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I want to know if the following proposition is true or false :

Let $p$ and $q$ $\in \mathbb Q[X]$ two polynomials of degree $\geq 1$ and let $g$ $\in \mathbb Q[X]$ an irreducible unitary polynomial, with deg(g) $\geq 1$ such that $\exists$ two polynomials $a$ and $b$ $\in \mathbb Q[X]$ with $g = ap + bq$. Then $g = gcd(p,q)$.

I have the feeling that it is true with a few tests but I don't know how to prove it and I feel like something is wrong with the fact that we are $\mathbb Q[X]$. What if we are $\mathbb R[X]$ ?

(I also have that $x-\frac74=(\frac18x+\frac14)(x-1)-(\frac18x-\frac12)(x-3)$ but $x-\frac74$ does not seem to be $(x-1)$ and $(x-3)$'s gcd).

Bill Dubuque
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Kilkik
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  • By here, $,py + qz = g,$ is solvable for $,y,z\in \Bbb Q[x]!\iff! \gcd(p,q)\mid g,,$ which is not equivalent to $,\gcd(p,q) = g,,$ e.g. it's solvable when $,\gcd(p,q)=1\neq g\ $ (here we assume for convenience both $g$ and gcds are normalized to be monic, i.e. lead coef $=1,,$ else you need to compare gcds not by equality but by associateness). – Bill Dubuque Apr 09 '21 at 08:52

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You have a good counterexample, so the proposition doesn't hold.

There's no particular reason to expect that it would hold, either. The claim is a converse of Bézout's identity for $\mathbb Q[X]$, but nobody says such a converse has to hold anywhere.

In fact, the corresponding converse doesn't even hold in $\mathbb Z$: There we know well that $ap+bq$ can be any multiple of the gcd of $p$ and $q$, and just because $ap+bq$ is prime (that is, irreducible) doesn't mean that this multiple needs to be the gcd itself. Namely, it might be that $p$ and $q$ are coprime, in which case anything in $\mathbb Z$ can be written as $ap+bq$.

That's essentially what goes on in your counterexample too: $\gcd_{\mathbb Q[X]}(x-1,x-3)=1$, so you can get anything you want as a sum of polynomial multiples of those two factors. Just pick any $g$ and let $a=\frac12g$ and $b=-\frac12g$ ...

Working over $\mathbb R[X]$ wouldn't make a difference here.

Troposphere
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    +1. If we amend the proposition to include $\deg (\gcd (p,q))>0$ and $g\ne 0$ then it is true: Let $h=\gcd (p,q).$ Let $p=(p^)(h)$ and $q=(q^)(h).$ Then $g=(ap^+bq^)(h)$ is (by assumption) non-$0$ and irreducible but $\deg (h)>0$ so $ap^+bq^$ must be a non-$0$ constant. – DanielWainfleet Apr 09 '21 at 04:34