MA522, Fall 2021, HOMEWORK 4: (due Thur Oct 21, 11:55pm, by email or upload on Moodle; you can send/upload cell phone photos of your handwritten solution). Problem 1 (the discriminant of a polynomial): Consider a polynomial f(x) = (x - y_1) ... (x - y_n) = x^n + a_{n-1} x^{n-1} + ... + a_0, where a_i \in K[y_1,...,y_n] with K a field. Note that the a_i are plus/minus the symmetric functions in y_i. The discriminant of f is defined as disc_x(f) = \prod_{1 <= i < j <= n} ( (y_i - y_j) (y_j - y_i) ) i) Show that disc_x(f) is a symmetric polynomial in the y_i, that is, there is a polynomial \delta(z_{n-1},...,z_0) in K[z_{n-1},...,z_0] such that \delta(a_{n-1},...,a_0) = disc_x(f). Note that if coefficient values in the field K are substituted in \delta for the variables, \delta evaluates to 0 precisely when the corresponding polynomial has a multiple root. ii) Show that \delta is irreducible in K[z_{n-1},...,z_0] for fields of characteristic not equal 2, and a square of an irreducible polynomial for fields of characteristic equal 2. iii) Consider the (2n-1) by (2n-1) Sylvester-like matrix M formed with the coefficient vector of f and f', the derivative of f w.r.t. x: f'(x) = n x^{n-1} + (n-1) a_{n-1} x^{n-2} + ... + a_1 Note that if the characteristic of the field K divides n, the first coefficient, n, is equal 0, hence the matrix is not a true Sylvester matrix. Prove that det(M) = disc_x(f). Note that the identity is one in K[y_1,...,y_n]. Hint: prove that f'(x) = sum_i (x-y_1)...(x-y_{i-1})(x-y_{i+1})...(x-y_n). Please note that this sum_i identity is with respect to the above definition of f' and unrelated to any properties from analysis. Bonus Problem (higher discriminant): The third discriminant of f is defined as disc3_x(f) = \prod_{1 <= i < j < k < \ell <= n} ( ( y_i + y_j - y_k - y_\ell) ( y_i - y_j + y_k - y_\ell) ( y_i - y_j - y_k + y_\ell) ); there are 3*(n choose 4) linear factors. Note: the disc3(f) is zero if the average of two roots is equal the average of another two distinct roots. i) Show that disc3_x(f) is a symmetric polynomial in the y_i, that is, there is a polynomial \delta3(z_{n-1},...,z_0) in K[z_{n-1},...,z_0] such that \delta3(a_{n-1},...,a_0) = disc3_x(f). ii) Show that \delta3 is irreducible in K[z_{n-1},...,z_0] for fields of characteristic not equal 2. Note: a formula in terms of resultants and squareroots of polynomials is in https://users.cs.duke.edu/~elk27/bibliography/21/Ka21.pdf ======================================================================