How to prove that $\frac{1}{2} \left( 1 - \sqrt{3} \pm i \sqrt{2} \sqrt[4]{3} \right)$ is not a root of unity?

Question

I have a proof for the complex roots $\sqrt{2} - 1 \pm i \sqrt{2\sqrt{2}-2}$ which involve a contradiction using the minimal polynomial of the real part of the root, but the proof does not work on the requested number, as the minimal polynomial of $\frac{1 - \sqrt{3}}{2}$ is $t^2 - t - \frac{1}{2}$. (not integral, so a lemma I used does not work).

The lemma:

If $\alpha$ is a root of unity whose real part is an algebraic integer, then $\alpha^4 = 1$.

Perhaps such a proof is impossible?

If this unfortunately turns out to be the case, the most elementary proof possible will be accepted.

Answer 1

Let $\alpha:=\frac12(1-\sqrt{3}+i\sqrt{2}\sqrt[4]3)$, then $|\alpha|=1$, whence $\overline{\alpha}=\frac1\alpha$. Therefore, $$1-\sqrt{3}=2\operatorname{Re}(\alpha)=\alpha+\overline{\alpha}=\alpha+\frac1\alpha.$$ We conclude that $$3\alpha^2=\left(\alpha^2-\alpha+1\right)^2=\alpha^4+\alpha^2+1+2(-\alpha^3+\alpha^2-\alpha),$$ which may be simplified to $$\alpha^4-2\alpha^3-2\alpha+1=0.$$ Now, we use @JyrkiLahtonen's remark that the above polynomial is irreducible because $\alpha+\overline{\alpha}\not\in\mathbb{Q}$, and it has two real zeroes, at approximately $0.435$ and $2.297$. Those are clearly not roots of unity, and if $\alpha$ were a root of unity, all its conjugates would be too.

Alternatively, once you know the polynomial is (monic) irreducible, it can only have a root of unity as one of its roots if it is cyclotomic. Take a quick look at the first $20$ or so cyclotomic polynomials, and you'll see that's not the case.

Answer 2

I am going to "piggyback" on Mastrem's answer. The object is to show explicitly that the minimal polynomial derived there has roots not on the unit circle.

Quote the minimal polynomial equation:

$\alpha^4-2\alpha^3-2\alpha+1=0$

Now we have a little fun with this. Suppose $\alpha$ is on the unit circle. If we render this as the ratio $(1+z)/(1-z)$ then $|1+z|=|1-z|$ and thus $z$ is pure imaginary. Let us make this substitution and determine whether all roots for $z$ meet this requirement.

$(1+z)^4-2(1+z)^3(1-z)-2(1+z)(1-z)^3+(1-z)^4=0$

$2(3z^4+6z^2-1)=0$

We have good news and bad news. The good news is that both roots for $z^2$ are real. The bad news is that they are opposite in sign. The negative root for $z^2$, properly corresponding to pure imaginary $z$, comes from both complex conjugates for $\alpha$ that lie on the unit circle. But... the positive root for $z^2$ represents nonzero real roots for $z$, and these correspond to real reciprocal roots for $\alpha$ that do not lie on the unit circle.

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Upon further review, there actually is an approach to make the proof with the OP's quoted lemma. Simply square the given value of $\alpha$. Using standard complex-number multiplication gives

$\alpha^2=\color{blue}{(1-\sqrt3)}+i[(1-\sqrt3)\sqrt2\sqrt[4]3/2]$

where now, the real part is an algebraic integer but the fourth power is not $1$ (the real part does not match with any fourth root of unity). So $\alpha^2$ is not a root of unity and $\alpha$ can't be one either.