Let $R$ be any ring. The total quotient ring of $R$, denoted $K(R)$, is $R$ localized at the set of non zero-divisors.
I am trying to solve Exercise 3.15 in Eisenbud's Commutative Algebra:
a) Let $R$ be a Noetherian reduced ring and let $U$ be a multiplicative subset. Then $K(R[U^{-1}])=K(R)[U^{-1}]$.
b) If $R$ is a Noetherian ring, then $K(R[U^{-1}])=K(K(R)[U^{-1}])$.
For a), I was able to show that all prime ideals in $K(R)$ are maximal. Further, there are finitely many of them. Let these ideals be denoted $P_1,\ldots ,P_k$. By the Chinese Remainder Theorem, we have that
$$K(R)=\prod_{i=1}^k K(R/P_i)$$
How should I proceed? Also, can I have a hint to start part b)? Should I use the universal property of the localization somehow?
Best Answer
Note that your $P_1, \dotsc, P_k$ are the minimal primes of $R$. Note that they are only finite if $R$ is noetherian, you have to add this assumption as the other answer points out.
For $a)$ you should show the following:
Using this, we get:
$K(R)[U^{-1}]$ is the product of all $K(R/P_i)$, where $U \cap P_i=\emptyset$.
Now we start with the reduced ring $R[U^{-1}]$. The minimal primes of this ring are all $P_i[U^{-1}]$ with $P_i \cap U = \emptyset$, i.e. its total quotient ring is the product of all $K(R[U^{-1}]/P_i[U^{-1}])$, where $U \cap P_i=\emptyset$.
Localization commutes with quotients, hence we have
$$K(R[U^{-1}]/P_i[U^{-1}])=K((R/P_i)[U^{-1}])=K(R/P_i),$$ which gives you the result. The latter equality holds, since we have $(R/P_i)[U^{-1}]=R/P_i$ if $U \cap P_i=\emptyset$.