Let $R$ be an integral domain with field of quotients $K$. For $q\in K$ let $\phi: R[X] \to K,\,f \mapsto f(q)$ be the evaluation at $q$.
Question: Can the generators of $\ker(\phi)$ be explicitly written down ?
In particular, I wonder, if
$$\ker(\phi)=(\,bX-a\mid q=\frac{a}{b},\,a,b \in R\,)\qquad ?$$
Of course, if $q \in R$, then $\ker(\phi)=(X-q)$. But I don't know what happens in the more general case given in the question. If $f \in \ker(\phi)$, I tried to utilize that there are $c_i \in K$ such that
$$f(X) = (X-q)\sum_{i=0}^nc_iX^i=c_nX^{n+1}+ \sum_{i=1}^n(c_{i-1}-c_iq)X^i-c_0q$$
but I didn't manage to conclude how the $c_i$ are looking.
Added: In the special case $q=\frac{1}{b}$, the formula also holds, i.e. $\ker(\phi)=(bX-1)$.
For, we have $\frac{c_0}{b}\in R$, hence there is $r_0 \in R$ s.t. $c_0 = br_0$. By induction there are $r_i \in R$ s.t. $c_i =br_i$ for $i=0,\ldots,n$ and thus $f=(bX-1)\sum_i r_i X^i \in (bX-1)$.