1.1 A brief introduction to computational algebraic geometry  (Page 18/20)

Exercises

1. Compute the resultant of $f\left(x\right)=x^5-3x^4-2x^3+3x^2+7x+6$ and $g\left(x\right)=x^4+x^2+1$ . Do these polynomials have a common factor in $\mathbf{Q}\left[x\right]$ ?
2. Let $f,g\in \mathbf{C}\left[x\right]$ be polynomials of degree 3. Verify directly in this case that the following are equivalent:
   1. The polynomials $f$ and $g$ have a non-constant common factor.
   2. $\text{Res}(f,g,x)=0$ .
3. Resultants give us another method of finding an implicit polynomial equation for a parametric rational curve, in the following way: given a rational curve defined parametrically by
   $$(x,y)=\left(\frac{a\left(t\right)}{p\left(t\right)},,,\frac{b\left(t\right)}{q\left(t\right)}\right),$$
   we can find an implicit equation by calculating the resultant $\text{Res}(f,g,t)$ of the polynomials $f=a\left(t\right)-xp\left(t\right)$ and $g=b\left(t\right)-yq\left(t\right)$ , where we are regarding $f$ and $g$ as polynomials in $t$ with coefficients in $\mathbf{C}[x,y]$ .
   1. Show that $\text{Res}(f,g,t)$ , as a polynomial in $x$ and $y$ , vanishes on the parametric curve.
   2. Use this method to find an implicit equation for the following curves:
      1. $x=t^2$ ,    $y=t^2(t+1)$ .
      2. $x=\frac{t-1}{t^2}$ ,    $y=t-1$ .
      3. $x=\frac{t(t^2+1)}{t^4+1}$ ,    $y=\frac{t(t^2-1)}{t^4+1}$ .
   3. Use the Gröbner basis method to find implicit equations for the above parametric curves and check that they define the same curves.
4. Consider the following polynomials in $k[x,y]$ :
   $$\begin{array}{cc}\hfill f& =x^{2}y-3x{y}^2+x^2-3xy\hfill \\ \hfill g& =x^{3}y+x^3-4y^2-3y+1\hfill \end{array}$$
   1. Compute $\text{Res}(f,g,x)$ .
   2. Compute $\text{Res}(f,g,y)$ .
   3. What do your answers from (a) and (b) tell you about $f$ and $g$ ?

Resultants and symmetric polynomials

Last time we claimed (without proof) that the resultant of two monic polynomials could be defined in terms of the roots. Of course, we can do the same with polynomials which aren't monic: if $f\left(t\right)=a_n{t}^n+\cdots +a_0$ and $g\left(t\right)=b_m{t}^m+\cdots +b_0$ are polynomials with complex coefficients, then we may write

where the ${\alpha }_i$ are the roots of $f$ and the ${\beta }_j$ are the roots of $g$ , counted with multiplicities. In the Sylvester matrix, replacing $f$ and $g$ by the monic polynomials with the same roots corresponds to dividing the first $m$ columns by $a_n$ and dividing the first $n$ columns by $b_m$ . Thus the general formula for the resultant in terms of the roots and $a_n$ and $b_m$ should be

Now, consider the case of the discriminant, i.e. take $g=f^{\text{'}}$ . We have

by the product rule. Thus we have

and so the alternate definition of the discriminant

which explains the apparently unnecessary $\frac{{(-1)}^{n(n-1)/2}}{a_n}$ in defining the discriminant.

Of course, the discriminant ${\prod }_{i<j}{({\alpha }_i-{\alpha }_j)}^2$ of a monic polynomial $f$ with roots ${\alpha }_i$ , is the square of another polynomial in the roots. It turns out one of the two square roots is the Vandermonde determinant

so that the discriminant is the square of the Vandermonde determinant. Note that the Vandermonde determinant is not itself a symmetric polynomial in the ${\alpha }_i$ : interchanging ${\alpha }_i$ with ${\alpha }_j$ interchanges two rows of the matrix, multiplying the determinant by $-1$ . This means that it can't possibly be a polynomial in the coefficients of $f$ .