I'm now reading Eisenbud's, Commutative Algebra, p.132~133, Nullstellensatz, General form.
My question is,
Q. Let $R$ be a integral domain and $Q$ a prime ideal of $R[x]$, and $S:=R[x]/Q$.
Let $0\neq b:=g+Q \in S$ be a nonzero element and $K$ the fraction field of $R$.
Then the localization $S[b^{-1}]$ is isomorphic to $K[x]/QK[x]$ ?
This question originates from following underlined statement.
Since localization commutes with quotient, $S[b^{-1}] \cong R[x][b^{-1}]/Q[b^{-1}]$. But this ring is isomorphic to $K[x]/QK[x]$?
My first attempt to construct such an isomorphism is $(1/b^{n})f(x) + Q[b^{-1}] \mapsto (1/b^{n})f(x) + QK[x]$. This map is well defined but I stucked at showing injectivity.
Or is there any other way to show the isomorphism?
