Let $f, g$ be two polynomials in $S[t]$ where the coefficient ring is $S = \mathbb{C}[a_1..a_n]$. The resultant of $R(f,g)$ gives some measure as to whether or not $f$ and $g$ share a common factor.

My question is what happens once we set a factor of the resultant equal to zero.

For example, suppose that for some nonconstant polynomials $c, p, q$ (with $p$ and $q$ relatively prime) $f = cp + a_1F$

$g = cq + a_1G$.

Clearly, if we set $a_1 = 0$, then $f$ and $g$ contain a common factor $c$. So $a_1$ divides the resultant. But now if we set $a_1 = 0$, and then cancel the common factor, it is reasonable to next study the resultant of $p$ and $q.$

Question: What is the relationship between $R(p,q)$ and $R(f,g)$?

We first thought that an irreducible factor of $R(p,q)$ must be [(some factor of $R(f,g)$) modulo $a_1$]. This is not true (see example below), however I hope some version of it will be true. The biggest difficulty, is that once we set $a_1$ to zero, we have to divide our polynomials by a common factor, and it's very difficult to say what happens to either the roots of the polynomials, or to the Sylvester matrix - the main tools we have to study resultants.

In even the simplest case: If $R(f,g)$ is a monomial in $S$, is $R(p,q)$ a monomial?

Thanks for your help!

$f = t*t + at^3;$

$g = t*(t+b) + a(t^2 + 1);$

Then $R(f,g) = a^2(a^3-ba+a+1)$, but upon setting $a=0$, these factors become $0$ and $1$ respectively, whereas the resultant $R(p,q) = R(t,t+b) = b$.


It's true if the polynomials are monic. This is because the resultant of two monic polynomials is the product of the differences of their roots. The resultant of a subset of the roots divides the resultant of all the roots.

(We work over a ring extension in which $f$ and $g$ factors so we can define $p$ and $q$ before setting $a=0$.)

Your problem occurs when the polynomials are not monic.

EDIT: As Gerry Myerson shows, this is not actually true. The trouble is that the concept of dividing by the highest power of $a$ behaves badly over a ring extension, because $a$ can split into $f\bar{f}$ with $f$ not zero when $a=0$. His example comes from $f=\frac{b+a+\sqrt{(b+a)^2-4a}}{2}$, $\bar{f}= \frac{b+a-\sqrt{(b+a)^2-4a}}{2}$, which have $f\bar{f}=a$, yet $f=b$ and $\bar{f}=0$ when $a=0$. Thus $f\bar{f}$, divided by the highest power of $a$, is $1$, which $f$ does not divide when $a=0$, despite $f$ not being a power of $a$.

For any such conjugate pair, to realize it as a resultant, you need only consider the polynomials $t$ and $t^2-(f+\bar{f})t+f\bar{f}$.

Hopefully the way in which this argument fails is illuminating to the reason why the divisibility statement is not, in fact, true.

| cite | improve this answer | |
  • $\begingroup$ I'm not sure what monic means in this context. I think the polynomials $f(t)=t^2+at$, $g(t)=t(t+b)+a(t+1)$ should qualify as monic. I get $R(f,g)=a^2(1-b)$, $R(p,q)=R(t,t+b)=b$, and $b$ is not a factor of $1-b$. Am I misunderstanding something? $\endgroup$ – Gerry Myerson Jun 6 '12 at 3:25
  • $\begingroup$ You are not. My argument is wrong. I can even simplify your example by choosing $f(t)=t$, making the resultant $a$. $\endgroup$ – Will Sawin Jun 6 '12 at 6:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.