Discriminant - Discriminant of A Polynomial

Discriminant of A Polynomial

To find the formula for the discriminant of a polynomial in terms of its coefficients, it is easiest to introduce the resultant. Just as the discriminant of a single polynomial is the product of the squares of the difference between the distinct roots of a polynomial, the resultant of two polynomials is the product of the differences between their roots, and just as the discriminant vanishes if and only if the polynomial has a repeated root, the resultant vanishes if and only if the two polynomials share a root.

Since a polynomial has a repeated root if and only if it shares a root with its derivative the discriminant and the resultant both have the property that they vanish if and only if p has a repeated root, and they have almost the same degree (the degree of the resultant is one greater than the degree of the discriminant) and thus are equal up to a factor of degree one.

The benefit of the resultant is that it can be computed as a determinant, namely as the determinant of the Sylvester matrix, a (2n − 1)×(2n − 1) matrix, whose n − 1 first rows contain the coefficients of p and the n last ones the coefficients of its derivative.

The resultant of the general polynomial

is, up to a factor, equal to the determinant of the (2n − 1)×(2n − 1) Sylvester matrix:

$\left[\begin{matrix} & a_n & a_{n-1} & a_{n-2} & \ldots & a_1 & a_0 & 0 \ldots & \ldots & 0 \\ & 0 & a_n & a_{n-1} & a_{n-2} & \ldots & a_1 & a_0 & 0 \ldots & 0 \\ & \vdots\ &&&&&&&&\vdots\\ & 0 & \ldots\ & 0 & a_n & a_{n-1} & a_{n-2} & \ldots & a_1 & a_0 \\ & na_n & (n-1)a_{n-1} & (n-2)a_{n-2} & \ldots\ & 1a_1 & 0 & \ldots &\ldots & 0 \\ & 0 & na_n & (n-1)a_{n-1} & (n-2)a_{n-2} & \ldots\ & 1a_1 & 0 & \ldots & 0 \\ & \vdots\ &&&&&&&&\vdots\\ & 0 & 0 & \ldots & 0 & na_n & (n-1)a_{n-1} & (n-2)a_{n-2}& \ldots\ & 1a_1 \\ \end{matrix}\right].$

The discriminant of is now given by the formula

For example, in the case n = 4, the above determinant is

$\begin{vmatrix} & a_4 & a_3 & a_2 & a_1 & a_0 & 0 & 0 \\ & 0 & a_4 & a_3 & a_2 & a_1 & a_0 & 0 \\ & 0 & 0 & a_4 & a_3 & a_2 & a_1 & a_0 \\ & 4a_4 & 3a_3 & 2a_2 & 1a_1 & 0 & 0 & 0 \\ & 0 & 4a_4 & 3a_3 & 2a_2 & 1a_1 & 0 & 0 \\ & 0 & 0 & 4a_4 & 3a_3 & 2a_2 & 1a_1& 0 \\ & 0 & 0 & 0 & 4a_4 & 3a_3 & 2a_2 & 1a_1 \\ \end{vmatrix}.$

The discriminant of the degree 4 polynomial is then obtained from this determinant upon dividing by .

In terms of the roots, the discriminant is equal to

where r₁, ..., r_n are the complex roots (counting multiplicity) of the polynomial p(x):

$\begin{matrix}p(x)&=&a_n x^n+a_{n-1}x^{n-1}+\ldots+a_1 x+a_0\\ &=&a_n(x-r_1)(x-r_2)\ldots (x-r_n).\end{matrix}$

This second expression makes it clear that p has a multiple root if and only if the discriminant is zero. (This multiple root can be complex.)

The discriminant can be defined for polynomials over arbitrary fields, in exactly the same fashion as above. The product formula involving the roots r_i remains valid; the roots have to be taken in some splitting field of the polynomial. The discriminant can even be defined for polynomials over any commutative ring. However, if the ring is not an integral domain, above division of the resultant by should be replaced by substituting by 1 in the first column of the matrix.

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