I have a question.
Let $R$ be a jacobson ring, integral domain with quotient field $K$. And, let $S=R[t]$ (i.e., $R$-algebra generated by just one element). So we can write $S \cong R[x]/Q$ for some prime ideal $Q\subseteq R[x]$.
And assume furthur that there is a nonzero element $ 0\neq b := f+Q \in S$ such that $S[b^{-1}]$ is a field. Then note that $K \subseteq S[b^{-1}]$.
Also note that we can show that $QK[x]$ is a prime ideal of $K[x]$ and $K[x]/QK[x][b^{-1}]$ is also field(really?).
Then,
Q. From these data, can we show that $K[x]/QK[x][b^{-1}] \cong S[b^{-1}]$?
My question originates from following proof of the "nullstellansatz - General form" (Eisenbud's Commutative Algebra p.132, Theorem 4.19)
If our question is true, then first, by the lemma 4.20-b. (above image), $K[x]/QK[x]$ is a field.
And second, we have $$K[x]/QK[x] \subseteq K[x]/QK[x][b^{-1}] \subseteq \operatorname{Frac}(K[x]/QK[x]) = K[x]/QK[x] $$ So, $K[x]/QK[x] \cong K[x]/QK[x] [b^{-1}] = S[b^{-1}]$ and we are done.
So, is my question true?
One of possible attempt (I'm now trying), is to construct an explicit sujective homomorphism between the 'fields' $K[x]/QK[x][b^{-1}] \to (R[x]/Q) [b^{-1}]$ but I stuck at well-definess problem. (If needed, I will exhibit constructing homomorphism as a trial).
Second possible attempt: Since $K[x]/QK[x][b^{-1}]$ is field, $K[x]/QK[x][b^{-1}] =\operatorname{Frac}(K[x]/QK[x])$. So it suffices to show that $\operatorname{Frac}(K[x]/QK[x]) \cong S[b^{-1}]=(R[x]/Q)[b^{-1}]$. But since $S$ is a domain and $S[b^{-1}]$ is a field, $S[b^{-1}]=\operatorname{Frac}(S)$. So finally it suffices to show that
$$\operatorname{Frac}(K[x]/QK[x]) \cong \operatorname{Frac}(S)=\operatorname{Frac}(R[x]/Q).$$
(Correct argument?) And is it true?
Is there other approach's for our question?
Anyone helps?
