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 $R(p,q) = R(t,t+b) = b$.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Jun 6, 2012 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, 2012 at 6:22

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