Help me motivate a topic.

Question

I'm teaching a topic for the first time, and I'm struggling to motivate it. I usually know where things lead and what they're eventually used for, but in this case I'm a bit stumped. (I suspect Galois Theory?!)

Say you have a cubic equation $ax^3+bx^2+cx+d=0$ with roots $\alpha,\beta$ and $\gamma$.

We can show that $\displaystyle{\alpha+\beta+\gamma = -\frac{b}{a}}$, $\displaystyle{\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}}$ and $\displaystyle{\alpha\beta\gamma=-\frac{d}{a}}$.

Given a cubic, say $2x^3-3x^2+4x-1=0$, we're then asked to find things like $\alpha^2+\beta^2+\gamma^2$ or $\displaystyle{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}$. This is all very straight forward, but what're these quantities good for?

Given a cubic, say $2x^3-3x^2+4x-1=0$, with roots $\alpha,\beta$ and $\gamma$, we're asked to find another cubic with roots $2\alpha-1,2\beta-1$ and $2\gamma-1$. Substituting $w=\frac{1}{2}(x+1)$ to get $w^3+5w+2=0$. Again, this is quite straightforward, but where is this leading to? What comes next?

I'm a pure mathematician, so I don't need an application in the real world. I'd just like to know its context in Mathematics and, more importantly, some references/links to the next step.

Answer 1

The overall motivation is to solve the general polynomial equation.

For the first part you were already given links to symmetric functions. Vieta's formulas give that the coefficients of a polynomial equation are equal to certain symmetric polynomials on the roots. It happens that all symmetric polynomials (or rational functions) on that number of variables can be written in term of those few special symmetric functions. Those exercises that you got are particular cases of the general result.

A useful consequence is that all those values can be computed without having to compute the roots (without extending the field of coefficients), therefore avoiding the difficult problem.

For the second part, you can look at Tschirnhaus transformations (simplifying the equation to try to solve it).

The transformation $x=w-\frac{a_{n-1}}{na_n}$ eliminates the term of degree $n-1$ in the polynomial $$a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$

This transformation plus taking one square root is enough to solve the general quadratic equation. So, all is being done is trying the same trick in higher degree to see if it helps.

Other transformations can allow you to eliminate other terms.

How much can the polynomial equation be simplified in this way? some people call it Klein's resolvent problem.

Answer 2

As others have noted, a big part of this is the aim to solve polynomial equations (implicitly, usually, by radicals, as opposed to using elliptic functions or modular functions). Also as noted, an interesting tangential point is that symmetric functions in the roots can be expressed in terms of the "standard" ones (and/or the sums of powers, by the Girard-Newton identities).

This does predate Galois theory by many decades, if not a century or two.

An often-neglected point is that use of Lagrange resolvents (a late 18th-century idea) leads one to discover the formulas of del Ferri, Ferraro, Tartaglia, and Cardano. Knowing how to manipulate symmetric polynomials is essential.

Similarly, to express roots of unity in terms of radicals, although by now general Galois theory proves that this is possible, the usual incarnations of it nowadays do not mention any tangible device to do it. I.e., Lagrange resolvents are not usually high-lighted. But, again, if one uses Lagrange resolvents to set things up, and knows how to employ symmetric polynomials, one can obtain the expressions in radicals.

(The question of why we might care about complicated expressions in terms of radicals is not quite answered by this, but one might claim that the details and ideas in the very process itself are of surprising interest.)