Given any algebraic number $a\in \overline{\mathbb{Q}}$, let $f_a(x)$ denote its minimal polynomial. Let $R_a=\{r\in \mathbb{C}|f_a(r)=0\}$. The set in question is $$C:=\{a\in \overline{\mathbb{Q}}: |a|\le \min_{r\in R_a}\{|r|\}\}. $$ Is the set $C$ closed under multiplication?
This is a question concerning my research which is a little outside my area. Does anyone know if the following problem is already solved? My intuition(which has been wrong plenty of times before) tells me the answer is yes. Any insights/answers given will be greatly appreciated.
To see that $C$ is closed under multiplication, it is helpful to know that $R_a$ is exactly the orbit of $a$ under the Galois group $G=Gal(\overline{\mathbb{Q}}/\mathbb{Q})$. So, suppose $a,b\in C$ and let $c\in R_{ab}$. Then $c=g(ab)=g(a)g(b)$ for some $g\in G$. Since $a,b\in C$, we have $|g(a)|\geq|a|$ and $|g(b)|\geq |b|$ and so $|c|\geq |a||b|=|ab|$.