5
$\begingroup$

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 $I=(f(x),g(x))∩\mathbf{Z}$.

$\endgroup$
2
8
$\begingroup$

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 $\deg u<\deg g$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.