Let’s say that you have two polynomials, and you REALLY need to know if they have a common root. Now, if they’re quadratic, you’re in luck, because we can solve them both completely and just check. In fact, if you’re patient, you can do it whenever the Galois group is solvable, in particular, you can do it for cubics and quartics. But in general? What if I gave you a degree 100 polynomial and a degree 103 polynomial? Well, you can still do it, without having to solve anything.
The object that solves this problem is called the Resultant. Let and with and . Then the resultant is the determinant of the matrix
Now, this is actually a rather good thing to use for computations, though the determinant gets big, there’s a lot of zeros floating around, and computers can handle large determinants. Now, resultants have three really nice properties:
- The resultant is an integer polynomial in the coefficients of .
- iff have a common factor.
- There are polynomials such that .
The last property is called the Elimination property, and it ties in with the Elimination Theorem from last time. If , then we choose a variable and write as polynomials in that variable with coefficients polynomials in the others. Then (written to point out the dependence on the variable chosen) is in the ideal generated by , but also doesn’t depend on the variable , and so is in the elimination ideal.
Now, we’ve done resultants for one variable. But algebra in one variable is the same as algebra of homogeneous polynomials in two variables, so we extend the resultant to these in the most obvious way possible. Then the condition that the resultant is zero turns out to be equivalent to having a nontrivial solution. That is, one other than . But if they have one, then there’s a line worth, and so this turns out to determine if the two polynomials share any zeros on .
Now, we generalize to get some nice stuff. To start, we denote the above resultants, as integer polynomials, by . We can handle pairs of polynomials fairly well with these. But what about a whole pile of polynomials of various degrees in more variables? Say, take to be homogeneous polynomials in the variables .
Well, we can fix degrees , the degrees of the . Now, there is a unique integer polynomial satisfying the properties:
- has a nontrivial solution if adn only if .
- (This is just to normalize things, generally we’ll only care if a resultant is zero or not).
- is an irreducible polynomial over .
Now, proving existence is hard and long, but we can still look at these polynomials and see what we can say about them.
For instance, in the case that all the are linear, with , then . Also, the resultant of two polynomials is a special case here.
Now, these general resultants, when computable, are rather nice for working out if a hypersurface is smooth. Let be a homogeneous polynomial of degree on . Then a singular point is a common solution to the partial derivatives, which are in variables. So we look at , and it is zero if and only if is singular.
As an aside and an application, the hypersurfaces of degree on naturally form a projective space, and the zero locus of the resultant as above is precisely the set of singular hypersurfaces. As this locus is cut out by a single polynomial, the collection of singular degree hypersurfaces is a divisor on the space of all degree hypersurfaces.
Now, before stopping, we’ll define one more thing in terms of resultants, and we’re even going to go back to the case of just two polynomials in one variable. Let be a polynomial in , then we define the discriminant to be . This turns out to be wonderful, as it tells us if has a multiple root, because that’s the same thing as and to have a common root. Well, strictly, we define the discriminant to be , to make the signs all work out. So now, if you plug in a quadratic, you get the expected .