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Math 254A. the Discriminant of Xn + Ax + B

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Math 254A. the Discriminant of Xn + Ax + B Math 254A. The Discriminant of xn + ax + b The discriminant of a polynomial f of degree n is a homogeneous polynomial of degree n(n − 1) in the roots of f . By Lang's Algebra, Ch. IV Thm. 6.1, this discriminant is also a polynomial in the coefficients a , b of f . These coefficients are homogeneous polynomials of degrees n−1 and n in the roots of f , so the discriminant is a weighted homogeneous polynomial φ of degree n(n − 1) in a and b , where a has weight n − 1 and b has weight n . Therefore any monomial aibj occurring in φ must have (n − 1)i + nj = n(n − 1) . There are only two possibilities with i ≥ 0 and j ≥ 0 , namely (i; j) = (0; n − 1) and (i; j) = (n; 0) , so the discriminant is of the form D = αan + βbn−1 : (*) We first restrict to working over C . We start by finding the coefficient β . This can be found by computing the discrim- inant of the polynomial xn − 1 , which has roots 1; ζ; : : : ; ζn−1 , where ζ is a primitive nth root of unity. We have Y i j 2 Dxn−1 = (ζ − ζ ) 0≤i<j<n n−1 Y Y = (−1)n(n−1)=2 (ζi − ζj) i=0 0≤j<n j6=i n−1 n−1 Y Y = (−1)n(n−1)=2 (ζi − ζi+`) i=0 `=1 n−1 ! n−1 !n Y Y = (−1)n(n−1)=2 ζi(n−1) (1 − ζ`) : i=0 `=1 2 The first product in this expression is ζn(n−1) =2 . If n is odd, then (n − 1)2=2 2 Z , and the product is 1 ; if n is even, then this is an odd power of ζn=2 = −1 . Thus, the first product is (−1)n−1 . The second product equals n , because it is the value of the polynomial xn − 1 = xn−1 + ··· + 1 ; x − 1 evaluated at x = 1 . Thus n(n−1)=2 n−1 n Dxn−1 = (−1) (−1) n ; and since this equals (−1)n−1β , we have β = (−1)n(n−1)=2nn : 2 Next we find the coefficient α , this time using the polynomial xn − x . This polynomial has roots 0 , ζ0 , ζ1 , ::: , ζn−2 , where ζ is now a primitive (n − 1)st root of unity. The discriminant of this polynomial is therefore n−2 Y i 2 Y i j 2 Dxn−x = (ζ − 0) · (ζ − ζ ) i=0 0≤i<n−1 0≤j<n−1 i<j = ζ(n−1)(n−2) · (discriminant of xn−1 − 1) = (−1)(n−1)(n−2)=2(−1)n−2(n − 1)n−1 = −(−1)n(n−1)=2(n − 1)n−1 : This equals (−1)nα , so α = (−1)n(n−1)=2(1 − n)n−1 : By (*), we then have D = (−1)n(n−1)=2(1 − n)n−1an + nnbn−1 : (**) The above computation determined the discriminant for polynomials of the above form over C . But the discriminant of a polynomial of the form xn + ax + b over an arbitrary commutative ring A is of the form φ(a; b) for some polynomial φ 2 A[t; u]; moreover, by functoriality, if α: A ! B is a homomorphism of commutative rings, then the polynomials φA(t; u) and φB(t; u) giving discriminants of polynomials of the α given form with coefficients in A and B , respectively, are related by φB = φA . Thus, (**) gives the discriminant for polynomials xn + ax + b in Z[x] and (via the unique map Z ! A ) it also gives the discriminant for arbitrary commutative rings A ..
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