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Let $R$ be an integral domain and denote its field of fractions by $R'$.

Let $\phi : R[x] \to R'[x]$ be the inclusion. If $I'$ is a (principal) ideal of $R'[x]$, is $\phi^{-1}(I')$ always principal?

The motivation for the problem is the following: You can show that the kernel of the evaluation map $\mathbb{Z}[x] \to \mathbb{C}$ is a principal ideal by factoring the map as $\mathbb{Z}[x] \to \mathbb{Q}[x] \to \mathbb{C}$, using the fact that the ring in the middle is a PID and using the primitivization of its generating element to generate the kernel.

A pretty messy argument overall, I'm wondering if there is a) a better way to prove this and b) if it holds generally between any integral domain $R$ to its field of fractions (or perhaps an even more/less general statement holds, e.g. need $R$ to be a UFD)?

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    @Watson Yeah, I was hoping you would be able to do something like this. Though I don't think your $c$ is correct -- I think you want the lcm of the $b_i$s divided by the gcd of the $a_i$s. So I guess we need to be able to take gcds. – 2017-02-13
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    This is true if $I' = (f)$ where $f \in R[X]$ and the dominant coefficient of $f$ is a unit in $R$ (see prop. 1 [here](http://www.math.uchicago.edu/~may/VIGRE/VIGRE2007/REUPapers/FINALFULL/Chariker.pdf)). Since any ideal $I'$ of $R'[X]$ can be generated by a polynomial $f$ in $R[X]$, the only issue would be when the dominant coefficient of any such $f$ is not a unit in $R$. – 2017-02-14
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    Suppose $R$ is a GCD domain (or a UFD if you like it). Then you may assume $f\in R[X]$ and $f$ *primitive*. Use Gauss' Lemma to show that $fR'[X]\cap R[X]=fR[X]$. – 2017-02-14
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    I'm pretty sure that if $R$ is the ring of integers in a number field, then any nonprincipal ideal $(a,b) \subseteq R$ can be used to produce a counterexample; I expect $I' = (a+bx)$ should do. – 2017-02-14

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