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?

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

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 $$P_i$$.

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 $$k_i=1$$.