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Let $R$ be an integral domain and $0 \neq a \in R$. I need to show that: $$\frac{R[T]}{(aT-1)} \cong R[a^{-1}].$$

My Works:

Let us define the ring homomorphism $\phi: R[T] \to R[a^{-1}]$ which is defined by $T \mapsto a^{-1}$. It is easy to show that $\phi$ is an epimorphism and $(aT-1) \subset \mathrm{ker} (\phi)$. I do not know how to get the inverse inclusion. In the case that $R$ is a field, we may use the Euclidean algorithm to show that every $g \in \mathrm{ker} (\phi)$ is a multiple of $aT-1$. But since $R$ is not a field, I do not know how to show that $g$ must be a multiple of $aT-1$.

Thanks for your help.

Ali
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  • Thanks to Bill Dubuque, this question is an special case of [Factorizing a polynomial f in A[x] (with A commutative), where f has a zero in its field of fractions] (http://math.stackexchange.com/questions/88810/factorizing-a-polynomial-f-in-ax-with-a-commutative-where-f-has-a-z) – Ali Feb 15 '17 at 08:42

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