If $\mathbb A$ is a principal ideal domain, Bézout's identity (for the gcd) reads as follows:

  • For any $a,b\in\mathbb A$, there exist $u,v\in\mathbb A$ such that $a\cdot u+b\cdot v= \gcd(a,b)$.
  • Conversely, if there exist $u,v,g\in\mathbb A$ such that $a\cdot u+b\cdot v=g$, then $\gcd(a,b)$ divides $g$.

This applies in particular to polynomial rings over a field.

Consider now a polynomial ring $\mathbb A[X]$ over some ring $\mathbb A$. For $f,g\in\mathbb A[X]$, let $\text{res}(f,g)$ denote their resultant, defined for instance as the determinant of their Sylvester matrix. We have:

For any $f,g\in\mathbb A[X]$, there exist $u,v\in\mathbb A[X]$ such that $f\cdot u+g\cdot v=\text{res}(f,g)$.

Does the converse of Bézout identity for the resultant holds? The answer of this question may of course depend a lot on $\mathbb A$. For instance, when $\mathbb A$ is a field, if $r=f\cdot u+g\cdot v$ is a constant, $\text{res}(f,g)$ divides $r$ by Bézout's identity and the fact that $\gcd(f,g)$ divides $\text{res}(f,g)$.

My question can be formulated more precisely as follows:

Let $f,g,u,v\in\mathbb A[X]$. What is the minimum requirement on $\mathbb A$ such that for all $u,v\in\mathbb A$ such that $r=f\cdot u + g\cdot v$ is a constant, $\text{res}(f,g)$ divides $r$?

Note that the proof sketch given above in the case where $\mathbb A$ is a field does not work when $\mathbb A$ is an integral domain only. For instance over the integers, $\text{res}(x+2,x+4)=2$ does not divide $\gcd(x+2,x+4)=1$. Yet in this particular case, if $(x+2)\cdot u + (x+4)\cdot v=r$, we have $2\cdot u(0)+4\cdot v(0)=r$, whence $r$ is even (and $\text{res}(x+2,x+4)$ divides $r$).

  • 1
    $\begingroup$ You mean "... such that for all $u, v \in \mathbb{A}\left[X\right]$ for which $fu+gv$ is a constant, the resultant $\operatorname{res}\left(f,g\right)$ divides $fu+gv$"? $\endgroup$ – darij grinberg Oct 12 '15 at 16:55
  • $\begingroup$ @darijgrinberg: Right! $\endgroup$ – Bruno Oct 13 '15 at 8:05

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