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A recent question tells us that even if two polynomials $f,g\in \mathbb Z[X]$ have no common factor as polynomials, their values at integer points may have common factors. That question gives this example: $$ f=x^3-x^2+3x-1, \qquad g=x^3+2, \qquad \gcd(f(27),g(27))=31 $$

The explanation I've given for this example is that even though $\gcd(f,g)=1$ in $\mathbb{Z}[x]$, we cannot always write $1=uf+vg$ with $u,v \in \mathbb{Z}[x]$ (because $\mathbb{Z}[x]$ is not a PID). But we can write $1=uf+vg$, if we allow $u,v \in \mathbb{Q}[x]$. In the example above, we get $$ 1 = \dfrac1{31} (-6 x^2-7 x-3)f(x)+\dfrac1{31}(6 x^2+x+14)g(x) $$

Now, clearing denominators, we get $d = uf+vg$ with $u,v \in \mathbb{Z}[x]$ and $d \in \mathbb{Z}$.

  • Is there a name for $d$ in terms of $f$ and $g$?

  • Can we compute $d$ without performing the entire extended Euclidean algorithm in $\mathbb{Q}[x]$?

  • When $d>1$, is it always true that some values of $f$ and $g$ (at the same point) are not coprime?

It seemed that $d$ is the resultant of $f$ and $g$, but perhaps not.

lhf
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  • See also https://mathoverflow.net/questions/248488/reduced-resultants-and-bezouts-identity – lhf Mar 10 '23 at 17:02

1 Answers1

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  • Regarding Question $1$: I think that $d$ in your post is -in the general case- some divisor of the resultant. However, any divisor $r>1$ of the resultant would do, in the sense that it may be the $\gcd\big(f(n), g(n)\big)$ for some $n$:

for any divisor $r$ of the resultant $Res(f,g)=d$, we can always find some integer $n$ such that $\gcd\big(f(n),g(n)\big)\geq r>1$, given that the following system of simultaneous equations $$ \begin{array}{c} f(x)\equiv 0\mod r \\ % \\ g(x)\equiv 0\mod r \end{array} $$ has a solution $x\equiv n\mod r$.

Then, the secuence $$ m_{n,r}(i)=n+r*i $$ for all $i\geq0$, has the property that: $$ \gcd\Big(f\big(m_{n,r}(i)\big),g\big(m_{n,r}(i)\big)\Big)\geq r>1 $$

  • Regarding Question $2$: The resultant $Res(f,g)$ of two polynomials over a field, is by definition the determinant of the Sylvester matrix of the polynomials, and it can be shown to be equal to the product between the differences of their roots (some of them may lie in some field extension) times suitable powers of the leading coefficients of $f,g$.

  • Regarding Question $3$: The answer is negative in general:

For example consider the polynomials $f(x)=5x+1$, $g(x)=5x+6$. Then $Res(f,g)=25$ but the simultaneous equations $$ \begin{array}{c} 5x+1\equiv 0\mod 5 \\ % \\ 5x+6\equiv 0\mod 5 \end{array} $$ have no common solutions: $f-g=5$ thus any common divisor of $f(n), g(n)$ must divide $5$. However, none of $f(n), g(n)$ is divisible by $5$. Thus, $\gcd(5n+1,5n+6)=1$ for any positive integer $n$. Notice also that $$ u(x)f(x)+w(x)g(x)=5 $$ for $u(x)=-1$ and $w(x)=1$.

KonKan
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  • $d$ is not the resultant: try $f=x^2-4$ and $g=x^2+1$. You'll get $d=5$ but a resultant of $25$. See http://mathoverflow.net/a/17506/532. – lhf Apr 07 '16 at 09:54
  • you are right. interesting counterexample and nice reference! I need to think some more on your method to tell for sure, but it seems that in the general case $d$ is some divisor of the resultant. – KonKan Apr 07 '16 at 23:27