Solvability of quintics with complex coefficients?

Question

I was trying to explain Galois theory to a non-specialist and was given a question I couldn't solve to my satisfaction, apologies if I have missed something obvious.

By Galois theory, a polynomial $f(x) \in K[x]$ is solvable by radicals (which I take to mean: there is a closed-form expression only involving addition, subtraction, multiplication, division, and taking $n$th roots) if and only if the Galois group of $F/K$ (where $F$ is the splitting field of $f(x)$, some subfield of $\mathbb{C}$) is solvable.

I'm a little vague on the details of how $f(x)$ being solvable by radicals corresponds to the existence of a general equation, but my understanding is that because a polynomial of degree 4 or less always corresponds to a Galois group that is solvable, any such polynomial is solvable. This is why we can find a general formula for the roots. On the other hand, the splitting field of a quintic can sometimes have Galois group $S_5$, which is not solvable, so in general quintic polynomials can't be solved by radicals. Therefore there cannot be a general formula.

The question then, is what happens if $K = \mathbb{C}$? Then any polynomial $f(x) \in K[x]$ already splits, because $\mathbb{C}$ is algebraically closed, and the Galois group is trivial and solvable. Or, if $K = \mathbb{R}$ then either $f(x)$ splits, or has complex roots in which case the relevant extension is $\mathbb{C}/\mathbb{R}$ with Galois group $C_2$, also solvable. In either case, it seems that Galois theory should imply that $f(x)$ is solvable by radicals. This must be wrong, because otherwise we could just embed $\mathbb{Q}[x]$ into $\mathbb{C}[x]$ and the quintic with rational coefficients would be solvable.

Putting aside rational polynomials, if $f(x) \in \mathbb{C}[x]$ with complex coefficients, is this now solvable by radicals?

Answer 1

A polynomial $f\in F[x]$ being solvable by radicals over $F$ means that there is some radical extension of $F$ where $f$ splits. Note, this is dependent on the field $F$. For example, for $a,b,c\in\mathbb{Q}$, the equation $ax^2+bx+c=0$ is solvable by radicals over $\mathbb{Q}$, because its roots are: (assume $a\ne 0$)

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Which are elements in a radical extension of $\mathbb{Q}$. Note that this element contains only scalars from the field (like $2,4$) and roots of some order.

Now, not every quintic polynomial in $\mathbb{Q}[x]$ is solvable by radicals over $\mathbb{Q}$. For example, $f(x)=x^5-6x+3$ is not solvable. You can take any expression as above, which contains rational scalars and roots of arbitrary order, and you will never get a root of this $f$. On the other hand, if you allow the expression to contain complex scalars then trivially you can get the roots of $f$, just take its roots (which are valid expressions if you allow complex scalars) and you are done. So yes, every polynomial is solvable by radicals over $\mathbb{C}$. But not necessary over $\mathbb{Q}$.

So this implies that there is no general formula over $\mathbb{Q}$, i.e a formula which contains the coefficients of the polynomial and rational scalars. Now of course we can ask the question - is there a general formula over $\mathbb{C}$, i.e such that contains complex scalars. (that is, it might contain scalars like $i$ or $2-\pi i$, unlike the quadratic formula which only contains rational scalars like $2$ and $4$) We know every specific polynomial is solvable over $\mathbb{C}$, but is there a general formula? The answer is still no. If you are familiar with transcendental extensions, you can read my answer here:

implication of the Abel–Ruffini theorem

Answer 2

In either case, it seems that Galois theory should imply that $f(x)$ is solvable by radicals. This must be wrong...

What Galois theory tells you is that an individual $f \in K[X]$ is solvable by radicals if and only if its Galois group is solvable. (I'll work in characteristic zero for simplicity.)

By definition, an individual $f \in K[X]$ is solvable by radicals over $K$ iff its splitting field $F$ is contained inside an extension of the form $K(\alpha_1, \dots, \alpha_n)$, where each $\alpha_i$ is an $k$th root of some element of $K(\alpha_1, \dots, \alpha_{i-1})$.

Now take $f$ to be some quintic polynomial in $\mathbb R[X]$, and for the sake of illustration, let's suppose that $f$ has some non-real roots. Then the splitting field of $f$ is $\mathbb C$. The field $\mathbb C$ is an extension of $\mathbb R$ of the form $\mathbb C = \mathbb R(i)$, and $i$ is the square root of the element $-1 \in \mathbb R$. So according to our definition above, $f$ is solvable by radicals over $\mathbb R$. As you observed, the Galois group in this example is the solvable group $\mathbb Z_2$, and there is no contradiction.

...because a polynomial of degree 4 or less always corresponds to a Galois group that is solvable, any such polynomial is solvable. This is why we can find a general formula for the roots.

All I've said is that each individual quintic $f \in K[X]$ is solvable by radicals over $K$, when $K = \mathbb R$ or $\mathbb C$, according to the definition of what it means for an individual polynomial to be solvable by radicals. This observation has no bearing on whether there is a general formula that expresses the roots of a general quintic in terms of applying arithmetic operations and $k$th root operations to the coefficients of the quintic together with and to constant numbers in $K$. I must admit that before I saw Mark's answer, I had no idea whether such a formula exists.

Note that when $K = \mathbb Q$, we can find a quintic $f \in K[X]$ whose Galois group is not solvable. This $f$ is therefore not solvable by radicals. From this, we validly can infer that there cannot exist a general formula for the roots of general rational quintics.

But this logic doesn't work the other way round. When $K = \mathbb R$ or $\mathbb C$, each polynomial $f \in K[X]$ is solvable by radicals, but this does not imply that there exists a general formula for the roots.