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I'm thinking about the relation between the GCD, the resultant, and the Bezout lemma/identity for the ring $\mathbb{Z}[X]$. Take two elements $f,g \in \mathbb{Z}[X]$. We know:

  • the GCD exists, because we are in a UFD
  • the resultant exists
  • we generally do not have a Bezout identity for the gcd

(The last part can be seen by taking for example $f=2$ and $g=X$. The GCD is 1 but $1 \notin (2,X)$.)

However, we do have a natural equation that looks like a Bezout identity: View $f,g$ as elements of $\mathbb{Q}[X]$. Then we have a bezout identity $$\frac{a}{b}f + \frac{c}{d}g = r$$ where $r$ is the GCD of $f,g$ in $\mathbb{Q}[X]$ and $a,b,c,d \in \operatorname{Frac}(\mathbb{Z}[X])$. Now multiply out denominators to get: $$a f + cg = bdr$$ Then $bdr$ is a natural object, and my question is what is it?

I think that if $f,g$ are very nice - say monic+irreducible+coprime - then it will be the resultant. However, in the same case, taking $f$ and $g^2$ instead I think we will get the same result, but the resultant squares. (Hence it's not the resultant anymore.)

Johann Birnick
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    An MO page on this topic is https://mathoverflow.net/questions/248488/reduced-resultants-and-bezouts-identity – KCd Mar 10 '23 at 13:36
  • Thanks, that will help! –  Mar 10 '23 at 14:18
  • See also https://math.stackexchange.com/questions/1591430/denominator-in-rational-gcd-of-integer-polynomials – lhf Mar 10 '23 at 17:02

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