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Suppose I have two monic polynomials $f$ and $g$ with coefficients in $\mathbb{Z}$. I also suppose that $f$ and $g$ are coprime as polynomials over $\mathbb{Q}$. In particular, there exists a Bezout relation $f(x)u(x)+g(x)v(x)=1$ with $u$ and $v$ two rational polynomials. Is it true that I can find such a relation with $u$ and $v$ polynomials with coefficients in $\mathbb{Z}$ ?

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  • $\begingroup$ You can, but the right-hand side will be an integer (which anyway is a unit in $\mathbf Q[x]$). $\endgroup$
    – Bernard
    May 2, 2016 at 19:51
  • $\begingroup$ Of course but that's not very interesting. Are you claiming that I can't do it with the right-hand side being 1 ? $\endgroup$
    – Geoffroy
    May 2, 2016 at 20:02
  • $\begingroup$ I'm afraid not. $\mathbf Z[x]$ is not a PID. $\endgroup$
    – Bernard
    May 2, 2016 at 20:08
  • $\begingroup$ Of course you are right. I was hoping the hypothesis that f and g are monic could make this true but f=x+2, g=x+4 is a counterexample. For any u and v with integral coefficients fu+gv will take an even value at 0. Turns out it was a rather stupid question... $\endgroup$
    – Geoffroy
    May 2, 2016 at 21:09
  • $\begingroup$ It's not stupid at all. Everyone would like to have a simpler results. The fact is that, when performing the extended Euclidean algorithm, it's very easy, starting from innocent-looking polynomials of low degree, to obtain polynomials that have coefficients numerators and denominators with > 50 digits… $\endgroup$
    – Bernard
    May 2, 2016 at 21:13

1 Answer 1

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The question has been answered in the comments, but the site prefers to have answers posted as answers, so here goes. The answer is that it is not true, and a simple counterexample is given by $f(x)=x+1,g(x)=x-1$. We have $f(1)u(1)+g(1)v(1)=2u(1)$, so $f(x)u(x)+g(x)v(x)$ identically equal to $1$ is impossible for $u,v$ with integer coefficients.

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