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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 $(f)+(g) = (\mathrm{res}(f,g))$?

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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 $R$.

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  • $\begingroup$ 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

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