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I am posting the following question after posting a similar question: What is the kernel of $K[x^2,x^3][T] \to K[x]$, defined by $T \mapsto x$?

If $R$ is an integral domain, $w=u/v$, where $u,v \in R$ (namely, $w$ is in the field of fractions on $R$), and $R$ is integrally closed, then the kernel of $R[T] \to R[w]$, $T \mapsto w$, equals $(vT−u)$?

(Of course, one does not assume that $w$ is integral over $R$, since if it was, by the assumption that $R$ is integrally closed, we would immediately get $w \in R$, which is not interesting).

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user237522
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  • I think you need some sort of "no common divisors" hypothesis between $u$ and $v$, otherwise take $R = \mathbb{Z}$ and $u = 2$, $v = 4$. The kernel here is generated by $2T - 1$, not $4T - 2$. – Cihan Jun 10 '15 at 04:34

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Actually the kernel is generated by all polynomials of the form $aX-b$ with $b=aw$.

Let $f\in R[X]$, $f\ne0$ which belongs to the kernel of the $R$-homomorphism $\phi:R[X]\to R[w]$ defined by $\phi(X)=w$. Write $f(X)=a_0+a_1X+\cdots+a_nX^n$. From $f(w)=0$ we get that $a_nw$ is integral over $R$ hence $b_n=a_nw\in R$. Now let $f_1(X)=f(X)-X^{n-1}(a_nX-b_n)$. Notice that $f_1\in\ker\phi$, and $\deg f_1<\deg f$ and proceed by induction on $n$.

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