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To prove that the Vandermonde determinant is equal to the product above, we can apply row and column operations to compute the determinant:

det 1 α 1 α 1 2 α 1 n - 1 1 α 2 α 2 2 α 2 n - 1 1 α n α n 2 α n n - 1 = det 1 α 1 α 1 2 α 1 n - 1 0 α 2 - α 1 α 2 2 - α 1 2 α 2 n - 1 - α 1 n - 1 0 α n - α 1 α n 2 - α 1 2 α n n - 1 - α n n - 1 = det α 2 - α 1 α 2 2 - α 1 2 α 2 n - 1 - α 1 n - 1 α n - α 1 α n 2 - α 1 2 α n n - 1 - α n n - 1 = i = 2 n ( α i - α 1 ) det 1 α 2 + α 1 α 2 n - 2 + + α 1 n - 2 1 α n + α 1 α n n - 2 + α n n - 2 = i = 2 n ( α i - α 1 ) det 1 α 2 α 2 n - 2 1 α n α n n - 2

where in the last step we are subtracting from each column multiples of earlier columns. The desired formula for the Vandermonde determinant then follows by induction on n .

Exercises

  1. If f , g C [ x ] are non-constant polynomials over the complex numbers, show that f and g have a common root in C if and only if Res ( f , g , x ) = 0 . [Hint: Use the “fundamental theorem of algebra”: the fact that any non-constant polynomial in C [ x ] factors completely into linear factors.]
  2. If f ( x ) = a n x n + a n - 1 x n - 1 + + a 1 x + a 0 k [ x ] , where a n 0 and n > 0 , then the discriminant of f is defined to be
    disc ( f ) = ( - 1 ) n ( n - 1 ) / 2 a n Res ( f , f ' , x )
    Prove that f has a multiple factor (that is, f is divisible by g 2 for some non-constant g k [ x ] ) if and only if disc ( f ) = 0 .
    1. Compute the discriminant of the polynomial x 2 + b x + c for b , c k
    2. Compute the discriminant of the polynomial x 3 + p x + q for p , q k .
  3. An alternative definition of the discriminant of a polynomial f ( x ) = a n x n + a n - 1 x n - 1 + + a 1 x + a 0 C [ x ] of degree n is
    D ( f ) = a n 2 n - 2 i < j ( α i - α j ) 2
    where α 1 , ... , α n are the roots of f , counted with multiplicities, i.e. f ( x ) = a n i ( x - α i ) .
    1. Show that D ( f ) doesn't depend on the choice of ordering on the roots α i .
    2. Show that i < j ( α i - α j ) 2 = ( - 1 ) n ( n - 1 ) / 2 i j ( α i - α j ) .
  4. For f , g Z [ x ] , show that Res ( f , g , x ) Z .
  5. Let f , g k [ x ] have degrees n and m . Show that if the ( m + n ) × ( m + n ) Sylvester matrix has rank m + n - 1 , then f and g share a common linear factor, but not a common quadratic factor.

Quotient rings and the tjurina number

Starting with the integers Z , if we fix a positive integer n , we can construct the integers modulo n as follows: we let

Z / n = { 0 , 1 , 2 , ... , n - 1 }

be the set of possible remainders upon dividing an integer by n . To add or multiply x , y Z / n , we add or multiply them in Z and then take the remainder upon division by n , e.g. we would write x y = q n + r , and the product of x and y in Z / n would be r . Alternatively, one says that x and y are congruent modulo n and write x y ( mod n ) if y - x is divisible by n . One can check directly from the definition this relation satisfies the following properties (where denotes congruence modulo a fixed number n ):

x x for all x Z x y y x for all x , y Z x y and y z x z for all x , y , z Z x 1 x 2 and y 1 y 2 x 1 + y 1 x 2 + y 2 for all x 1 , y 1 , x 2 , y 2 Z x 1 x 2 and y 1 y 2 x 1 y 1 x 2 y 2 for all x 1 , y 1 , x 2 , y 2 Z
Properties 1-3 above show that congruence modulo n is an equivalence relation on Z , which thus partitions the set Z into equivalence classes
C x = { a Z : a x ( mod n ) } .
We can then define
Z / n = { C x : x Z }
to be the set of equivalence classes, and define addition and multiplication on Z / n by setting C x + C y = C x + y and C x · C y = C x y ; this is well-defined by properties 4 and 5 above. This is equivalent to the definition in terms of remainders: Z / n has n elements C 0 , C 1 , ... , C n - 1 corresponding to the n possible remainders upon division by n and the addition and multiplication operations correspond to those defined above. This definition has one advantage over the definition in terms of remainders: it generalizes (with no need for Gröbner bases) to the case where we replace Z with Q [ x 1 , ... , x k ] (or in fact any ring) and replace n with any ideal I . We define f g ( mod I ) to mean that g - f I . Then we can again show that properties 1-5 above hold for this relation, and we set
Q [ x 1 , ... , x k ] / I = { C f : f Q [ x 1 , ... , x k ] }
with addition and multiplication defined in the same way.

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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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