Let's consider $A\subset \mathbb{C}$ which posses the following nice property: If a given irreducible polynomial $f\in\mathbb{Q}$ has one root in $A$, then all of its roots must lie in $A$.
Examples
Given an irreducible polynomial $f\in\mathbb{Q}[X]$, if $f$ has one root in $\mathbb{Q}$, then $f$ has al of its roots in $\mathbb{Q}$, since it has to be of the form $f=X-a$ for some $a\in\mathbb{Q}$.
The same goes (trivially) for $\emptyset$. If $f$ has one root in $\emptyset$, all if its roots are in $\emptyset$. The other trivial example is $\mathbb{C}$.
Now the somewhat more complicated example which prompted me to ask this question. Let $$A_q=\left\{2\cos\left(\frac{2\pi p}{q}\right):p\in\mathbb{Z}\right\}.$$ If an irreducible $f\in\mathbb{Q}$ has one root in $A_q$, then all of its roots are in $A_q$. A proof of this can be found below.
Questions
- Do you have more examples of such $A$?
- Is my proof correct?
- Is there anything to say in general about these $A$?
- Do they have any use?
Proof of earlier statement
I'll now prove my claim about
$$$A_q=\left\{2\cos\left(\frac{2\pi p}{q}\right):p\in\mathbb{Z}\right\}.$$
Let $p/q\in\mathbb{Q}$ be in its lowest terms and define:
$$\alpha:=2\cos\left(\frac{2\pi p}{q}\right)\in A_q$$
If $f$ is an irreducible polynomial in $\mathbb{Q}[X]$ with $f(\alpha)=0$, then $f$ is some rational number times $f^{\alpha}_{\mathbb{Q}}$ and thus has the same roots as $f^\alpha_\mathbb{Q}$. We conclude it's enough to prove that all of $f^\alpha_\mathbb{Q}$'s roots lie in $A_q$. If $\alpha=0$, then $f^\alpha_{\mathbb{Q}}=X$ and we are done. For the rest of this proof, assume $\alpha\neq 0$.
Define the polynomial sequence $(g_{n})_{n=0}^{\infty}\subset\mathbb{Q}[X]$ by: $$g_0=2,g_1=X$$ $$g_{n+1}=Xg_n-g_{n-1}$$ and the sequence $(a_n)_{n=0}^{\infty}$ by: $$a_0=2,a_1=\alpha$$ $$a_{n+1}=\alpha a_n-a_{n-1}$$ So that $a_n=g_n(\alpha)$. We have $$a_1=\alpha=\exp\left(\frac{2i\pi p}{q}\right)+\exp\left(-\frac{2i\pi p}{q}\right)$$ and by induction: $$a_n=\exp\left(\frac{2i\pi pn}{q}\right)+\exp\left(-\frac{2i\pi pn}{q}\right)$$ Which is clearly periodic with period $q$. In other words: For all non-negative integers $k$ and $m$, we have: $$a_k=a_{k+qm}\iff g_k(\alpha)=g_{k+qm}(\alpha)\implies (g_{k+qm}-g_k)(\alpha)=0$$ Which means that $(g_n)_{n=1}^{\infty}$ is periodic modulo $f^\alpha_{\mathbb{Q}}$ with period $q$.
Now, let $\beta\in\mathbb{C}$ be some other root of $f^\alpha_\mathbb{Q}$. We can take $\beta\neq 0$. Define the sequence $(b_n)_{n=1}^{\infty}$ by: $$b_0=2,b_1=\beta$$ $$b_{n+1}=\beta b_n-b_{n-1}$$ So that $b_n=g_n(\beta)$. Since $(g_{n})_{n=1}^{\infty}$ is periodic modulo $f^\alpha_{\mathbb{Q}}$ with period $q$ and $f^\alpha_\mathbb{Q}(\beta)=0$, we have that $(b_n)_{n=1}^{\infty}$ is periodic with a period dividing $q$.
Since $\beta\neq 0$, we can write $\beta=\gamma+\gamma^{-1}$ for some $\gamma\in\mathbb{C}$ and show by induction that: $$b_n=\gamma^n+\gamma^{-n}.$$ Suppose that $|\gamma|\neq 1$. If $|\gamma|<1$, switch $\gamma$ and $\gamma^{-1}$, so that you can be sure that $|\gamma|>1$. Now: $$2=\lim_{k\to\infty}|b_{kd}|=\lim_{k\to\infty}|\gamma^{kd}+\gamma^{-kd}|=\infty$$ which is a contradiction. Therefore, $|\beta|=1$ and $\beta=e^{ic}$ for some $c\in\mathbb{R}$ and: $$b_n=e^{nic}+e^{-nic}=2\cos(nc)$$ Since $(b_n)_{n=1}^{\infty}$ is periodic with period $d$, we know $c$ is a rational number and $c=r/q$ for some $r\in\mathbb{Z}$, so $\beta\in A_q$.