If the magnitude of a root of an irreducible polynomial can be expressed via radicals, can the root?

Question

Let $f$ be an irreducible polynomial with coefficients in $\mathbb{Q}$, with roots $c_1,...,c_n \in \mathbb{C}$. I have two related but slightly different questions:

If $|c_1|$ lies in a radical extension of $\mathbb{Q}$, does that imply that $c_1$ lies in a radical extension of $\mathbb{Q}$?

If $|c_1|,...,|c_n|$ all lie in a radical extension of $\mathbb{Q}$, does that imply that $c_1,...,c_n$ all lie in a radical extension of $\mathbb{Q}$?

Obviously if the answer to the first question is yes, then so is the answer to the second question. However since I'm not sure of how to tackle either one, can someone with more expertise point me in the right direction?

Answer 1

Answer to Question 1 is NO. The main reason is that there exist algebraic numbers with modulus $1$, but whose minimal polynomial have non solvable Galois group.

For example:

The irreducible polynomial $X^{10}-X^6-X^5-X^4+1$ has a unique root $c$ such that $\vert c\vert=1$, so $\vert c\vert=1$ obviously belong to a radical extension of $\mathbb{Q}$. However, its Galois group is a non solvable group of order 1920 (according to LMFDB, for example), hence $c$ cannot belong to a radical extension of $\mathbb{Q}$.

Here is a way to construct many examples.Start with an algebraic real number $0<a<1$ such that the Galois group of its minimal polynomial is not solvable. Set $c=a+i\sqrt{1-a^2}$, so that $c$ is algebraic with modulus $1$. Then $c$ does not belong to a radical extension of $\mathbb{Q}$. Otherwise, it is not difficult to show that $\bar{c}$ would belong to a radical extension to (the image of a radical extension by complex conjugation should be radical, if i am not mistaken). Since the compositum of radical extensions is radical, $a=\dfrac{c+\bar{c}}{2}$ would belong to a radical extension, contradiction.

I don't know about Question 2.

Answer 2

Thank you very much to GreginGre for the help with my first question. Their answer was enough to help me to strengthen the argument to prove that my second question is also false, I'll provide my answer to that here:

Start with an irreducible polynomial with all real roots, with insoluble Galois group. For example:$p(x) := x^5-5x^3+4x+1$.

Now rescale this polynomial so that all of the roots lie within $(-1,1)$. For the example above this can be done with $q(x) := p(3x)$.

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Next we want to use the identity GreginGre uses, namely $y = x \pm i \sqrt{1-x^2}$. This rearranges to the following:

$$\frac{y + y^{-1}}{2} = x$$

Then the polynomial $r(y) := (2y)^{\deg(q)} q(\frac{y + y^{-1}}{2})$ satisfies $r(y) = 0 \iff y = x \pm i \sqrt{1-x^2}$ with $q(x) = 0$.

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For the running example we have:

$$r(y) = 243y^{10} + 675y^8 + 1002y^6 + 32y^5 + 1002y^4 + 675y^2 + 243$$

No roots of this polynomial lie in a radical extension of $\mathbb{Q}$, yet every root has absolute value one.