# Can Abel-Ruffini be "seen" by looking at the graphs of polynomials?

The Abel-Ruffini theorem states that

there is no algebraic solution – that is, solution in radicals – to the general polynomial equations of degree five or higher with arbitrary coefficients.

The standard proof of this uses Galois theory and is almost purely algebraic, uses abstract permutations, and thus is not very visual.

I wonder, if the fact that polynomials of degree five or higher have no general solution in radicals can somehow be seen by looking at the graphs of the polynomials:

Which visual properties do graphs of polynomials of degree five or higer have (if any) that are somehow "responsible" for the fact of not having radical solutions and that graphs of polynomials of degree less than five don't have?

Two sorts of graphs might be considered;

• the function graphs $$P:\mathbb{R}\rightarrow \mathbb{R}$$ sending $$x$$ to $$P(x)$$

$$P(x) = x^5 -4x^3 + 3x$$

$$P(x) = x^5 -2x^3 -x^2+ 3x$$

• I'd be very surprised if this question has a "yes" answer, and it would be hard to prove the answer is "no". Commented Oct 19, 2018 at 12:04

As far as the graphs of polynomial functions from $$\mathbb R$$ into $$\mathbb R$$ is concerned, the answer is negative. In fact, we can solve by radicals every polynomial equation of the typ $$x^3+x=a$$, where $$a$$ is an algebraic number, but we cannot solve by radicals most equations of the form $$x^5+x=a$$. However, their graphs are very similar visualy. Just translate up or down the graphs of $$x^3+x$$ and of $$x^5+x$$ respectively. And they look like this:

In the complex version, I don't know the answer, but I would be very much surprised if it turned out to be affirmative.

• Even if the answer is negative, I find it interesting: That there can be so profound and intricate differences between two seemingly very similar objects. Commented Oct 19, 2018 at 12:50
• Maybe you want to play around with $x^5 + x = a$ compared to $x^3 + x = a$. (You are right: there are no obvious differences between the two.) Commented Oct 19, 2018 at 13:10

Here is a partial "yes" answer:

If $$f(x)$$ is a polynomial of degree $$p$$ for an odd prime $$p$$, and if $$f(x)$$ has exactly two nonreal roots (which is to say it has $$p-2$$ real roots, which you can "see" on the graph), then the Galois group (of the splitting field of) $$f$$ over $$\mathbf{Q}$$ is isomorphic to the symmetric group $$S_p$$. In particular, if $$p > 3$$, then this group isn't solvable, so you can't solve your polynomial using radicals.

To see this, just note that:

1. Complex conjugation will be a transposition in the Galois group (as it transposes the two nonreal roots and fixes the rest.

2. The Galois group acts transitively on the roots. A transitive subgroup of $$S_n$$ has order divisible by $$n$$; in this case it has order divisible by $$p$$ and thus must contain a $$p$$-cycle.

3. $$S_p$$ is generated by any $$p$$-cycle and any transposition.