§ Discriminant and Resultant
I had always seen the definition of a discriminant of a polynomial as:
While it is clear why this tracks if a polynomial has repeated roots
or not, I could never motivate to myself or remember this definition.
I learnt that in fact, this comes from a more general object, the resultant
of two polynomials , which provides a new polynomial
which is zero iff share a common root. Then, the discriminant is
defined as the resultant of a polynomial and its derivative. This makes far more
- If a polynomial has a repeated root , then its factorization willbe of the form . The derivative of the polynomial will have an term that can be factored out.
This cleared up a lot of the mystery for me.
- On the contrary, if a polynomial only has a root of degree 1, then thefactorization will be , where is not divisible by .Then, the derivative will be . We cannot take commonfrom this, since is not divisible by .
§ How did I run into this? Elimination theory.
I was trying to learn how elimination theory works: Given a variety
, how does one find a rational parametrization
such that , and are
rational functions? That is, how do we find a rational parametrization of the
locus of a polynomial ? The answer is: use resultants!
- We have two univariate polynomials , where the notation means that we have a polynomial .The resultant isa polynomial which is equal to when and share a common root.
- We can use this to eliminate variables. We can treat a bivariate polynomial as a univariate polynomial over the ring . This way, given twobivariate polynomial , , we can compute their resultant,giving us conditions to detect for which values of , there existsa common such that and share a root. If are constants, then we get a polynomial that tracks whether and share a root.
- We can treat the implicit equation above as two equations, , . We can apply the method of resultants to project out from the equations.
§ 5 minute intro to elimination theory.
Recall that when we have a linear system , the system has a non-trivial
solution iff . Formally: . Also, the
ratio of solutions is given by:
If we have two polynomials , and
, then the system , has
a simeltaneous zero iff:
§ Big idea
The matrix is setup in such a way that any solution vector such that
will be of the form . That is,
the solution vector is a polynomial, such that . Since ,
we have that , and .
- Necessity is clear. If we have some non trivial vector such that, then we need .
- Sufficiency: Since , there is some vector such that . We need to show that this is non-trivial. If the polynomials , are not equal, then we have that the rows which have coefficientsfrom and are linearly independent. So, the pair of rows ,and the pair are linearly independent. This means thatthe linear system:
Since the coefficients of the two systems are the same, we must have that
and are linearly dependent. That is:
We can take arbitrarily, giving us a vector of the form
, which is the structure
of the solution we are looking for!