1

I have learned the definition of the Resultant of two polynomials Resultant of two polynomials. In most places, it is defined over a field. Can we similarly define it for the general ring, for example, $R = \mathbb{Z}/(p^{k}\mathbb{Z})$, where $p$ is any prime and $k \in \mathbb{N}$?

Assume $f(x), g(x) \in R[x]$. Suppose we define the function $\mathcal{R}(y) = \mathrm{Resultant}(f(x), y - g(x))$. Now, I think this function is valid because to compute this function, we need to calculate the determinant of the Sylvester matrix of $f(x), y - g(x)$ (assume $y$ like a constant).

My second question is: Can we prove that $f(x)$ divides $\mathcal{R}(g(x))$?

For example let $f(x) = f_0 + f_1 x + f_2 x^2, g(x) = g_{0} + g_{1}x$, then $~S = \begin{pmatrix}f_{0} & y - g_{0} & 0 \\f_{1} & - g_{1} & y - g_{0} \\f_{2} & 0 & - g_{1} \end{pmatrix}$ Then $\mathcal{R}(y) = \det(S)$ and $\mathcal{R}(g_{0} + g_{1}x) \\= \det \begin{pmatrix}f_{0} & g_{1}x & 0 \\f_{1} & - g_{1} & g_{1}x \\f_{2} & 0 & - g_{1} \end{pmatrix} \\= g_{1}^{2}(f_0 + f_1 x + f_2 x^2) \\= g_{1}^{2}f(x)$.

So $f(x)$ divides $\mathcal{R}(g(x))$, but $\mathcal{R}(g(x)) \neq 0$.

But I couldn't able to prove the general statement.

D.W.
  • 5,958
Afntu
  • 2,393

1 Answers1

2

To answer the first part in your question, resultants can be defined over commutative rings. In fact it is often the case that one is interested in polynomials $R[x]$, where $R$ itself is a polynomial ring in some other indeterminate(s), with the aim of eliminating $x$. In that case one would be interested in the resultant over $R$.

In the Wikipedia article that you've linked, in the section Properties, you can see that a resultant is defined over a commutative ring, but some important properties like uniqueness are valid only for integral domains.

However, always check against a more "traditional" source, as I have previously found some issues with the properties listed in that wiki article that were swiftly corrected.

About your second question:

Let $z$ be an indeterminate independent of $x$ and $y$ and consider $f$ as a polynomial in $z$. Let $h$ be a polynomial in $z$ given by: $$h(z) = \sum_{i=1}^k g_i x^i - \sum_{i=1}^k g_i z^i,$$ where $k$ is the degree of $g$. With this notation in mind, your matrix $\mathcal R$ is recovered as the Sylvester matrix of $f$ and $h$ with respect to $z$, that is $\mathcal R(g) = \operatorname{Res}_z(f,h)$.

Now $h$ factors nicely as $h(z)=(x-z)\sum_{i=1}^kg_i(x^{k-1}+x^{k-2}z+\cdots+xz^{k-2}+z^{k-1})$. To simplify notation a bit, I'll write $h(z)=(x-z)H(x,z)$.

Using the multiplicative property of the resultant, we have $$\operatorname{Res}_z(f,h) = \operatorname{Res}_z(f,(x-z)H(x,z))=\operatorname{Res}_z(f,x-z)\cdot \operatorname{Res}_z(f,H(x,z)).$$ The only thing that $\operatorname{Res}_z(f,x-z)$ is doing is "renaming" $f$ from a polynomial in $z$ to a polynomial in $x$, therefore $$\operatorname{Res}_z(f,x-z)\cdot \operatorname{Res}_z(f,H(x,z))=f(x)\cdot \operatorname{Res}_z(f,H(x,z))$$

Randy Marsh
  • 3,786