§ Discriminant and Resultant

I had always seen the definition of a discriminant of a polynomial p(x)p(x) as:
Disc(p(x))≡an(2n−n)∏i<j(ri−rj)2 Disc(p(x)) \equiv a_n^{(2n - n)} \prod_{i< j} (r_i - r_j)^2
While it is clear why this tracks if a polynomial has repeated roots or not, I could never motivate to myself or remember this definition. I learnt that in fact, this comes from a more general object, the resultant of two polynomials P(x),Q(x)P(x), Q(x), which provides a new polynomial Res(P(x),Q(x)Res(P(x), Q(x) which is zero iff P,QP, Q share a common root. Then, the discriminant is defined as the resultant of a polynomial and its derivative. This makes far more sense:
  • If a polynomial has a repeated root rr, then its factorization will be of the form p(x)=(x−r)2q(x)p(x) = (x - r)^2 q(x). The derivative of the polynomial will have an (x−r)(x-r) term that can be factored out.
  • On the contrary, if a polynomial only has a root of degree 1, then the factorization will be p(x)=(x−r)q(x)p(x) = (x - r) q(x), where q(x)q(x) is not divisible by (x−r)(x-r). Then, the derivative will be p′(x)=1⋅q(x)+(x−r)q′(x)p'(x) = 1 \cdot q(x) + (x - r) q'(x). We cannot take (x−r)(x - r) common from this, since q(x)q(x) is not divisible by (x−r)(x-r).
This cleared up a lot of the mystery for me.

§ How did I run into this? Elimination theory.

I was trying to learn how elimination theory works: Given a variety V={(x,y):Z(x,y)=0}V = \{ (x, y) : Z(x, y) = 0 \}, how does one find a rational parametrization (p(t),q(t))(p(t), q(t)) such that Z(p(t),q(t))=0Z(p(t), q(t)) = 0, and p(t),q(t)p(t), q(t) are rational functions? That is, how do we find a rational parametrization of the locus of a polynomial Z(x,y)Z(x, y)? The answer is: use resultants!
  • We have two univariate polynomials p(a;x),p(b;x)p(a; x), p(b; x), where the notation p(a;x)p(a; x) means that we have a polynomial p(a;x)≡∑ia[i]xip(a; x) \equiv \sum_i a[i] x^i. The resultant isa polynomial Res(a;b)Res(a; b) which is equal to 00 when p(a;x)p(a; x) and p(b;x)p(b; x) share a common root.
  • We can use this to eliminate variables. We can treat a bivariate polynomial p(x,y)p(x, y)as a univariate polynomial p′(y)p'(y) over the ring R[X]R[X]. This way, given two bivariate polynomial p(a;x,y)p(a; x, y), q(b;x,y)q(b; x, y), we can compute their resultant, giving us conditions to detect for which values of a,b,xa, b, x, there exists a common yy such that p(a;x,y)p(a; x, y) and (q,x,y)(q, x, y) share a root. If (a,b)(a, b)are constants, then we get a polynomial Res(x)Res(x) that tracks whether p(a;x,y)p(a; x, y)and q(a;x,y)q(a; x, y) share a root.
  • We can treat the implicit equation above as two equations, x−p(t)=0x - p(t) = 0, y−q(t)=0y - q(t) = 0. We can apply the method of resultants to project out ttfrom the equations.

§ 5 minute intro to elimination theory.

Recall that when we have a linear system Ax=0Ax = 0, the system has a non-trivial solution iff ∣A∣=0|A| = 0. Formally: x≠0  ⟺  ∣A∣=0x \neq 0 \iff |A| = 0. Also, the ratio of solutions is given by:
xi/xj=(−1)i+j∣Ai∣/∣Aj∣x_i / x_j = (-1)^{i+j} |A_i|/|A_j|
If we have two polynomials p(a;x)=a0+a1x+a2x2p(a; x) = a_0 + a_1 x + a_2 x^2, and q(b;x)=b0+b1x+b2x2q(b; x) = b_0 + b_1x + b_2 x^2, then the system p(a;x)p(a; x), q(b;x)q(b; x) has a simeltaneous zero iff:
[a2a1a000a2a1a0b2b1b000b2b1b0][1xx2x3]=0Ax=0 \begin{aligned} &\begin{bmatrix} a_2 & a_1 & a_0 & 0 \\ 0 & a_2 & a_1 & a_0 \\ b_2 & b_1 & b_0 & 0\\ 0 & b_2 & b_1 & b_0\\ \end{bmatrix} \begin{bmatrix} 1 \\ x \\ x^2 \\ x^3 \end{bmatrix} = 0 \\ &A x = 0 \end{aligned}

§ Big idea

The matrix is setup in such a way that any solution vector vv such that Qv=0Qv = 0 will be of the form v=(α3,α2,α,1)v = (\alpha^3, \alpha^2, \alpha, 1). That is, the solution vector is a polynomial , such that Qv=0Qv = 0. Since Qv=0Qv = 0, we have that a2α2+a1α+a0=0a_2 \alpha^2 + a_1 \alpha + a_0 = 0, and b2α2+b1α+b0=0b_2 \alpha^2 + b_1 \alpha + b_0 = 0.

§ Proof

  • Necessity is clear. If we have some non trivial vector v≠0v \neq 0 such that Qv=0Qv = 0, then we need ∣Q∣=0|Q| = 0.
  • Sufficiency : Since ∣Q∣=0|Q| = 0, there is some vector v=(w,x,y,z)v = (w, x, y, z)such that Qv=0Qv = 0. We need to show that this vv is non-trivial. If the polynomials p(a;x)p(a;x), q(b;x)q(b;x) are not equal, then we have that the rows which have coefficients from pp and qq are linearly independent. So, the pair of rows (1,3)(1, 3), and the pair (2,4)(2, 4) are linearly independent. This means that the linear system:
a2w+a1x+a0y=0b2w+a1x+a0y=0 a_2 w + a_1 x + a_0 y = 0 \\ b_2 w + a_1 x + a_0 y = 0 \\
Similarly:
a2x+a1y+a0z=0b2x+a1y+a0z=0 a_2 x + a_1 y + a_0 z = 0 \\ b_2 x + a_1 y + a_0 z = 0 \\
Since the coefficients of the two systems are the same, we must have that (w,x,y)(w, x, y) and (x,y,z)(x, y, z) are linearly dependent. That is:
(w,x,y)=α(x,y,z)w=αx=α2y=α3z (w, x, y) = \alpha (x, y, z) \\ w = \alpha x = \alpha^2 y = \alpha^3 z \\
We can take z=1z = 1 arbitrarily, giving us a vector of the form (w,x,y,z)=(α3,α2,α,1)(w, x, y, z) = (\alpha^3, \alpha^2, \alpha, 1), which is the structure of the solution we are looking for!

§ References