Is there a proof that this polynomial solvable by radicals?

Question

If $f(x)$ is the minimal polynomial of $t$, a constructible number, over $\mathbb{Q}$, then is $f(x)$ solvable by radicals?

It seems to be true, at least with the examples I came up with. Can it be proved though?

Answer 1

Actually, the algebraic characterization of a constructible number $t$ is that there exists a sequence of field extensions $F_0=\mathbb Q, F_1,\ldots, F_n$ such that $ [F_i:F_{i-1}]=2$ and $t\in F_n$. This can be easily proved starting from the geometric definition of constructibility, considering that the elementary ruler and compass operations correspond, from the point of view of the analytical geometry, to solving quadratic equations. The fact that $ [F_i:F_{i-1}]=2$ implies that any $x\in F_{i} $ can be expressed in terms of square roots of elements in $F_{i-i}$. In particular $t$ is a combination of radicals of radicals ... of the rational numbers.