Roots for quintic equations

Question

I have been pondering over this question for a few months now. Why exactly do quintic equations have no closed general expression for their roots? Looking at graphs and reading about it hasn't really convinced me.

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http://en.wikipedia.org/wiki/Quintic_function

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Post scriptum: Pardon me if this question is too broad for the forum here but I do not know where else to expect a more convincing explanation from. I am putting it under the 'open questions' category for 'quintic enthusiasts' like myself. :) Thanks in advance.

Answer 1

The short answer, is why should they?

Radicals are solutions to special equations of the form $$ ?^n = x $$ which we commonly abbreviate as $\sqrt[n]{x}$, whereas you are looking at arbitrary equations of the form $$ \sum_{i=0}^n \lambda_n x^n = 0 \text{.} $$

Why would you even expect that you can find roots of arbitrary equations by repeatedly finding radics? Sure, it works for $n< 5$, but that only shows these equations are also somewhat special.

To actually prove that this doesn't work for $n \geq 5$, Galois theory basically looks at the numbers you get if you start out with some field, say the rational numbers $\mathbb{Q}$, and add to it a particular radical. If you pick $x^2 = 2$, you must thus add the "numbers" $\sqrt{2}$ and $-\sqrt{2}$, and the everything you can algebraically express, like $3 + \sqrt{2}$, $7\sqrt{2}$, $\frac{1}{\sqrt{2}}$ and so on. Then you pick a new radical, and do the same over again.

Finally, you compare that with the "numbers" you get if you add all the zeros of some polynomial $p = \lambda_0 + \lambda_1x + \ldots + \lambda_n x^n$ in one swoop. In algebraic terms, that means finding a field extensions where $p$ has $n$ zeros. And then you ask: Could I have gotten the same field by the process above, i.e. by repeatedly adding radical, i.e. zeros of $x^n = 0$? As it turns out, for $n \leq 4$ the answer is yes, but for $n \geq 5$, there are polynomials $p$ where the answer is no.

But if there was a formula using only $\cdot,/,+,-$ and radicals that expresses the roots of $p(x)$, then such roots would always have to be reachable by the iterative process above, since that why you do if you evaluate such a formula. Thus, start with $n=5$, there's no such general formula.

Answer 2

I am copying the answer I posted to the same question last year at Intuitive reasoning why are quintics unsolvable

Let the roots of an equation be A, B, C, etc. We are told that the unsolvability of the general quintic equation is related to the unsolvability of the associated Galois group, the symmetric group on five elements. I think I can tell you what this means on an intuitive level.

For three elements A, B, and C, you can create these two functions:

AAB + BBC + CCA ABB + BCC + CAA

These functions have the interesting property that no matter how you reshuffle the letters A, B and C, you get back the same functions you started with. You might reverse them (as you would if you just swap A and B) or they might both stay put (as they would if you rotate A to B to C) but either way you get them back.

For four elements, something similar happens with these three functions:

AB + CD AC + BD AD + BC

No matter how you reshuffle A, B, C and D, you get these three functions back. They might be re-arranged, or they might all stay put, but either way you get them back.

For five elements, there exists no such group of functions. Well, not exactly...there is a pair of huge functions consisting of sixty terms each that works, similar to the ones I drew out for the cubic equation...but that's it. There are no groups of functions with three or especially four elements, which is what you would actually want.

(There is also a set of six functions that map to each other under permutations, but these don't help you either. We had an intersting follow-up about Dummit's Resolvents in this discussion: Resolvent of the Quintic...Functions of the roots )

If you try to create functions on five letters with this symmetry property, you'll convince yourself that it's impossible. But how can you prove it's impossible? You probably need a little group theory for that.

EDIT: I'm going to expand on the answer I posted the other day, because I think I really have identified the "intuitive" reason the quintic is unsolvable, as opposed to the "rigorous" reason which involves a lot more group theory. For the third degree equation, I identified these functions:

AAB + BBC + CCA = p

ABB + BCC + CAA = q

A, B and C are the roots of a cubic, but p and q are the roots of a quadratic. You can see that because if you look at pq and (p+q), the elementary symmetric polynomials in p and q, you will see they are symmetric in A, B and C. So they are easily expressible in terms of the coefficients of our original cubic equation. And that's why p and q are the stepping stone which gets us to the roots of the cubic.

Similarly, for the fourth degree, we identified these functions:

AB + CD = p

AC + BD = q

AD + BC = r

You can rewrite the previous paragraph word for word but just take everything up a degree, and it remains true. A, B, C, and D are the roots of a quartic, but p,q and r are the roots of a cubic. You can see they must be because if you look at the elementary symmetric polynomials in p, q and r, you will see they are symmetric in A, B, C and D. So they are easily expressible in terms of the coefficients of our original quartic equation. And that's why they are the stepping stone which gets us to the roots of the quartic.

And the intuititve reason why the fifth degree equation is unsolvable is that there is no analagous set of four functions in A, B, C, D, and E which is preserved under permutations of those five letters. As I mentioned earlier, I think Lagrange understood this intuitively fifty years before Galois. You probably needed a little more group theory to make it completely rigorous, but that's another question.

I think Lagrange would have understood the algebraic tricks whereby you went from, say, A B and C to p and q. It involves taking linear functions which mix A B and C with the cube roots of unity and examining the cube of those functions. Its a reversible process, so you can work backward the other way (by taking cube roots of functions in p and q) to solve the cubic. A very similar trick works for the fourth degree. I think Lagrange was able to show conclusively that the same trick does not work for the fifth degree...that's the "intuitive" proof. The "rigorous" proof would have had to show that in the absence of the obvious tricks (analogous to the 3rd and 4th degree), there was no other possible tricks that you could come up with.

Answer 3

There are different reasons for insolvability of quinitics in radicals (or in elementary numbers): e.g. algebraic independence, symmetry restrictions, topological restrictions.

Chow [Chow 1999] gives his
Corollary 1:
"If Schanuel's conjecture is true, then the algebraic numbers in" the explicit elementary numbers "are precisely the roots of polynomial equations with integer coefficients that are solvable in radicals."
That means, the quintics that are not solvable in radicals cannot be solved by elementary numbers (means by applying elementary functions).

All solutions of polynomial equations can be represented with help of transcendental functions, e.g. theta functions, elliptic functions, Bring radicals, exponential/elliptic modular functions, Siegel modular forms, hyperelliptic integrals.

Here are general closed-form solution formulas for the quintics:
https://en.wikipedia.org/wiki/Bring_radical#Solution_of_the_general_quintic
How to solve fifth-degree equations by elliptic functions?
$\ $

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

Answer 4

Comment: Under certain conditons,

$$x^5-x=\rho\iff x=\sqrt[5]{\rho+\sqrt[5]{\rho+\cdots}}$$