I have a complete local Noetherian normal domain $A$ (commutative with 1) with fraction field $K$. I was trying to compute the pullback of the prime ideal $(t - \frac{a}{b}) \subset K[t]$ to $A[t]$. This is the same as the kernel of $\phi :A[t] \rightarrow K$, where $\phi:t \mapsto \frac{a}{b}$ and I came up with the following guess:
Let $\{(a_i, b_i)\} \subset A \times A$ be the set of pairs such that $\frac{a_i}{b_i} = \frac{a}{b}$. Then $I = \ker(\phi) = (b_i t - a_i)$, the ideal generated by all the linear polynomials that have $\frac{a}{b}$ as a root.
I am looking for a proof or a counterexample of the above statement.
What I tried so far: Take any polynomial $f \in I$, i.e. $f(\frac{a}{b}) = 0$. If $f = \sum_{i = 0}^d \alpha_i t^i$ is linear then certainly it lies in $(b_i t - a_i)$. If it has higher degree I want to reduce it as follows: If $\alpha_d \in \{b_i\}$ I subtract $\alpha_d t^d - a_{i_{\alpha_d}} t^{d - 1}$ from $f$ and obtain a polynomial of lower degree that still has $\frac{a}{b}$ as a root. If $\alpha_0 \in \{a_i\}$ I subtract $b_{i_{\alpha_0}} t - \alpha_0$ and divide by $t$ to again get a polynomial of lower degree with $\frac{a}{b}$ as a root. This means that we have reduced the problem to showing that $\alpha_0 \in \{a_i\}$ or $\alpha_n \in \{b_i\}$. This statement reminds me of the rational root theorem, but I am particularly interested in the case where $A$ is not regular, so not a UFD.
I also noticed that sets $\{a_i\}$ and $\{b_i\}$ form ideals, since $\frac{a_1 + a_2}{b_1 + b_2} = \frac{a}{b}$. This means they are finitely generated, which might be useful for something.
I am also unsure how many of the assumptions on $A$ are actually needed, so I am also interested in counterexamples when some of the assumptions are removed.
Thank you for your time!