# Bezout relation with integral coefficients

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 $\mathbb{Z}$ ?

• You can, but the right-hand side will be an integer (which anyway is a unit in $\mathbf Q[x]$). May 2, 2016 at 19:51
• Of course but that's not very interesting. Are you claiming that I can't do it with the right-hand side being 1 ? May 2, 2016 at 20:02
• I'm afraid not. $\mathbf Z[x]$ is not a PID. May 2, 2016 at 20:08
• 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... May 2, 2016 at 21:09
• 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… May 2, 2016 at 21:13

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.