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Let $R$ be an integral domain, not necessarily integrally closed. Let $0 \neq v,u \in R$. Assume that $vT-u$ is a prime element of $R[T]$ (then, by Exercise 4 page 102 in Kaplansky's book "Commutative rings", $-u,v$ is an $R$-sequence).

What is the kernel of $R[T] \to R[w]$, where $w=u/v$? (I do not mind to further assume that $R$ is Noetherian).

Some remarks:

(1) This question is similar: Instead of "$vT-u$ is prime" it assumes "$R$ is integrally closed".

(2) On the one hand, perhaps, in view of this question, there is no hope to prove that an element of the kernel has the form $bT-a$, with $a/b=u/v$.

(3) On the other hand, in remark (2), the minimal polynomial $x^2T-x^3$ is not prime ($x^3,x^2$ is not a $k[x^2,x^3]$-sequence), so maybe there is still some hope to have a nice answer to my current question.

(4) I wonder if the theory of anti-integral elements is relevant here.

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user237522
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    It looks like the ideal you are looking for might be a contraction of an ideal. The containment $R \subset R[w]$ induces $R[T] \subset R[w][T]$, and the morphism $R[T] \to R[w]$ is the composition of $R[T] \subset R[w][T]$ and $R[w][T] -> R[w][T]/(w-T)$. So, $(w-T)R[w,t] \cap R$ is the ideal you are looking for. – Youngsu Aug 02 '16 at 01:46
  • Thanks! I am not sure yet I completely understood the last line in your argument. (I will let you know later if I still have problems understanding it. You can post your comment as an answer, if you wish). – user237522 Aug 02 '16 at 02:04
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    I think this answer could be helpful. – user26857 Aug 02 '16 at 14:22
  • Thank you very much! The answers to that question are indeed helpful! – user237522 Aug 02 '16 at 16:07
  • The hint in Exercise 3 page 102 in Kaplansky's book "Commutative rings", is that the kernel of $R[T] \to R[w]$ is a principal ideal generated by $vT-u$, which answers my question-- the details of the proof can be found in the answers that @user26857 mentioned. What about assuming that $vT-u$ is irreducible instead of prime? Now instead of $-u,v$ a regular sequence, is it a "weak" regular sequence? – user237522 Aug 02 '16 at 21:18
  • The above definition of "weak regular" is because we do not demand that $(u,v)\neq R$? (as Kaplansky noted in his exercises)? But the assumption that $(v):(u)=(v)$ is equivalent to $u+(v)$ not being a zero-divisor in $R/(v)$? – user237522 Aug 02 '16 at 21:32
  • Since $\deg(vT-u)=1$, irreducible is equivalent to $u,v$ have no common factor other than units. In general, this seems to be weaker than $u,v$ regular sequence. – user26857 Aug 02 '16 at 21:49
  • $x^2T-x^3 \in k[x^2,x^3][T]$ is irreducible, not prime, and the kernel is NOT a principal ideal http://math.stackexchange.com/questions/1319223/what-is-the-kernel-of-kx2-x3t-to-kx-defined-by-t-mapsto-x – user237522 Aug 02 '16 at 22:09

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