Is the discriminant of a polynomial always a polynomial in the coefficients? What does this have to do with symmetric polynomials?
Solution
Yes. For a fixed degree, the discriminant is a polynomial with integer coefficients in the coefficients of the polynomial.
1. First suppose\[ f(x)=x^n+c_1x^{n-1}+\cdots+c_n \]is monic, with roots \(r_1,\ldots,r_n\) in a splitting field. Its discriminant is\[ \Delta(f)=\prod_{i<j}(r_i-r_j)^2. \]
This is the usual root definition of the discriminant of a monic polynomial.
2. The polynomial\[ D(r_1,\ldots,r_n)=\prod_{i<j}(r_i-r_j)^2 \]is symmetric in the roots.
Permuting the roots permutes the unordered pairs \(\{i,j\}\) and therefore leaves the product unchanged. Equivalently, the unsquared Vandermonde changes by the sign of the permutation, so its square is invariant.
3. Hence \(D\) is a polynomial in the elementary symmetric polynomials\[ e_1(r),\ldots,e_n(r). \]
By the fundamental theorem of symmetric polynomials, every symmetric polynomial with integer coefficients is a polynomial with integer coefficients in the elementary symmetric polynomials. Thus there exists \[ P\in\ZZ[y_1,\ldots,y_n] \] such that \[ D=P(e_1,\ldots,e_n). \]
4. Therefore the discriminant is a polynomial in the coefficients \(c_i\).
Viète’s formulas give \[ e_i(r_1,\ldots,r_n)=(-1)^i c_i. \] Substituting these into 3 yields \[ \Delta(f)=P(-c_1,c_2,\ldots,(-1)^n c_n), \] a polynomial in the coefficients.
5. For a general degree-\(n\) polynomial\[ f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0, \qquad a_n\ne0, \]the usual discriminant\[ \Delta(f)=a_n^{2n-2}\prod_{i<j}(r_i-r_j)^2 \]is likewise a polynomial with integer coefficients in \(a_0,\ldots,a_n\).
One equivalent formula is \[ \Delta(f)=(-1)^{n(n-1)/2}a_n^{-1}\operatorname{Res}(f,f'). \] The resultant is the determinant of the Sylvester matrix, hence a polynomial with integer coefficients in the coefficients of \(f\) and \(f'\). For the universal degree-\(n\) polynomial, that resultant is divisible by the leading coefficient \(a_n\), and the quotient is precisely the displayed discriminant. Equivalently, homogenizing the monic symmetric-polynomial expression in 4 by the factor \(a_n^{2n-2}\) clears all denominators. Thus \(\Delta(f)\in\ZZ[a_0,\ldots,a_n]\).
6. The unsquared Vandermonde product \(V=\prod_{i<j}(r_i-r_j)\), which is alternating rather than symmetric, need not be a polynomial in the coefficients.
For \(f(x)=x^2-2\) with roots \(r_1=\sqrt2\), \(r_2=-\sqrt2\), one has \(V=2\sqrt2\notin\QQ\), while \(\Delta(f)=V^2=8\).