqrt{3}}{2}i$...

I don't see how to get $\frac{1}{2}+\frac{\sqrt{3}}{2}i$ from $\sqrt[6]{3}e^{\pi i/2}$, for example.

Let $\alpha$ be such a root, then $\beta=\alpha^3$ satisfies $\beta^2+3=0$, hence ${1+\beta\over 2}={1+\alpha^3\over 2}$ is a $6^{th}$ root of $1$.

You can even verify this directly:

$$(1+\beta)^6 = 1+6\beta+15\beta^2+20\beta^3+15\beta^4+6\beta^5+\beta^6$$

Now use that $\beta^2=-3$ to reduce this to:

$$(1+5\cdot(-3)+15\cdot(-3)^2+(-3)^3)+\beta(6+20\cdot(-3)+6\cdot(-3)^2)$$ $$=(1-45+135-27)+\beta(6-60+54)= 64+0\beta$$ $$= 2^6.$$

And this is primitive as if we only cube things we get

$$(1+\beta)^3 = 1+3\beta + 3\beta^2+\beta^3 = 1-9 + \beta(3-3) = -8=-2^3.$$

• $(1+\beta)^3=-8$ is easier and proves primitivity.
– lhf
Dec 2, 2016 at 17:15
• @lhf true, but if I'm already verifying, I find it's satisfying to see it all the way when it's not extremely cumbersome. Thanks for the feedback! Dec 2, 2016 at 17:19
• @lhf ah, good point. That is an excellent reason to change it. Much appreciated! Dec 2, 2016 at 17:21
• I don't understand. Why do we automatically have $\frac{1+\beta}{2}$ a $6^{th}$ root of $1$. And why does cubing show it is primitive?
– FTem
Dec 2, 2016 at 17:24
• @FTem I verified it above, so there's your proof it's a $6^{th}$ root. But it only cubes to $-1$ so the smallest power we can raise it to and get $1$ is 6, that is definition primitive. Dec 2, 2016 at 17:25