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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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    $\begingroup$ Hint: A Noetherian ring has only finitely many prime ideals belonging to $0$. $\endgroup$ Jul 14, 2011 at 9:21
  • $\begingroup$ Is the converse true? $\endgroup$ Nov 6, 2018 at 0:22

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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. By 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 $P_i$.

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The total ring of fraction $Q(A)$ is artinian (its non invertible elements are zero divisors so that all its minimal primes are maximal): it is the product $$ A=\prod_i A/\mathfrak{m}_i^{k_i} $$ with $\mathfrak{m}_i$ its minimal/maximal prime ideals (see for example Atiyah MacDonald 8.7)

The total ring of fraction $Q(A)$ is still reduced (commutation of $S^{-1}$ and nilradical - Atiyah MacDonald 3.11 - or by hand $(a/s)^k=0\implies ta^k=0\implies (ta)^k=0\implies A/s=0$): all the factor $A/\mathfrak{m}_i^{k_i}$ must be with $k_i=1$.

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