More facts about the resultant over a UFD

For this post, let be a unique factorization domain (UFD).

Let and be two polynomials in of degrees and respectively. Writing for the -module of polynomials in having degree strictly less than , we define the linear map by the equation

For , we may choose the ordered basis while for we may choose In these bases, the map has a matrix representation that is called the Sylvester matrix of and . Now you can look elsewhere on the Internet to find what this matrix looks like. Here’s an example of what it looks like when and : Very nice. Obviously always a square matrix.

Now the resultant of and , denoted , is defined to be the determinant of the Sylvester matrix, or equivalently the determinant of the linear map . For a good guess at where all of this comes from, look at p.24 of Walker’s Algebraic Curves (1991).

From now on, suppose and are non-constant polynomials (in particular they are both non-zero). The proof of the following lemma is simple and uses the fact is a UFD:

Lemma 1. The polynomials and share a non-constant factor if and only if there exists two non-zero polynomials and , with and , such that .

Corollary. The polynomials and share a non-constant factor if and only if .

Proof. Suppose and share a non-constant factor. The previous lemma gives us such and . Now clearly , hence is in the kernel. Conversely, suppose there exists some non-zero tuple in the kernel of . If we had , then we would have , whence because is not zero and is an integral domain. Similarily, it is impossible that is zero. Therefore both and are non-zero polynomials, with and . They are in the kernel of , so . We win by applying the previous lemma with and .

We will need the following technical result:

Lemma 2. Let be any square matrix with coefficients lying in an integral domain . Then if and only if there exists some non-zero vector in the kernel of . Said differently, an endomorphism on a free -module has zero determinant if and only if it kills some non-zero vector.¹

Proof. Suppose we are given a non-zero vector in the kernel of . The equation also holds in , where is the fraction field of . Because is a field, we obtain that the determinant of over is zero. This determinant is an algebraic expression in terms of elements of only, so it is zero in as well. Conversely, suppose . Again, seeing this as a fact in the field , we know there must exists some non-zero vector in the kernel of over , i.e. the vector has coefficients in . This is no problem since we can simply clear the denominators of each component by multiplying by an appropriate element . Then is a vector with components in . Moreover, in , whence in . This shows is a non-zero element in the kernel of .

By combining everything we have, we obtain this really useful theorem:

Theorem. Two non-constant polynomials and with coefficients in a unique factorization domain share a non-constant factor if and only if their resultant is zero.

Proof. The polynomials and share such a factor if and only if (by the corollary), if and only if (by the previous lemma).