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 $I=(f(x),g(x))∩\mathbf{Z}$.


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 $\deg u<\deg g$.


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