# The total ring of fractions of a reduced Noetherian ring is a direct product of fields

This is question 6.5 in Matsumura's "Commutative ring theory":

How can I prove that the total ring of fractions of a reduced Noetherian ring is a direct product of fields?

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Hint: A Noetherian ring has only finitely many prime ideals belonging to $0$. – Amitesh Datta Jul 14 '11 at 9:21

Since $A$ is a reduced Noetherian ring, the minimal prime ideals $P_1, \dots, P_n$ are exactly the primes belonging to $(0)$. Pick $S = A \setminus (P_1 \cup \dots \cup P_n)$ the multiplicative set of non-zero-divisors of $A$.
Observe that the only primes of $S^{-1}A$ are $S^{-1}P_1, \dots, S^{-1}P_n$, so they are pairwise coprime. Since chinese remainder theorem, the product $$f \ \colon \ S^{-1} A \longrightarrow \prod_{i=1}^n S^{-1}A / S^{-1} P_i$$ of the quotient projections $S^{-1} A \to S^{-1}A / S^{-1} P_i$ is surjective. On the other hand, $f$ is injective because $S^{-1} P_1 \cap \dots \cap S^{-1} P_n = 0$. It is clear that $S^{-1} A / S^{-1}P_i$ is the residue field $k(P_i)$ of $P_i$.