On the kernel of the surjective ring homomorphism $R[X] \to R[ab^{-1}]$ taking $X$ to $ab^{-1}$

Question

Let $R$ be a GCD domain (https://en.wikipedia.org/wiki/GCD_domain) . Let $0\ne a,b\in R$ be relatively prime elements i.e. with gcd 1. Let $K$ be the fraction field of $R$. Let $u:=ab^{-1} \in K$ . We have a surjective ring homomorphism $\phi : R[X] \to R[u]$ given by $\phi(f(X))=f(u),\forall f(X) \in R[X]$. I can see that $(bX-a)R[X] \subseteq \ker \phi$ .

My question is: Is it true that $(bX-a)R[X]=\ker \phi$ ?

Answer 1

Let us first prove the Lemma

Let $e = e_{n} x^{n} + e_{n-1} x^{n-1} + \dots + e_{1} x + e_{0} \in R[x]$, with $e_{n} \ne 0$. If $e(u) = 0$, then $b$ divides $e_{n}$.

In fact $$ e_{n} a^{n} + e_{n-1} a^{n-1} b + \dots + e_{1} a b^{n-1} + e_{0} b^{n} = 0. $$ Since $a$ and $b$ are coprime, and $b$ divides $e_{n} a^{n}$, we have that $b$ divides $e_{n}$.

Coming to the main argument, let $c = c_{n} x^{n} + c_{n-1} x^{n-1} + \dots + c_{1} x + c_{0} \in R[x]$, with $c_{n} \ne 0$, be such that $c(u) = 0$.

We claim that $c$ is divisible by $b x - a$. Let us show that we can do long division of $c$ by $b x - a$.

Since $c(u) = 0$, we have $b \mid c_{n}$ by the Lemma, so let $c_n = b d_n$. .

The first step of long division is obtained by taking $$ c - d_{n} x^{n-1} (b x - a) = (c_{n-1} + a d_{n}) x^{n-1} + \dots + c_{1} x + c_{0}. $$ But clearly the resulting polynomial also vanishes when evaluated in $u$, so by the Lemma $b$ divides its leading coefficient, and we can continue long division.