Let $f(x)$ and $g(x)$ be coprime monic polynomials in $\mathbf{Z}[X]$ of positive degrees $m$ and $n$ respectively. It seems that in this case their reduced resultant can be obtained from the expression $uf + vg = 1$ over $\mathbf{Q}[X]$ with $\deg u < n$ and $\deg v < m$. Namely - reduced resultant in this case is the smallest natural $D$ such that $Du$ and $Dv$ are in $\mathbf{Z}[X]$? For the notion of reduced resultant see: [this MO question of Felipe Voloch](https://mathoverflow.net/questions/17501/). Shortly it is the generator $D>0$ of the ideal $I=(f(x),g(x))∩\mathbf{Z}$.



Reduced resultant of monic polynomials
The ideal generated by $f$ and $g$ in ${\bf Z}[x]$ is the set of all $uf+vg$ with $u,v$ in ${\bf Z}[x]$. 


Now suppose $uf+vg=d$ for some integer $d$. Assuming $f$ and $g$ are monic, we have $$v=fq\_1+r\_1,\qquad u=gq\_2+r\_2$$ with $q\_i$ and $r\_i$ in ${\bf Z}[x]$, $\deg r\_1<\deg f$, $\deg r\_2<\deg g$. Then $$fg(q\_1+q\_2)+r\_2f+r\_1g=d$$ But $\deg (fg)>\deg(r\_2f+r\_1g)$, so we must have $q\_1+q\_2=0$, and $$r\_2f+r\_1g=d$$ It follows that the smallest positive integer in the ideal generated by $f$ and $g$ (that is, the reduced resultant of $f$ and $g$) can be expressed as $uf+vg$ with $\deg v<\deg f$ and $\deg u<\deg g$. 


