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Suppose $R$ is a UFD and $p(x)\in R[x]$ a polynomial of degree $\ge 1$. Suppose $\frac ab$ is in the fraction field $K$ of $R$, with $a$ and $b\in R$ and $\text{gcd}(a,b)=1$, and such it is a root of $p(x)$, so $p(\frac ab)=0$.

Question: Is it true that then we have $p(x)=(bx-a)q(x)$ with $q(x)\in R[x]$?

It is clear that $q(x)\in K[X]$ exists and it is unique. Since the highest degree coefficient of $p(X)$ is divisible by $b$ and the constant term is divisible by $a$ (e.g. the rational root theorem ) one can easily show that the highest degree coefficient and the constant term of $q(x)$ are in $R$. This shows the case $p(x)$ has degree 2.

A suspect it should a consequence of the so called Gauss Lemma, but I was not able to find a convincing argument.

Bill Dubuque
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Nulhomologous
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2 Answers2

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Note: $\,p(x) = (bx\!-\!a)q(x) = (bx-a)\frac{c}d\bar q(x)\,$ by Factor Theorem in $K[x],\,$ $\color{#0a0}{{\rm primitive} \ \bar q}\in R[x]$

thus $\, d p(x) = (bx\!-\!a)\, c\,\color{#0a0}{\bar q(x)},\,$ so taking $C = $ content of this

$\Rightarrow\, d\,C(p(x))\ =\ 1\cdot c\cdot\color{#0a0} 1,\, $ i.e. $\,d\mid c\,$ in $\,R,\,$ so $\,q = \frac{c}d \bar q\in R[x].\ \ $ QED

Or by nonmonic Division: $\, b^k p(x) = (bx\!-\!a) q(x) + r,\ r\in R,\,$ so $\,r = 0\,$ by eval at $\,x = a/b.\,$ Note $\color{#c00}{(b,bx\!-\!a)}=\color{#0a0}{(b,a)}=\color{#c00}1,\,$ so $\,\color{#c00}{b\mid (bx\!-\!a)}q(x)\,\overset{\rm Euclid}\Longrightarrow\,b\mid q(x),\,$ thus $\,b^k\mid q(x)\,$ by induction [or we can use localization or the AC method as described in the link]. This proof works not only when $R$ is a gcd domain, but also when the ideal $\,\color{#0a0}{(b,a)}=(1)\,$ in any commutative ring $R.\,$ See also here.

Bill Dubuque
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  • The "nonmonic division" answer is the best because it makes clear that this only depends on the root $a/b$ in question having a gcd between $a$ and $b$; the answer does not depend on any global property of $R$. – Badam Baplan May 21 '20 at 16:18
  • @Badam Indeed, that was part of my motivation for adding that method even though the OP leaned towards Gauss. A warm welcome back to you. – Bill Dubuque May 21 '20 at 16:23
  • Thank you Bill, I appreciate that! Warm regards to you as well. As for the comment, I only meant it as an endorsement of your answer, which I feel is the pedagogically 'right' one. – Badam Baplan May 22 '20 at 02:50
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By the Factor Theorem, since $f(a/b)=0$, we have $$f(x)=\left(x-{\small{\frac{a}{b}}}\right)v(x)$$ for some $v\in K[x]$.

Letting $g(x)=v(x)/b$, we have $$f(x)=(bx-a)g(x)$$ where $g\in K[x]$.

Our goal is to show $g\in R[x]$.

Letting $d\in R$ be the least common multiple of the denominators of the fractionally reduced coefficients of $g$, it follows that $$g(x)=G(x)/d$$ where $G\in R[x]$ is primitive.

Then from $f(x)=(bx-a)g(x)$, we get $d{\,\cdot}f(x)=(bx-a)G(x)$.

By Gauss' lemma, since $bx-a$ and $G$ are primitive in $R[x]$, the product $(bx-a)G(x)$ is primitive in $R[x]$.

Hence $d{\,\cdot}f(x)$ is primitive in $R[x]$, so $d$ is a unit of $R$.

Finally, since $G\in R[x]$ and $d$ is a unit of $R$, the equation $g(x)=G(x)/d$ implies $g\in R[x]$, as was to be shown.

quasi
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