Why do you need to introduce complex numbers in order to solve cubics by radicals?

Solution 1:

For degree greater than $2$, the answer is generally "no".

Recall that a Galois extension can be obtained by extracting roots iff its Galois group is solvable. As a corollary, every Galois subextension of a Galois extension obtained by adjoining roots is also obtained by adjoining roots, since any quotient of a solvable group is solvable.

Suppose $F$ is the splitting field of a real polynomial and $F\subseteq K_n$. Then $F$ is a Galois subextension of the Galois closure $K_n(\zeta_m)$ of $K_n$, where $m=\mathrm{lcm}(m_1,\ldots,m_{n-1})$. Thus $F$ is obtained by adjoining roots. But $F$ is real and Galois, so by your observation it must be obtained by adjoining square roots! It follows that $F$ has degree a power of $2$, so $\mathrm{Gal}(F/K)$ is a $2$-group. In particular, this rules out generic polynomials of degree greater than $2$ and irreducible polynomials of degree not a power of $2$.