# Computing the kernel of $A[t] \rightarrow K : t \mapsto \frac{a}{b}$.

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.

• The result holds for any normal domain. Set $x=a/b$. From $f(x)=0$ we get that $b_n=a_nx$ is integral over $A$, so $b_n\in A$. Now let $$f_1(t)=f(t)-t^{n-1}(a_nt-b_n).$$ Notice that $f_1\in\ker\phi$ and $\deg f_1<\deg f$ which allows you to proceed by induction on $\deg f$. – user26857 Apr 15 at 19:47
• There are counterexamples for $A$ not normal. – user26857 Apr 15 at 19:48