Prime ideal in $R[x]$ lying above $P$

Question

I am trying to understand this proof, I think "$Q$ is lying above $P$" means $Q \cap P = R$, but I don't know why we can assume "$P=0$" or "$R$ is a field", can someone explain it to me?

Answer 1

That is because the prime ideals in $R[X]$ which lie above $P$ are exactly those which contain $PR[X]$, hence considering the commutative diagram: $$\begin{array}{cl} R\hskip{-2em}&\longrightarrow S=R[X]\\\downarrow\hskip{-2em}&\hskip{3.5em} \downarrow\\R/P\hskip{-0.5em}&\longrightarrow S/PS\simeq R/P[X] \end{array}$$ we may replace $R$ with $R/P$, i.e. suppose $P=0$, which means $R$ is an integral domain.

Similarly, supposing $R$ is an integral domain with fraction field $K$, the prime ideals of $R[X]$ which lie above $0$ correspond bijectively to the ideals of $K[X]$, so considering the commutative diagram: $$\begin{array}{cl} R\hskip{-2em}&\longrightarrow S=R[X]\\\downarrow\hskip{-2em}&\hskip{3.5em} \downarrow\\K\hskip{-0.5em}&\longrightarrow S\otimes_R K\simeq K[X] \end{array}$$ we may replace $R$ with $K$, i.e. suppose $R$ is a field.