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. Shortly it is the generator $D>0$ of the ideal $I=(f(x),g(x))∩\mathbf{Z}$.

$\begingroup$ Are you asking the same questions as in here: mathoverflow.net/questions/17501/… ? $\endgroup$ – Felipe Voloch Dec 29 '15 at 18:23

$\begingroup$ Yes, about that notion  the reduced resultant $\endgroup$ – Algirdas Rugys Jan 12 '16 at 12:21
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 $\deg u<\deg g$.