Let $R[X]$ be the polynomial ring over a commutative ring $R$, $f$ and $g \in R[X]$ two coprime polynomials, and $\mathrm{res}(f,g)\in R$ their resultant.

Because $f$ and $g$ are coprime, $\mathrm{res}(f,g)\neq 0$. But does this determinant, which the resultant is, measure something more?

Is it true that there are $a, b \in R[X]$ such that $af + bg = \mathrm{res}(f,g)$? If R is a Bézout domain, is $(f)+(g) = (\mathrm{res}(f,g))$?


That's true for every field. For ex. Prop. 9, p. 157. of Cox Little O'Shea (Ideals, algorithms, varieties...) old edition, or page 164, prop 5 of 2015 edition, states that; Even more, the coefficients of such a and b are pol in terms of the coefficients of f and g. I think it is true for the rings as well: Considering the field of fractions of $R$, we will be allowed to use the Cramer rule/division; At the last step we multiply everything by $Res$ and get something in $R$.

  • $\begingroup$ page 157, prop. 9. Sorry for the mistake....am going to edit it $\endgroup$ – araz-panther Feb 3 '17 at 15:59
  • $\begingroup$ As far as I know in the referenced book all polynomials have coefficients in a field. $\endgroup$ – user26857 Feb 3 '17 at 20:33
  • $\begingroup$ Yes, you are right; However, I think the proposition is true for a ring as well: In the proof, in order to use the division in Cramer's rule, consider the fraction field, do the division, and at the last step multiply everything by the resultant. $\endgroup$ – araz-panther Feb 5 '17 at 0:21
  • $\begingroup$ As far as I know only the integral domains have a field of fractions. $\endgroup$ – user26857 Feb 5 '17 at 8:13
  • $\begingroup$ You are right; I was sloppy on that. $\endgroup$ – araz-panther Oct 10 '17 at 4:41

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.