I was thinking of a way to map any polynomial $P$ with at least one real root onto some polynomial $Q$, s.t. the real roots of $P$ are exactly the real fixed points of $Q$, (There could be many, so we could choose some ordering to pick the least one).
My question is in three parts:
- Can any algebraic number be the fixed point of some polynomial?
- Does there exist a mapping of $P\to Q$, with the constraints as described above
- Can we go the other way? Map any polynomial $Q$ not equal to the identity polynomial with at least one fixed point to another polynomial $P$, where the fixed points of $Q$ are the roots of $P$?
The answer to 1 is yes, because every algebraic number $a$ is the root of a polynomial $p(x)$ by definition. To make $a$ a fixed point, we just need to take the polynomial $p(x) + a$.
The answer to #2 is yes, because for $p(x)$ with roots $a_i$, we can map to $p(x) + x$.
The answer to #3 is yes, because we cam map $q(x)$ to $q(x) -x$.