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

Historically, complex numbers were first introduced (by Bombelli) in order to provide a method for solving the cubic by radicals. As far as I can tell, previous solutions of the cubic by radicals didn't use complex numbers, but instead used some incredibly complicated reasoning that is basically equivalent to introducing complex numbers.

Later on, it turned out that cubics over the complex numbers had particularly nice properties; in particular, they always had three roots in the complex plane (up to multiplicity).

But the original guys weren't interested in cubics over the complex plane. They wanted to find real roots of cubics with real coefficients, and intermediate complex numbers turned out to be a way to do that. Indeed, the cubic formula uses complex numbers like $-\frac12+\frac{\sqrt3}2i$, but it still gives you three real numbers if you substitute in the coefficients of a real cubic with three real roots.

So my question is: why is this necessary? To be more specific:

Let $K\subset\mathbb R$ be a field, and let $P$ be a polynomial in $K[X]$. Do there exist fields $K=K_1\subset\dots\subset K_n\subset\mathbb R$, real numbers $\beta_i$ and natural numbers $m_i$ such that $\beta_i^{m_i}\in K_i$, $K_{i+1}=K_i(\beta_i)$ for each $i$, and all real roots of $P$ are contained in $K_n$?

[Note that the extensions $K_{i+1}/K_i$ are not necessarily Galois extensions (indeed, they are not Galois extensions if they have degree greater than $2$).]

If $P$ is a quadratic rather than a cubic, then the answer is 'yes'. I imagine that the answer for a cubic or quartic is probably 'no' (and it is definitely 'no' for polynomials of higher degree). But why?

Before anyone says it: yes, I know you can solve cubics over the real numbers by using trigonometric or hyperbolic substitutions. I'm only interested in solution by radicals.

• Yhere is a sketch of a proof in the Wikipedia article. I have seen a proof that the casus irreducibilis cannot be solved using real radicals in more than one Galois Theory book. Memory fails about explicit reference. – André Nicolas Sep 18 '13 at 18:04
• @AndréNicolas One such book is Dummit and Foote's algebra textbook, as I recall. – Ryan Reich Dec 25 '13 at 8:02

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