Contraction of an ideal in the polynomial ring over the fraction field of a UFD Let $K$ be the fraction field of a UFD $R$. Let $0\ne f(X)\in R[X]$. Let $I=f(X)R[X]$ and $J= f(X)K[X] \cap R[X]$. Then how to show that $I=aJ$ for some $0\ne a \in R$ ?
I think $a$ should be the gcd of the coefficients of $f(X)$, but I'm not sure. Please help. 
 A: The approach that I outline uses the notation and a result from the wonderful notes of Paul Garrett, found here. 
For any $g(X) \in K[X]$, write $c(g)$ for the content of $g(X)$ (see the material from the paragraph preceding Proposition 2.0.1 to the sentence after it in the notes linked above). From the definition of content, it is clear that $c(g) \in R$ if and only if $g(X) \in R[X]$. Indeed, if $c(g)$ is the greatest common divisor of the coefficients of $g$, then for any coefficient $\alpha \in K$ of $g$, there is some $r \in R$ such that $c(g) \cdot r = \alpha$. Since $c(g), r \in R$, this implies $\alpha \in R$. 
Now, let $f(X) \in R[X]$, and let $I = f(X)R[X]$. Note that $f'(X) := (1/c(f))f(X) \in R[X]$, and $f'(X)K[X] = f(X)K[X]$. I claim that, if $J := f(X)K[X] \cap R[X] = f'(X)K[X] \cap R[X]$, then $J = f'(X)R[X]$, whence 
$$I = f(X)R[X] = c(f)(f'(X)R[X]) = c(f)J.$$
Clearly, $J \supseteq f'(X)R[X]$, so it remains to show that $J \subseteq f'(X)R[X]$. Suppose that $g(X) \in J$. Then $g(X) = f'(X)p(X)$ for some $p(X) \in K[X]$. By Lemma 2.0.2 of the notes above, we must have $c(g) = c(f')c(p)$. Since $g \in R[X]$ and $c(f') = 1$, we get $c(p) = c(g) \in R$, whence $p(X) \in R[X]$, and the claim follows. 
