# Precise relationship between “finite” Fourier analysis and Galois theory in solving the cubic?

Suppose we want to solve$$x^3 - ax^2 + bx - c = 0.$$We know a priori that this can be factored as $(x - r_0)(x - r_1)(x - r_2)$; by Vieta's formulas, we know$$a = r_0 + r_1 + r_2,\quad b = r_0r_1 + r_1 r_2 + r_2r_0,\quad c = r_0r_1r_2.$$These expressions are invariant under three-cycles. Now, we make the substitution$$r_0 = u_0 + u_1 + u_2,\quad r_1 = u_0 + u_1\omega + u_2\omega^2,\quad r_2 = u_0 + u_1\omega^2 + u_2\omega^4,$$where $\omega$ is a primitive cube root of unity. Explicitly, we have$$u_0 = {1\over3}(r_0 + r_1 + r_2),\quad u_1 = {1\over3}(r_0 + r_1\omega^{-1} + r_2 \omega^{-2}),\quad u_2 = {1\over3}(r_0 + r_1 \omega^{-2} + r_2\omega^{-4}).$$Conceptually,$$\text{if }F(z) = u_0 + u_1z + u_2z^2,\text{ then }r_i = F(\omega^i).$$The first of Vieta's relations now reads$$a = F(1) + F(\omega) F(\omega^2) = 3u_0,$$which is a roots of unity filter on $F$. The second reads$$b = F(1)F(\omega) + F(\omega)F(\omega^2) + F(\omega^2)F(1),$$which is a roots of unity filter on$$F(z)F(\omega z) = (u_0 + u_1 z + u_2z^2)(u_0 + u_1z\omega + u_2z^2\omega^2).$$Since in a filter, we only care abou tthe cubic terms – here $u_0^2$ and $-u_1u_2z^3$ – we find that $b = 3u_0^2 - 3u_1u_2$. If we repeat the same thing for $c$ and compile everything together, we arrive at$$a = 3u_0,\quad b = 3u_0^2 - 3u_1u_2,\quad c = u_0^3 + u_1^3 + u_2^3 - 3u_0u_1u_2.$$The first equation gives us $u_0 = a/3$ for free, and we can then obtain $u_1u_2$ in terms of $a$, $b$, $c$. Finally, the last equation tells us what $u_1^3 + u_2^3$ are. So, we can compute the values of $u_1^3 + u_2^3$, $u_1^3 \cdot u_2^3$; this reduces to a quadratic, which we can solve.

My question is, what is the precise underlying interplay going on between the Galois theory and the "finite" Fourier analysis here, if there is any? Am I shooting in the dark for something that may possibly not exist? Or is there something deep behind all this?

The idea is that you can extract the explicit Kummer extension directly from the Fourier transform of the roots. If you have an element of the Galois group that cyclically permutes the roots, by weighting with appropriate roots of unity, i.e., taking a Fourier transform, the automorphism is diagonalized and acts on each $u_i$ via multiplication by a root of unity. Now to determine the roots, you just have to find $u_i^3$.
Usually, one first simplifies by shifting all of the roots by $-a/3$, i.e. letting $x_{new} = x - a/3$, thereby setting $a_{new}=u_{0,new}=0$ (I'll omit the "new" from now on). Then $b = u_1 u_2$ and $c = u_1^3 + u_2^3$. This makes the necessary steps easier to see.