Exercises
- Compute the resultant of
and
. Do these polynomials have a common factor in
?
- Let
be polynomials of degree 3. Verify directly in this case that the following are equivalent:
- The polynomials
and
have a non-constant common factor.
-
.
- 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
we can find an implicit equation by calculating the resultant
of the polynomials
and
, where we are regarding
and
as polynomials in
with coefficients in
.
- Show that
, as a polynomial in
and
, vanishes on the parametric curve.
- Use this method to find an implicit equation for the following curves:
-
,
.
-
,
.
-
,
.
- Use the Gröbner basis method to find implicit equations for the above parametric curves and check that they define the same curves.
- Consider the following polynomials in
:
- Compute
.
- Compute
.
- What do your answers from (a) and (b) tell you about
and
?
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
and
are polynomials with complex coefficients, then we may write
where the
are the roots of
and the
are the roots of
, counted with multiplicities. In the Sylvester matrix, replacing
and
by the monic polynomials with the same roots corresponds to dividing the first
columns by
and dividing the first
columns by
. Thus the general formula for the resultant in terms of the roots and
and
should be
Now, consider the case of the discriminant, i.e. take
. We have
by the product rule. Thus we have
and so the alternate definition of the discriminant
which explains the apparently unnecessary
in defining the discriminant.
Of course, the discriminant
of a monic polynomial
with roots
, 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
: interchanging
with
interchanges two rows of the matrix, multiplying the determinant by
. This means that it can't possibly be a polynomial in the coefficients of
.