Let $Q$ be a prime ideal in polynomial ring $R[X]$ and $P=Q\cap R$ be a prime ideal in $R$. Prove that $R[X]_Q$ is a localization of $R_P[X]$.
I'm looking for some hints that is which's the multiplicatively closed set $S$ such that $R_P[X]_{S}=R[X]_Q$ ?
Consider the natural homomorphisms $\phi: R\to R_P$ and $\imath': R_P\to R_P[X]$ (the latter is just inclusion). Try showing that this induces a homomorphism $\psi: R[X]\to R_P[X]$ such that if $\imath: R\to R[X]$ is the natural inclusion, then $\psi\circ \imath=\imath'\circ \phi$.
Now consider the homomorphism $\psi: R[X]\to R_P[X]$ and let $S=R[X]-Q$ (the original multiplicative set), and $T=\psi(S)$. Now there is a natural map $R[X]\to R_P[X]\to T^{-1}(R_P[X])$. Show that under this latter homomorphism the image of $S$ consists of units only. This gives you a natural map $\rho: R[X]_Q\to T^{-1}(R_P[X])$. Finally, show that the natural homomorphism $\rho$ is a bijection and hence an isomorphism.
If you think this way, you will also immediately see that this isomorphism is a natural one.
