If $R$ is a reduced Noetherian ring and $U \subset R$ is a multiplicatively closed set, how do we get $K(R[U^{-1}]) = K(R)[U^{-1}]$?

Question

If $R$ is a reduced Noetherian ring and $U \subset R$ is a multiplicatively closed set, how do we get $K(R[U^{-1}]) = K(R)[U^{-1}]$?

This post asks the same question I have but I don't see how the accepted answer answers the OP's question. I got up to the point where we use the Chinese remainder theorem just like in the post, but do not see where to go from here.

Can someone further elaborate the answer given?

More specifically:

1. How do we have $(R/P_i)[U^{-1}]=R/P_i$ if $U \cap P_i=\emptyset$?
2. How are we concluding $K(R[U^{-1}])=K(R)[U^{-1}]$?

Answer 1

(1) is false. The intuition is as follows. The primes in $R/I$ "are" the primes in $R$ containing $I$. The primes in $R[U^{-1}]$ similarly "are" the primes in $R$ which are disjoint from $U$ [since otherwise they include 1 after localization]. This is made precise e.g. here: there's explicit natural bijections/homeomorphisms/lattice isomorphisms between them.

Intuitively then, the primes in $R/P$ are the primes in $R$ containing $P$, and the primes in $R/P_i[U^{-1}]$ are the primes in $R$ which (a) contain $P$ and (b) are disjoint from $U$. Assuming $U \cap P = \varnothing$ does not make (b) trivial. Just pick another prime $Q \supsetneq P$ in $R$ and let $U = R - P$, so $U \cap P = \varnothing$ while $U \cap Q \neq \varnothing$. Then $R/P$ has $Q$ as a prime, while $R/P[U^{-1}]$ does not, so they can't be equal.

An explicit instance is $R = \mathbb{Z}[x]$, $P = (2)$, $Q = (2, x)$, $U = R - P$. Then $R/P \cong \mathbb{Z}/2[x]$ has no inverse for $x$ while $R/P[U^{-1}]$ will invert $x \in R - P$.

As user26857 said in the chat (modulo some missing parens), the linked post only wanted $$K((R/P_i)[U^{-1}])= K(R/P_i) \qquad\text{if }U \cap P_i = \varnothing. \qquad (*)$$ The claim without the $K$'s was a mistake. As user26857 said in the chat, (*) is pretty clear from the field of fractions perspective.

Actually, (*) is unnecessary anyway; the rest of the original answer is fine. Here's the corrected conclusion written out, answering your (2). First, if $P_1, \ldots, P_k$ are the minimal primes of $R$, then $$K(R) = \prod_{i=1}^k K(R/P_i) \qquad \Rightarrow \qquad K(R)[U^{-1}] = \prod_{i=1}^k K(R/P_i)[U^{-1}] = \prod_{\substack{i=1 \\ P_i \cap U = \varnothing}}^k K(R/P_i)[U^{-1}].$$

Second, the minimal primes of $R[U^{-1}]$ are $P_i[U^{-1}]$ where $P_i \cap U = \varnothing$. Hence $$K(R[U^{-1}]) = \prod_{\substack{i=1 \\ P_i \cap U = \varnothing}}^k K(R[U^{-1}]/P_i[U^{-1}])) = \prod_{\substack{i=1 \\ P_i \cap U = \varnothing}}^k K((R/P_i)[U^{-1}]) = \prod_{\substack{i=1 \\ P_i \cap U = \varnothing}}^k K(R/P_i)[U^{-1}].$$