First of all I know that by Picking two polynomials $f,g$ where $g$ is not $0$ there exists exactly two polynomials $q,r$ such that $f=qg+r$.

We say $g$ divides $f$ if $r=0$.

For example $x^3=x\cdot x^2$

So $x^2$ is a divisor of $x^3$

We say two polynomials $x,y$ are coprime to eachother if there exists no polynomial $g$ such that $g$ is a divisor of $x$ and $y$ and $g$ has at least Degree $1$.

$$x,y \text{ coprime } \iff x=qg\wedge\deg(g)\geq 1\Longrightarrow y=q'g+r \wedge r\neq 0$$

I was Looking for an easy example for such a pair of polynomials. I have tried it with $x^3$ and $x^2+1$

And assumed there exists a polynomial $g$ with

$$x^3=qg\wedge x^2+1=q'g$$

I played around a Little bit with this equation but I don't know where this is going. Hope somebody can help me not necessarily with this pair of polynomials but only to find an easy example for two coprime polynomilas. Maybe there is a generic formula for those Kind of Proofs it would be Wonderful if you could share it with me.

  • $\begingroup$ We always have $\,1+fg\,$ is coprime to $f$ since any common divisor must divide $\, 1+fg -f\cdot g = 1.\,$ Similary $\,f,g$ are coprime if $\,a f + b g = 1\,$ for some $a,b$ (and conversely (Bezout) by the Euclidean algorithm - just as for integers, assuming the coefficient ring is a field and the polynomials are univariate) $\endgroup$ – Bill Dubuque Mar 8 at 14:32

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